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How To… Given two complex numbers, find the sum or difference. 1. Identify the real and imaginary parts of each number. 2. Add or subtract the real parts. 3. Add or subtract the imaginary parts. Example 3 Adding and Subtracting Complex Numbers Add or subtract as indicated. a. (3 βˆ’ 4i) + (2 + 5i) b. (βˆ’5 + 7i) βˆ’ (βˆ’11 + 2i) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 114 CHAPTER 2 EQUATIONS AND INEQUALITIES Solution We add the real parts and add the imaginary parts. a. (3 βˆ’ 4i) + (2 + 5i) = 3 βˆ’ 4i + 2 + 5i = 3 + 2 + (βˆ’4i) + 5i = (3 + 2) + (βˆ’4 + 5)i = 5 + i b. (βˆ’5 + 7i) βˆ’ (βˆ’11 + 2i) = βˆ’5 + 7i + 11 βˆ’ 2i = βˆ’5 + 11 + 7i βˆ’ 2i = (βˆ’5 + 11) + (7 βˆ’ 2)i = 6 + 5i Try It #3 Subtract 2 + 5i from 3 βˆ’ 4i. Multiplying Complex Numbers Multiplying complex numbers is much like multiplying binomials. The major difference is that we work with the real and imaginary parts separately. Multiplying a Complex Number by a Real Number Lets begin by multiplying a complex number by a real number. We distribute the real number just as we would with a binomial. Consider, for example, 3(6 + 2i): 3(6 + 2i) = (3 Β· 6) + (3 Β· 2i) = 18 + 6i Distribute. Simplify. How To… Given a complex number and a real number, multiply to find the product. 1. Use the distributive property. 2. Simplify. Example 4 Multiplying a Complex Number by a Real Number Find the product 4(2 + 5i). Solution Distribute the 4. 4(2 + 5i) = (4 Δ‹ 2) + (4 Δ‹ 5i) = 8 + 20i Try It #4 1 _ (5 βˆ’ 2i). Find the product: 2 Multiplying Complex Numbers Together Now, let’s multiply two complex numbers. We can use either the distributive property or more specifically the FOIL
method because we are dealing with binomials. Recall that FOIL is an acronym for multiplying First, Inner, Outer, and Last terms together. The difference with complex numbers is that when we get a squared term, i2, it equals βˆ’1. (a + bi)(c + di) = ac + adi + bci + bdi2 = ac + adi + bci βˆ’ bd = (ac βˆ’ bd) + (ad + bc)i i2 = βˆ’1 Group real terms and imaginary terms. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.4 COMPLEX NUMBERS 115 How To… Given two complex numbers, multiply to find the product. 1. Use the distributive property or the FOIL method. 2. Remember that i2 = βˆ’1. 3. Group together the real terms and the imaginary terms Example 5 Multiplying a Complex Number by a Complex Number (4 + 3i)(2 βˆ’ 5i) = 4(2) βˆ’ 4(5i) + 3i(2) βˆ’ (3i)(5i) = 8 βˆ’ 20i + 6i βˆ’ 15(i2) = (8 + 15) + (βˆ’20 + 6)i = 23 βˆ’ 14i Multiply: (4 + 3i)(2 βˆ’ 5i). Solution Try It #5 Multiply: (3 βˆ’ 4i)(2 + 3i). Dividing Complex Numbers Dividing two complex numbers is more complicated than adding, subtracting, or multiplying because we cannot divide by an imaginary number, meaning that any fraction must have a real-number denominator to write the answer in standard form a + bi. We need to find a term by which we can multiply the numerator and the denominator that will eliminate the imaginary portion of the denominator so that we end up with a real number as the denominator. This term is called the complex conjugate of the denominator, which is found by changing the sign of the imaginary part of the complex number. In other words, the complex conjugate of a + bi is a βˆ’ bi. For example, the product of a + bi and a βˆ’ bi is (a + bi)(a βˆ’ bi) = a2 βˆ’ abi + abi βˆ’ b2i2 = a2 + b2 The result is a real number. Note that complex conjugates have an opposite
relationship: The complex conjugate of a + bi is a βˆ’ bi, and the complex conjugate of a βˆ’ bi is a + bi. Further, when a quadratic equation with real coefficients has complex solutions, the solutions are always complex conjugates of one another. Suppose we want to divide c + di by a + bi, where neither a nor b equals zero. We first write the division as a fraction, then find the complex conjugate of the denominator, and multiply. Multiply the numerator and denominator by the complex conjugate of the denominator. c + di ______ a + bi where a β‰  0 and b β‰  0 (c + di) _______ Δ‹ (a + bi) (a βˆ’ bi) _______ = (a βˆ’ bi) (c + di)(a βˆ’ bi) _____________ (a + bi)(a βˆ’ bi) Apply the distributive property. Simplify, remembering that i2 = βˆ’1. = ca βˆ’ cbi + adi βˆ’ bdi2 __ a2 βˆ’ abi + abi βˆ’ b2 i2 = ca βˆ’ cbi + adi βˆ’ bd(βˆ’1) ___ a2 βˆ’ abi + abi βˆ’ b2(βˆ’1) = (ca + bd) + (ad βˆ’ cb)i___ a2 + b2 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 116 CHAPTER 2 EQUATIONS AND INEQUALITIES the complex conjugate The complex conjugate of a complex number a + bi is a βˆ’ bi. It is found by changing the sign of the imaginary part of the complex number. The real part of the number is left unchanged. β€’ When a complex number is multiplied by its complex conjugate, the result is a real number. β€’ When a complex number is added to its complex conjugate, the result is a real number. Example 6 Finding Complex Conjugates Find the complex conjugate of each number. a. 2 + i √ β€” 5 Solution 1 _ b. βˆ’ i 2 a. The number is already in the form a + bi. The complex conjugate is a βˆ’ bi, or 2 βˆ’ i √ 1 1 _ _ i. The complex conjugate is a βˆ’ bi, or 0 + b. We can rewrite this number in the form a + bi as 0 βˆ’ i. 2 2
β€” 5. 1 _ i. This can be written simply as 2 Analysis Although we have seen that we can find the complex conjugate of an imaginary number, in practice we generally find the complex conjugates of only complex numbers with both a real and an imaginary component. To obtain a real number from an imaginary number, we can simply multiply by i. Try It #6 Find the complex conjugate of βˆ’3 + 4i. How To… Given two complex numbers, divide one by the other. 1. Write the division problem as a fraction. 2. Determine the complex conjugate of the denominator. 3. Multiply the numerator and denominator of the fraction by the complex conjugate of the denominator. 4. Simplify. Example 7 Dividing Complex Numbers Divide: (2 + 5i) by (4 βˆ’ i). Solution We begin by writing the problem as a fraction. (2 + 5i) _ (4 βˆ’ i) Then we multiply the numerator and denominator by the complex conjugate of the denominator. (2 + 5i) _ Δ‹ (4 βˆ’ i) (4 + i) _ (4 + i) To multiply two complex numbers, we expand the product as we would with polynomials (using FOIL). (4 + i) _ (4 + i) (2 + 5i) _ Δ‹ (4 βˆ’ i) = 8 + 2i + 20i + 5i2 __ 16 + 4i βˆ’ 4i βˆ’ i2 8 + 2i + 20i + 5(βˆ’1) __________________ 16 + 4i βˆ’ 4i βˆ’ (βˆ’1) 3 + 22i _______ 17 = = Because i2 = βˆ’1. Note that this expresses the quotient in standard form. = + 3 ___ 17 22 ___ i 17 Separate real and imaginary parts. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.4 COMPLEX NUMBERS 117 Simplifying Powers of i The powers of i are cyclic. Let’s look at what happens when we raise i to increasing powers. i1 = i i2 = βˆ’1 i3 = i2 Δ‹ i = βˆ’1 Δ‹ i = βˆ’i i4 = i3 Δ‹ i = βˆ’i Δ‹ i = βˆ’i2 = βˆ’ (βˆ’1) = 1 i5 = i4 Δ‹ i =
1 Δ‹ i = i We can see that when we get to the fifth power of i, it is equal to the first power. As we continue to multiply i by increasing powers, we will see a cycle of four. Let’s examine the next four powers of i. i6 = i5 Δ‹ i = i Δ‹ i = i2 = βˆ’1 i7 = i6 Δ‹ i = i2 Δ‹ i = i3 = βˆ’i i8 = i7 Δ‹ i = i3 Δ‹ i = i4 = 1 i9 = i8 Δ‹ i = i4 Δ‹ i = i5 = i The cycle is repeated continuously: i, βˆ’1, βˆ’ i, 1, every four powers. Example 8 Simplifying Powers of i Evaluate: i35. Solution Since i4 = 1, we can simplify the problem by factoring out as many factors of i4 as possible. To do so, first determine how many times 4 goes into 35: 35 = 4 Δ‹ 8 + 3. i35 = i4 Δ‹ 8 + 3 = i4 Δ‹ 8 Δ‹ i3 = (i4) Δ‹ i3 = 18 Δ‹ i3 = i3 = βˆ’i 8 Try It #7 Evaluate: i 18 Q & A… Can we write i35 in other helpful ways? As we saw in Example 8, we reduced i35 to i3 by dividing the exponent by 4 and using the remainder to find the simplified form. But perhaps another factorization of i35 may be more useful. Table 1 shows some other possible factorizations. Factorization of i35 i34 Δ‹ i Reduced form Simplified form 17 (i2) Δ‹ i (βˆ’1)17 Δ‹ i Table 1 i33 Δ‹ i2 i33 Δ‹ (βˆ’1) βˆ’i33 i31 Δ‹ i4 i31 Δ‹ 1 i31 i19 Δ‹ i16 4 i19 Δ‹ (i4) i19 Each of these will eventually result in the answer we obtained above but may require several more steps than our earlier method. Access these online resources for additional instruction and practice with complex numbers. β€’ Adding and Subtracting Complex Numbers (http://openstaxcollege.org/l/addsubcomplex) β€’ Multiply Complex Numbers (http://openstaxcollege.org/l/multiplycomplex) β€’ Multiplying Complex Conjugates (http://openstaxcollege.
org/l/multcompconj) β€’ Raising i to Powers (http://openstaxcollege.org/l/raisingi) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 118 CHAPTER 2 EQUATIONS AND INEQUALITIES 2.4 SECTION EXERCISES VERBAL 1. Explain how to add complex numbers. 2. What is the basic principle in multiplication of complex numbers? 3. Give an example to show that the product of two imaginary numbers is not always imaginary. 4. What is a characteristic of the plot of a real number in the complex plane? ALGEBRAIC For the following exercises, evaluate the algebraic expressions. 5. If y = x2 + x βˆ’ 4, evaluate y given x = 2i. 7. If y = x2 + 3x + 5, evaluate y given x = 2 + i. 6. If y = x3 βˆ’ 2, evaluate y given x = i. 8. If y = 2x2 + x βˆ’ 3, evaluate y given x = 2 βˆ’ 3i. 9. If y =, evaluate y given x = 5i. x + 1 _ 2 βˆ’ x 10. If y =, evaluate y given x = 4i. 1 + 2x _ x + 3 GRAPHICAL For the following exercises, plot the complex numbers on the complex plane. 11. 1 βˆ’ 2i 12. βˆ’2 + 3i 13. i 14. βˆ’3 βˆ’ 4i NUMERIC For the following exercises, perform the indicated operation and express the result as a simplified complex number. 15. (3 + 2i) + (5 βˆ’ 3i) 16. (βˆ’2 βˆ’ 4i) + (1 + 6i) 17. (βˆ’5 + 3i) βˆ’ (6 βˆ’ i) 18. (2 βˆ’ 3i) βˆ’ (3 + 2i) 19. (βˆ’4 + 4i) βˆ’ (βˆ’6 + 9i) 20. (2 + 3i)(4i) 21. (5 βˆ’ 2i)(3i) 22. (6 βˆ’ 2i)(5) 23. (βˆ’2 + 4i)(8) 24. (2 + 3i)(4 βˆ’ i) 25. (βˆ’1 + 2i)(βˆ’2 + 3i) 26. (4 βˆ’ 2i)(4 + 2i) 27. (3 + 4i)(3 βˆ’ 4i) 28. 31.
6 + 4i_ i 3 + 4i_ 2 2 βˆ’ 3i _ 4 + 3i 32. 35. √ β€” βˆ’9 + 3 √ β€” βˆ’16 36. βˆ’ √ β€” βˆ’4 βˆ’ 4 √ β€” βˆ’25 39. i8 40. i15 29. 6 βˆ’ 2i_ 3 3 + 4i_ 2 βˆ’ i 33. 37. β€” 2 + √ βˆ’12 __________ 2 41. i22 30. 34. βˆ’5 + 3i_ 2i 2 + 3i _ 2 βˆ’ 3i 38. β€” 4 + √ βˆ’20 __________ 2 TECHNOLOGY For the following exercises, use a calculator to help answer the questions. 42. Evaluate (1 + i)k for k = 4, 8, and 12. Predict the 43. Evaluate (1 βˆ’ i)k for k = 2, 6, and 10. Predict the value if k = 16. value if k = 14. 44. Evaluate (l + i)k βˆ’ (l βˆ’ i)k for k = 4, 8, and 12. Predict the value for k = 16. 45. Show that a solution of x6 + 1 = 0 is β€” √ 3 ____ + 2 1_ i. 2 46. Show that a solution of x8 βˆ’1 = 0 is β€” √ 2 ____ + 2 β€” √ 2 ____ i. 2 EXTENSIONS For the following exercises, evaluate the expressions, writing the result as a simplified complex number. 47. 51. 55. 4 1 __ __ + i3 i (2 + i)(4 βˆ’ 2i) ____________ (1 + i) 4 + i _____ i + 3 βˆ’ 4i ______ 1 βˆ’ i 48. 52. 56. 1 __ βˆ’ i21 1 __ i11 (1 + 3i)(2 βˆ’ 4i) _____________ (1 + 2i) 49. i7(1 + i2) 53. (3 + i)2 _______ (1 + 2i)2 50. iβˆ’3 + 5i7 54. 3 + 2i ______ 2 + i + (4 + 3i) 3 + 2i ______ 1 + 2i βˆ’ 2 βˆ’ 3i ______ 3 + i Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.5 QUADRATIC EQUATIONS 119 LEARNING OBJECTIVES
In this section you will: β€’ Solve quadratic equations by factoring. β€’ Solve quadratic equations by the square root property. β€’ Solve quadratic equations by using the quadratic formula. 2.5 QUADRATIC EQUATIONS Figure 1 The computer monitor on the left in Figure 1 is a 23.6-inch model and the one on the right is a 27-inch model. Proportionally, the monitors appear very similar. If there is a limited amount of space and we desire the largest monitor possible, how do we decide which one to choose? In this section, we will learn how to solve problems such as this using four different methods. Solving Quadratic Equations by Factoring An equation containing a second-degree polynomial is called a quadratic equation. For example, equations such as 2x2 + 3x βˆ’ 1 = 0 and x2 βˆ’ 4 = 0 are quadratic equations. They are used in countless ways in the fields of engineering, architecture, finance, biological science, and, of course, mathematics. Often the easiest method of solving a quadratic equation is factoring. Factoring means finding expressions that can be multiplied together to give the expression on one side of the equation. If a quadratic equation can be factored, it is written as a product of linear terms. Solving by factoring depends on the zero-product property, which states that if a Δ‹ b = 0, then a = 0 or b = 0, where a and b are real numbers or algebraic expressions. In other words, if the product of two numbers or two expressions equals zero, then one of the numbers or one of the expressions must equal zero because zero multiplied by anything equals zero. Multiplying the factors expands the equation to a string of terms separated by plus or minus signs. So, in that sense, the operation of multiplication undoes the operation of factoring. For example, expand the factored expression (x βˆ’ 2)(x + 3) by multiplying the two factors together. (x βˆ’ 2)(x + 3) = x2 + 3x βˆ’ 2x βˆ’ 6 = x2 + x βˆ’ 6 The product is a quadratic expression. Set equal to zero, x2 + x βˆ’ 6 = 0 is a quadratic equation. If we were to factor the equation, we would get back the factors we multiplied. The process of factoring a quadratic equation depends on
the leading coefficient, whether it is 1 or another integer. We will look at both situations; but first, we want to confirm that the equation is written in standard form, ax2 + bx + c = 0, where a, b, and c are real numbers, and a β‰  0. The equation x2 + x βˆ’ 6 = 0 is in standard form. We can use the zero-product property to solve quadratic equations in which we first have to factor out the greatest common factor (GCF ), and for equations that have special factoring formulas as well, such as the difference of squares, both of which we will see later in this section. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 120 CHAPTER 2 EQUATIONS AND INEQUALITIES the zero-product property and quadratic equations The zero-product property states If a Δ‹ b = 0, then a = 0 or b = 0, where a and b are real numbers or algebraic expressions. A quadratic equation is an equation containing a second-degree polynomial; for example where a, b, and c are real numbers, and if a β‰  0, it is in standard form. ax 2 + bx + c = 0 Solving Quadratics with a Leading Coefficient of 1 In the quadratic equation x2 + x βˆ’ 6 = 0, the leading coefficient, or the coefficient of x2, is 1. We have one method of factoring quadratic equations in this form. How To… Given a quadratic equation with the leading coefficient of 1, factor it. 1. Find two numbers whose product equals c and whose sum equals b. 2. Use those numbers to write two factors of the form (x + k) or (x βˆ’ k), where k is one of the numbers found in step 1. Use the numbers exactly as they are. In other words, if the two numbers are 1 and βˆ’2, the factors are (x + 1)(x βˆ’ 2). 3. Solve using the zero-product property by setting each factor equal to zero and solving for the variable. Example 1 Factoring and Solving a Quadratic with Leading Coefficient of 1 Factor and solve the equation: x 2 + x βˆ’ 6 = 0. Solution To factor x 2 + x βˆ’ 6 = 0, we look for two numbers whose product
equals βˆ’6 and whose sum equals 1. Begin by looking at the possible factors of βˆ’6. 1 Δ‹ (βˆ’6) (βˆ’6) Δ‹ 1 2 Δ‹ (βˆ’3) 3 Δ‹ (βˆ’2) The last pair, 3 Δ‹ (βˆ’2) sums to 1, so these are the numbers. Note that only one pair of numbers will work. Then, write the factors. (x βˆ’ 2)(x + 3) = 0 To solve this equation, we use the zero-product property. Set each factor equal to zero and solve. (x βˆ’ 2)(x + 3) = 0 (x βˆ’ 2) = 0 x = 2 (x + 3) = 0 x = βˆ’3 The two solutions are 2 and βˆ’3. We can see how the solutions relate to the graph in Figure 2. The solutions are the x-intercepts of x 2 + x βˆ’ 6 = 0. x2 + x βˆ’ 6 = 0 y (βˆ’3, 0) –4 –5 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 (2, 0) 21 3 4 5 x Figure 2 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.5 QUADRATIC EQUATIONS 121 Try It #1 Factor and solve the quadratic equation: x 2 βˆ’ 5x βˆ’ 6 = 0. Example 2 Solve the Quadratic Equation by Factoring Solve the quadratic equation by factoring: x2 + 8x + 15 = 0. Solution Find two numbers whose product equals 15 and whose sum equals 8. List the factors of 15. 1 Δ‹ 15 3 Δ‹ 5 (βˆ’1) Δ‹ (βˆ’15) (βˆ’3) Δ‹ (βˆ’5) The numbers that add to 8 are 3 and 5. Then, write the factors, set each factor equal to zero, and solve. (x + 3)(x + 5) = 0 (x + 3) = 0 x = βˆ’3 (x + 5) = 0 x = βˆ’5 The solutions are βˆ’3 and βˆ’5. Try It #2 Solve the quadratic equation by factoring: x 2 βˆ’ 4x βˆ’ 21 = 0. Example 3 Using the Zero-Product Property to Solve a Quadratic Equation Written as the Difference of Squares Solve the difference of squares equation using the zero-product property: x 2 βˆ’ 9
= 0. Solution Recognizing that the equation represents the difference of squares, we can write the two factors by taking the square root of each term, using a minus sign as the operator in one factor and a plus sign as the operator in the other. Solve using the zero-factor property. x 2 βˆ’ 9 = 0 (x βˆ’ 3)(x + 3) = 0 (x βˆ’ 3) = 0 x = 3 (x + 3) = 0 x = βˆ’3 The solutions are 3 and βˆ’3. Try It #3 Solve by factoring: x 2 βˆ’ 25 = 0. Factoring and Solving a Quadratic Equation of Higher Order When the leading coefficient is not 1, we factor a quadratic equation using the method called grouping, which requires four terms. With the equation in standard form, let’s review the grouping procedures: 1. With the quadratic in standard form, ax2 + bx + c = 0, multiply a Δ‹ c. 2. Find two numbers whose product equals ac and whose sum equals b. 3. Rewrite the equation replacing the bx term with two terms using the numbers found in step 1 as coefficients of x. 4. Factor the first two terms and then factor the last two terms. The expressions in parentheses must be exactly the same to use grouping. 5. Factor out the expression in parentheses. 6. Set the expressions equal to zero and solve for the variable. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 122 CHAPTER 2 EQUATIONS AND INEQUALITIES Example 4 Solving a Quadratic Equation Using Grouping Use grouping to factor and solve the quadratic equation: 4x2 + 15x + 9 = 0. Solution First, multiply ac : 4(9) = 36. Then list the factors of 36. 1 Δ‹ 36 2 Δ‹ 18 3 Δ‹ 12 4 Δ‹ 9 6 Δ‹ 6 The only pair of factors that sums to 15 is 3 + 12. Rewrite the equation replacing the b term, 15x, with two terms using 3 and 12 as coefficients of x. Factor the first two terms, and then factor the last two terms. 4x 2 + 3x + 12x + 9 = 0 x(4x + 3) + 3(4x + 3) = 0 (4x + 3)(x + 3) = 0 Solve using the zero-product
property. (4x + 3)(x + 3) = 0 (4x + 3x + 3) = 0 x = βˆ’3 3 _ and βˆ’3. See Figure 3. The solutions are βˆ’ 4 2 + 15x + 9 = 0 4x –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 x Figure 3 Try It #4 Solve using factoring by grouping: 12x 2 + 11x + 2 = 0. Example 5 Solving a Higher Degree Quadratic Equation by Factoring Solve the equation by factoring: βˆ’3x 3 βˆ’ 5x 2 βˆ’ 2x = 0. Solution This equation does not look like a quadratic, as the highest power is 3, not 2. Recall that the first thing we want to do when solving any equation is to factor out the GCF, if one exists. And it does here. We can factor out βˆ’x from all of the terms and then proceed with grouping. Use grouping on the expression in parentheses. βˆ’3x 3 βˆ’ 5x 2 βˆ’ 2x = 0 βˆ’x(3x 2 + 5x + 2) = 0 βˆ’x(3x 2 + 3x + 2x + 2) = 0 βˆ’x[3x(x + 1) + 2(x + 1)] = 0 βˆ’x(3x + 2)(x + 1) = 0 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.5 QUADRATIC EQUATIONS 123 Now, we use the zero-product property. Notice that we have three factors. βˆ’x = 0 x = 0 3x + 1 2 _, and βˆ’1. The solutions are 0, βˆ’ 3 Try It #5 Solve by factoring: x 3 + 11x 2 + 10x = 0. Using the Square Root Property When there is no linear term in the equation, another method of solving a quadratic equation is by using the square root property, in which we isolate the x 2 term and take the square root of the number on the other side of the equals sign. Keep in mind that sometimes we may have to manipulate the equation to isolate the x2 term so that the square root property can be used. the square root property With the x 2 term isolated, the square root property states that: if x 2 = k, then
x = Β± √ β€” k where k is a nonzero real number. How To… Given a quadratic equation with an x 2 term but no x term, use the square root property to solve it. 1. Isolate the x 2 term on one side of the equal sign. 2. Take the square root of both sides of the equation, putting a Β± sign before the expression on the side opposite the squared term. 3. Simplify the numbers on the side with the Β± sign. Example 6 Solving a Simple Quadratic Equation Using the Square Root Property Solve the quadratic using the square root property: x 2 = 8. Solution Take the square root of both sides, and then simplify the radical. Remember to use a Β± sign before the radical symbol2 √ β€” 8 β€” 2 The solutions are 2 √ β€” 2 and βˆ’2 √ β€” 2. Solving a Quadratic Equation Using the Square Root Property Example 7 Solve the quadratic equation: 4x 2 + 1 = 7. Solution First, isolate the x 2 term. Then take the square root of both sides. 4x 2 + 1 = 7 4x The solutions are and βˆ’ β€” Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 124 CHAPTER 2 EQUATIONS AND INEQUALITIES Try It #6 Solve the quadratic equation using the square root property: 3(x βˆ’ 4)2 = 15. Completing the Square Not all quadratic equations can be factored or can be solved in their original form using the square root property. In these cases, we may use a method for solving a quadratic equation known as completing the square. Using this method, we add or subtract terms to both sides of the equation until we have a perfect square trinomial on one side of the equal sign. We then apply the square root property. To complete the square, the leading coefficient, a, must equal 1. If it does not, then divide the entire equation by a. Then, we can use the following procedures to solve a quadratic equation by completing the square. We will use the example x2 + 4x + 1 = 0 to illustrate each step. 1. Given a quadratic equation that cannot be factored, and with a = 1, first add or subtract the constant term to the right sign of the equal sign. 1 _ and square it. 2.
Multiply the b term by 2 x 2 + 4x = βˆ’1 1 _ (4) = 2 2 22 = 4 1 b ) 3. Add ( _ 2 2 to both sides of the equal sign and simplify the right side. We have x 2 + 4x + 4 = βˆ’1 + 4 x 2 + 4x + 4 = 3 4. The left side of the equation can now be factored as a perfect square. 5. Use the square root property and solve. x 2 + 4x + 4 = 3 (x + 2)2 = 3 √ β€” (x + 2)2 Β± √ β€” 3 6. The solutions are βˆ’2 + √ β€” 3 and βˆ’2 βˆ’ √ β€” 3. Example 8 Solving a Quadratic by Completing the Square Solve the quadratic equation by completing the square: x 2 βˆ’ 3x βˆ’ 5 = 0. Solution First, move the constant term to the right side of the equal sign. x 2 βˆ’ 3x = 5 1 _ of the b term and square it. Then, take 2 Add the result to both sides of the equal sign. 3 1 _ _ (βˆ’3 βˆ’ 3x + ( βˆ’ ) = βˆ’ 3x + 4 4 2 Factor the left side as a perfect square and simplify the right side. 2 3 ) ( x βˆ’ _ 2 = 29_ 4 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.5 QUADRATIC EQUATIONS 125 Use the square root property and solve. The solutions are β€” 29 3 + √ _ 2 and β€” 29 _ 3 βˆ’ √. 2 2 ___ _________ = Β± √ √ 29 3 ) ( x βˆ’ _ _ 4 2 β€” 29 √ √ 29 _ 3 _ Β± x = 2 2 β€” Try It #7 Solve by completing the square: x 2 βˆ’ 6x = 13. Using the Quadratic Formula The fourth method of solving a quadratic equation is by using the quadratic formula, a formula that will solve all quadratic equations. Although the quadratic formula works on any quadratic equation in standard form, it is easy to make errors in substituting the values into the formula. Pay close attention when substituting, and use parentheses when inserting a negative number. We can derive the quadratic formula by completing the square. We will assume that the leading coefficient is positive;
if it is negative, we can multiply the equation by βˆ’1 and obtain a positive a. Given ax 2 + bx + c = 0, a β‰  0, we will complete the square as follows: 1. First, move the constant term to the right side of the equal sign: ax 2 + bx = βˆ’c 2. As we want the leading coefficient to equal 1, divide through by a: b 1 1 ) of the middle term, and add ( __ __ _ 3. Then, find 2 2 a b _ a x + x2 + b2 _ 4a2 = b2 c _ _ 4a2 βˆ’ b2 _ 4a2 to both sides of the equal sign: = 2 4. Next, write the left side as a perfect square. Find the common denominator of the right side and write it as a single fraction: 2 ) ( x + b _ 2a = b2 βˆ’ 4ac _ 4a2 5. Now, use the square root property, which gives x + b _ 2a = Β± √ x + = Β± b _ 2a ________ b2 βˆ’ 4ac _ 4a2 b2 βˆ’ 4ac __ √ 2a β€” 6. Finally, add βˆ’ to both sides of the equation and combine the terms on the right side. Thus, b _ 2a x = β€” βˆ’b Β± √ b2 βˆ’ 4ac __ 2a the quadratic formula Written in standard form, ax2 + bx + c = 0, any quadratic equation can be solved using the quadratic formula: where a, b, and c are real numbers and a β‰  0. x = β€” βˆ’b Β± √ b2 βˆ’ 4ac __ 2a Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 126 CHAPTER 2 EQUATIONS AND INEQUALITIES How To… Given a quadratic equation, solve it using the quadratic formula 1. Make sure the equation is in standard form: ax 2 + bx + c = 0. 2. Make note of the values of the coefficients and constant term, a, b, and c. 3. Carefully substitute the values noted in step 2 into the equation. To avoid needless errors, use parentheses around each number input into the formula. 4. Calculate and solve. Example 9 Solve the Quadratic Equation Using the Quadratic Formula Solve the
quadratic equation: x 2 + 5x + 1 = 0. Solution Identify the coefficients: a = 1, b = 5, c = 1. Then use the quadratic formula. x = β€” βˆ’(5) Β± √ (5)2 βˆ’ 4(1)(1) ___ 2(1) β€” = βˆ’5 Β± √ 25 βˆ’ 4 __ 2 β€” = 21 __ βˆ’5 Β± √ 2 Example 10 Solving a Quadratic Equation with the Quadratic Formula Use the quadratic formula to solve x 2 + x + 2 = 0. Solution First, we identify the coefficients: a = 1, b = 1, and c = 2. Substitute these values into the quadratic formula. x = β€” b2 βˆ’ 4ac βˆ’b Β± √ __ 2a βˆ’(1) Β± √ β€” (1)2 βˆ’ (4) Δ‹ (1) Δ‹ (2) = ___ 2 Δ‹ 1 β€” = 1 βˆ’ 8 βˆ’1 Β± √ __ 2 β€” = βˆ’7 __ βˆ’1 Β± √ 2 β€” = βˆ’1 Β± i √ 7 _ 2 The solutions to the equation are β€” 7 βˆ’1 + i √ _ 2 and βˆ’1 βˆ’ i √ _ 2 7. β€” Try It #8 Solve the quadratic equation using the quadratic formula: 9x 2 + 3x βˆ’ 2 = 0. The Discriminant The quadratic formula not only generates the solutions to a quadratic equation, it tells us about the nature of the solutions when we consider the discriminant, or the expression under the radical, b2 βˆ’ 4ac. The discriminant tells us whether the solutions are real numbers or complex numbers, and how many solutions of each type to expect. Table 1 relates the value of the discriminant to the solutions of a quadratic equation. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.5 QUADRATIC EQUATIONS 127 Value of Discriminant Results b2 βˆ’ 4ac = 0 One rational solution (double solution) b2 βˆ’ 4ac > 0, perfect square Two rational solutions b2 βˆ’ 4ac > 0, not a perfect square Two irrational solutions b2 βˆ’ 4ac < 0 Two complex solutions Table 1 the discriminant For ax2 + bx + c = 0, where a, b, and c are real numbers, the discriminant
is the expression under the radical in the quadratic formula: b2 βˆ’ 4ac. It tells us whether the solutions are real numbers or complex numbers and how many solutions of each type to expect. Example 11 Using the Discriminant to Find the Nature of the Solutions to a Quadratic Equation Use the discriminant to find the nature of the solutions to the following quadratic equations: a. x2 + 4x + 4 = 0 b. 8x2 + 14x + 3 = 0 c. 3x2 βˆ’ 5x βˆ’ 2 = 0 d. 3x2 βˆ’ 10x + 15 = 0 Solution Calculate the discriminant b2 βˆ’ 4ac for each equation and state the expected type of solutions. a. x2 + 4x + 4 = 0 b2 βˆ’ 4ac = (4)2 βˆ’ 4(1)(4) = 0. There will be one rational double solution. b. 8x2 + 14x + 3 = 0 b2 βˆ’ 4ac = (14)2 βˆ’ 4(8)(3) = 100. As 100 is a perfect square, there will be two rational solutions. c. 3x2 βˆ’ 5x βˆ’ 2 = 0 b2 βˆ’ 4ac = (βˆ’5)2 βˆ’ 4(3)(βˆ’2) = 49. As 49 is a perfect square, there will be two rational solutions. d. 3x2 βˆ’10x + 15 = 0 b2 βˆ’ 4ac = (βˆ’10)2 βˆ’ 4(3)(15) = βˆ’80. There will be two complex solutions. Using the Pythagorean Theorem One of the most famous formulas in mathematics is the Pythagorean Theorem. It is based on a right triangle, and states the relationship among the lengths of the sides as a2 + b2 = c2, where a and b refer to the legs of a right triangle adjacent to the 90Β° angle, and c refers to the hypotenuse. It has immeasurable uses in architecture, engineering, the sciences, geometry, trigonometry, and algebra, and in everyday applications. We use the Pythagorean Theorem to solve for the length of one side of a triangle when we have the lengths of the other two. Because each of the terms is squared in the theorem, when we are solving for a side of a triangle, we have a quadratic equation. We can use the methods for solving quadratic equations that we learned in this section to solve for the missing
side. The Pythagorean Theorem is given as a2 + b2 = c2 where a and b refer to the legs of a right triangle adjacent to the 90Β° angle, and c refers to the hypotenuse, as shown in Figure 4. b c a Figure 4 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 128 CHAPTER 2 EQUATIONS AND INEQUALITIES Example 12 Finding the Length of the Missing Side of a Right Triangle Find the length of the missing side of the right triangle in Figure 5. 12 4 a Figure 5 Solution As we have measurements for side b and the hypotenuse, the missing side is a. a2 + b2 = c2 a2 + (4)2 = (12)2 a2 + 16 = 144 a2 = 128 a = √ = 8 √ β€” 128 2 β€” Try It #9 Use the Pythagorean Theorem to solve the right triangle problem: Leg a measures 4 units, leg b measures 3 units. Find the length of the hypotenuse. Access these online resources for additional instruction and practice with quadratic equations. β€’ Solving Quadratic Equations by Factoring (http://openstaxcollege.org/l/quadreqfactor) β€’ The Zero-Product Property (http://openstaxcollege.org/l/zeroprodprop) β€’ Completing the Square (http://openstaxcollege.org/l/complthesqr) β€’ Quadratic Formula with Two Rational Solutions (http://openstaxcollege.org/l/quadrformrat) β€’ Length of a Leg of a Right Triangle (http://openstaxcollege.org/l/leglengthtri) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.5 SECTION EXERCISES 129 2.5 SECTION EXERCISES VERBAL 1. 2. 3. 5. ax+bx +c = y =ax+bx +cx 4. bβˆ’ac ALGEBRAIC 6. x +xβˆ’= 10. x βˆ’x+= 14. x +x = 18. x x = βˆ’ 7. x βˆ’x += 11. x βˆ’= 15. x =x + 8. x +x βˆ’= 12. x +x βˆ’= 16. x =x 9. x +
x += 13. x = 17. x +x = 19. x= 23. x += 20. x = 24. x βˆ’= 21. x βˆ’= 22. x βˆ’= 25. xβˆ’x βˆ’= 28. x + = x βˆ’ 29. +z =z 26. x βˆ’x βˆ’= 30. p+p βˆ’= 27. x βˆ’x = 31. x βˆ’x βˆ’= 32. xβˆ’x += 36. xβˆ’x βˆ’= 33. x +x += 37. x βˆ’x βˆ’= 35. x βˆ’x += 34. x +x βˆ’= 38. x +x += 42. x +x += 46. ++= TECHNOLOGY 39. x +x = x βˆ’ x= 43. + += 47. 40. xβˆ’x βˆ’= 44. ++= 48. += 41. xβˆ’x += 45. += 49. ++= x-2nd CALC 2:zero,, 50. =x+x βˆ’ 51. =βˆ’x+x βˆ’ 52. =x+x βˆ’ Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 130 CHAPTER 2 EQUATIONS AND INEQUALITIES 53. x+x βˆ’= =x+x βˆ’= 2nd CALC 5:intersection 54. x+x βˆ’= =x+x βˆ’= 2nd CALC 5:intersection 56. b a βˆ’ 58. P =tβˆ’t + t t = EXTENSIONS 55. 57. 59. ax+bx +c = x 119 ft. p =βˆ’x +x βˆ’ x p fi fi 2nd CALC maximum x y REAL-WORLD APPLICATIONS 60. 62. 64. A P =Aβˆ’A + d =t +t t 61. 63. infl P e fl t ft P =βˆ’t +t +≀ t ≀ e fl t Th C =x + R =x βˆ’xfi x fi den. Th f 378 ft x x x x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.6 OTHER TYPES OF EQUATIONS 131 LEARNING OBJECTIVES In this section you will: β€’ ole equations that inole rational eponents. β€’ ole equations by factoring. β€’ ole radical equations. β€’ ole absolute alue equations. 2.
6 OTHER TYPES OF EQUATIONS We have solved linear equations, rational equations, and quadratic equations using several methods. However, there are many other types of equations, and we will investigate a few more types in this section. We will look at equations involving rational exponents, polynomial equations, radical equations, absolute value equations, equations in quadratic form, and some rational equations that can be transformed into quadratics. Solving any equation, however, employs the same basic algebraic rules. We will learn some new techniques as they apply to certain equations, but the algebra never changes. Solving Equations Involving Rational Exponents Rational exponents are exponents that are fractions, where the numerator is a power and the denominator is a root. 1 __ is another way of writing √ For example, 16 2 exponents is a useful skill, as it is highly applicable in calculus. 1 __ 16 ; 8 3 β€” is another way of writing 3 √ β€” 8. The ability to work with rational We can solve equations in which a variable is raised to a rational exponent by raising both sides of the equation to the reciprocal of the exponent. The reason we raise the equation to the reciprocal of the exponent is because we want to eliminate the exponent on the variable term, and a number multiplied by its reciprocal equals 1. 3 1 2 ) = 1, and so on. ) = 1, 3 ( ( _ _ _ For example, 3 2 3 rational exponents A rational exponent indicates a power in the numerator and a root in the denominator. There are multiple ways of writing an expression, a variable, or a number with a rational exponentam) _ n = n √ β€” am = ( √ n β€” m a ) Example 1 Evaluating a Number Raised to a Rational Exponent 2 _. Evaluate 8 3 Solution Whether we take the root first or the power first depends on the number. It is easy to find the cube root of 2 _ 8, so rewrite 8 3 as ( 2)2 ) Try It #1 1 _ βˆ’. Evaluate 64 3 = 4 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 132 CHAPTER 2 EQUATIONS AND INEQUALITIES Example 2 Solve the Equation Including a Variable Raised to a Rational Exponent 5_ 4 = 32. Solve the equation in which a variable is raised to a rational exponent: x Solution The way
to remove the exponent on x is by raising both sides of the equation to a power that is the reciprocal 5 4 _ _., which is of 5 4 Try It #2 3 _ = 125. Solve the equation x 2 5 _ = 32 32) 5 4 x = (2)4 The fifth root of 32 is 2. = 16 Example 3 Solving an Equation Involving Rational Exponents and Factoring 3 1 _ _ = x. Solve 3x 2 4 Solution This equation involves rational exponents as well as factoring rational exponents. Let us take this one step at a time. First, put the variable terms on one side of the equal sign and set the equation equal to zero 3x 3x 2 4 Now, it looks like we should factor the left side, but what do we factor out? We can always factor the term with the 1 2 2 _ _ _ from both terms on the left.. Then, factor out x as x lowest exponent. Rewrite 3x 4 4 2 1 βˆ’ 1 ) = 0 ( 3x _ _ x 4 4 1 _ Where did x 4 come from? Remember, when we multiply two numbers with the same base, we add the exponents. 2 _ back in using the distributive property, we get the expression we had before the factoring, Therefore, if we multiply x 4 3 2 _ _ which is what should happen. We need an exponent such that when added to. Thus, the exponent on x equals 4 4 1 _ in the parentheses is. 4 Let us continue. Now we have two factors and can use the zero factor theorem. 2 1 βˆ’ 1 ) = 0 ( 3x The two solutions are 0 and 1 _. 81 Try It #3 3 _ = 8. Solve: (x + 5 81 Divide both sides by 3. 1 _. Raise both sides to the reciprocal of 4 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.6 OTHER TYPES OF EQUATIONS 133 Solving Equations Using Factoring We have used factoring to solve quadratic equations, but it is a technique that we can use with many types of polynomial equations, which are equations that contain a string of terms including numerical coefficients and variables. When we are faced with an equation containing polynomials of degree higher than 2, we can often solve them by factoring. polynomial equations A polynomial of degree
n is an expression of the type anx n + an βˆ’ 1x n βˆ’ 1 + Δ‹ Δ‹ Δ‹ + a2x 2 + a1 x + a0 where n is a positive integer and an, …, a0 are real numbers and an β‰  0. Setting the polynomial equal to zero gives a polynomial equation. The total number of solutions (real and complex) to a polynomial equation is equal to the highest exponent n. Example 4 Solving a Polynomial by Factoring Solve the polynomial by factoring: 5x 4 = 80x 2. Solution First, set the equation equal to zero. Then factor out what is common to both terms, the GCF. 5x 4 βˆ’ 80x 2 = 0 5x 2(x 2 βˆ’ 16) = 0 Notice that we have the difference of squares in the factor x 2 βˆ’ 16, which we will continue to factor and obtain two solutions. The first term, 5x 2, generates, technically, two solutions as the exponent is 2, but they are the same solution. 5x 2 = 0 x = 0 x 2 βˆ’ 16 = 0 (x βˆ’ 4)(x + 4) = 0 x βˆ’ 4 = 0 or x + 4 = 0 x = 4 or x = βˆ’4 The solutions are 0 (double solution ), 4, and βˆ’4. Analysis We can see the solutions on the graph in Figure 1. The x-coordinates of the points where the graph crosses the x-axis are the solutionsβ€”the x-intercepts. Notice on the graph that at 0, the graph touches the x-axis and bounces back. It does not cross the x-axis. This is typical of double solutions. y 500 400 300 200 100 –1 –100 –200 –300 –400 –500 –5 –4 –3 –2 21 3 4 5 x Figure 1 Try It #4 Solve by factoring: 12x 4 = 3x 2. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 134 CHAPTER 2 EQUATIONS AND INEQUALITIES Example 5 Solve a Polynomial by Grouping Solve a polynomial by grouping: x 3 + x 2 βˆ’ 9x βˆ’ 9 = 0. Solution This polynomial consists of 4 terms, which we can solve by grouping. Grouping procedures require factoring the first two terms and then factoring the last two
terms. If the factors in the parentheses are identical, we can continue the process and solve, unless more factoring is suggested. x 3 + x 2 βˆ’ 9x βˆ’ 9 = 0 x 2(x + 1) βˆ’ 9(x + 1) = 0 (x 2 βˆ’ 9)(x + 1) = 0 The grouping process ends here, as we can factor x 2 βˆ’ 9 using the difference of squares formula. (x 2 βˆ’ 9)(x + 1) = 0 (x βˆ’ 3)(x + 3)(x + 1 or or x + 3 = 0 x = βˆ’3 or or x + 1 = 0 x = βˆ’1 The solutions are 3, βˆ’3, and βˆ’1. Note that the highest exponent is 3 and we obtained 3 solutions. We can see the solutions, the x-intercepts, on the graph in Figure 2. y 25 20 15 10 5 –1 –5 –10 –15 –20 –25 –5 –4 –3 –2 21 3 4 5 x Figure 2 Analysis We looked at solving quadratic equations by factoring when the leading coefficient is 1. When the leading coefficient is not 1, we solved by grouping. Grouping requires four terms, which we obtained by splitting the linear term of quadratic equations. We can also use grouping for some polynomials of degree higher than 2, as we saw here, since there were already four terms. Solving Radical Equations Radical equations are equations that contain variables in the radicand (the expression under a radical symbol), such as β€” 3x + 18 = βˆ’ √ √ Radical equations may have one or more radical terms, and are solved by eliminating each radical, one at a time. We have to be careful when solving radical equations, as it is not unusual to find extraneous solutions, roots that are not, in fact, solutions to the equation. These solutions are not due to a mistake in the solving method, but result from the process of raising both sides of an equation to a power. However, checking each answer in the original equation will confirm the true solutions. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.6 OTHER TYPES OF EQUATIONS 135 radical equations An equation containing terms with a variable in the radicand is called a radical equation. How To… Given a radical equation, solve it. 1. Isolate the radical expression on one side of the equal
sign. Put all remaining terms on the other side. 2. If the radical is a square root, then square both sides of the equation. If it is a cube root, then raise both sides of the equation to the third power. In other words, for an nth root radical, raise both sides to the nth power. Doing so eliminates the radical symbol. 3. Solve the remaining equation. 4. If a radical term still remains, repeat steps 1–2. 5. Confirm solutions by substituting them into the original equation. Example 6 Solving an Equation with One Radical Solve √ β€” 15 βˆ’ 2x = x. Solution The radical is already isolated on the left side of the equal side, so proceed to square both sides. β€” 15 βˆ’ 2x = x √ 2 15 βˆ’ 2x ) = (x)2 β€” ( √ We see that the remaining equation is a quadratic. Set it equal to zero and solve. 15 βˆ’ 2x = x2 0 = x2 + 2x βˆ’ 15 0 = (x + 5)(x βˆ’ 3) 0 = (x + 5) or 0 = (x βˆ’ 3) βˆ’5 = x or 3 = x The proposed solutions are βˆ’5 and 3. Let us check each solution back in the original equation. First, check βˆ’5. β€” 15 βˆ’ 2x = x √ √ β€” β€” 15 βˆ’ 2( βˆ’ 5) = βˆ’5 25 = βˆ’5 5 β‰  βˆ’5 √ This is an extraneous solution. While no mistake was made solving the equation, we found a solution that does not satisfy the original equation. Check 3. The solution is 3. Try It #5 Solve the radical equation: √ β€” x + 3 = 3x βˆ’ 1 √ β€” √ β€” 15 βˆ’ 2x = x 15 βˆ’ 2(3) = 3 9 = 3 3 = 3 √ β€” Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 136 CHAPTER 2 EQUATIONS AND INEQUALITIES Example 7 Solving a Radical Equation Containing Two Radicals Solve √ β€” 2x + 3 + √ β€” x βˆ’ 2 = 4. Solution As this equation contains two radicals, we isolate one radical, eliminate it, and then isolate the second radical. β€” √ 2x + 2x + βˆ’ √ 2x + ) β€” ( √ Subtract οΏ½
οΏ½ β€” x βˆ’ 2 from both sides. Square both sides. Use the perfect square formula to expand the right side: (a βˆ’ b)2 = a2 βˆ’2ab + b2x βˆ’ 2) x βˆ’ 2 β€” 2x + 3 = (4)2 βˆ’ 2(4) √ 2x + 3 = 16 βˆ’ 8 √ 2x + 3 = 14 + x βˆ’ 8 √ β€” x βˆ’ 11 = βˆ’8 √ (x βˆ’ 11)2 = ( βˆ’ ) Combine like terms. Isolate the second radical. Square both sides. x 2 βˆ’ 22x + 121 = 64(x βˆ’ 2) Now that both radicals have been eliminated, set the quadratic equal to zero and solve. x 2 βˆ’ 22x + 121 = 64x βˆ’ 128 x 2 βˆ’ 86x + 249 = 0 (x βˆ’ 3)(x βˆ’ 83) = 0 x βˆ’ 3 = 0 or x βˆ’ 83 = 0 x = 3 or x = 83 Factor and solve. The proposed solutions are 3 and 83. Check each solution in the original equation. √ √ One solution is 3. Check 83. β€” 2x + 2x + 3 = 4 βˆ’ √ √ 2(3 √ β€” √ β€” 2x + 2x + 3 = 4 βˆ’ √ √ 2(83) + 3 = 4 βˆ’ √ 169 = 4 βˆ’ √ 13 β‰  βˆ’383 βˆ’ 2) β€” 81 The only solution is 3. We see that 83 is an extraneous solution. Try It #6 Solve the equation with two radicals: √ β€” 3x + 7 + √ β€” x + 2 = 1. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.6 OTHER TYPES OF EQUATIONS 137 Solving an Absolute Value Equation Next, we will learn how to solve an absolute value equation. To solve an equation such as |2x βˆ’ 6| = 8, we notice that the absolute value will be equal to 8 if the quantity inside the absolute value bars is 8 or βˆ’8. This leads to two different equations we can solve independently. or 2x βˆ’ 6 = 8 2x = 14 x = 7 2x βˆ’ 6 = βˆ’8 2x = βˆ’2 x = βˆ’1 Knowing how to solve problems involving absolute value functions is useful. For example, we may need to identify numbers or points on a line that are at
a specified distance from a given reference point. absolute value equations The absolute value of x is written as |x|. It has the following properties: If x β‰₯ 0, then |x| = x. If x < 0, then |x| = βˆ’x. For real numbers A and B, an equation of the form |A| = B, with B β‰₯ 0, will have solutions when A = B or A = βˆ’B. If B < 0, the equation |A| = B has no solution. An absolute value equation in the form |ax + b| = c has the following properties: If c < 0, |ax + b| = c has no solution. If c = 0, |ax + b| = c has one solution. If c > 0, |ax + b| = c has two solutions. How To… Given an absolute value equation, solve it. 1. Isolate the absolute value expression on one side of the equal sign. 2. If c > 0, write and solve two equations: ax + b = c and ax + b = βˆ’ c. Example 8 Solving Absolute Value Equations Solve the following absolute value equations: b. |3x + 4| = βˆ’9 a. |6x + 4| = 8 Solution a. |6x + 4| = 8 Write two equations and solve each: 6x + 4 = 8 6x = 4 2 _ x = 3 2 _ and βˆ’2. The two solutions are 3 b. |3x + 4| = βˆ’9 c. |3x βˆ’ 5| βˆ’ 4 = 6 d. |βˆ’5x + 10| = 0 6x + 4 = βˆ’8 6x = βˆ’12 x = βˆ’2 There is no solution as an absolute value cannot be negative. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 138 CHAPTER 2 EQUATIONS AND INEQUALITIES c. |3x βˆ’ 5| βˆ’ 4 = 6 Isolate the absolute value expression and then write two equations. |3x βˆ’ 5| βˆ’ 4 = 6 |3x βˆ’ 5| = 10 3x βˆ’ 5 = 10 3x = 15 x = 5 5 _ There are two solutions: 5 and βˆ’. 3 d. |βˆ’5x + 10| = 0 3x βˆ’ 5 = βˆ’10 3x = βˆ’5 5 _ x = βˆ’ 3 The equation is set equal to zero,
so we have to write only one equation. βˆ’5x + 10 = 0 βˆ’5x = βˆ’10 x = 2 There is one solution: 2. Try It #7 Solve the absolute value equation: |1 βˆ’ 4x| + 8 = 13. Solving Other Types of Equations There are many other types of equations in addition to the ones we have discussed so far. We will see more of them throughout the text. Here, we will discuss equations that are in quadratic form, and rational equations that result in a quadratic. Solving Equations in Quadratic Form Equations in quadratic form are equations with three terms. The first term has a power other than 2. The middle term has an exponent that is one-half the exponent of the leading term. The third term is a constant. We can solve equations in this form as if they were quadratic. A few examples of these equations include x 4 βˆ’ 5x 2 + 4 = 0, x 6 + 7x 3 βˆ’ 8 = 0, 2 1 _ _ + 2 = 0. In each one, doubling the exponent of the middle term equals the exponent on the leading + 4x and x 3 3 term. We can solve these equations by substituting a variable for the middle term. quadratic form If the exponent on the middle term is one-half of the exponent on the leading term, we have an equation in quadratic form, which we can solve as if it were a quadratic. We substitute a variable for the middle term to solve equations in quadratic form. How To… Given an equation quadratic in form, solve it. 1. Identify the exponent on the leading term and determine whether it is double the exponent on the middle term. 2. If it is, substitute a variable, such as u, for the variable portion of the middle term. 3. Rewrite the equation so that it takes on the standard form of a quadratic. 4. Solve using one of the usual methods for solving a quadratic. 5. Replace the substitution variable with the original term. 6. Solve the remaining equation. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.6 OTHER TYPES OF EQUATIONS 139 Example 9 Solving a Fourth-degree Equation in Quadratic Form Solve this fourth-degree equation: 3x 4 βˆ’ 2x 2
βˆ’ 1 = 0. Solution This equation fits the main criteria, that the power on the leading term is double the power on the middle term. Next, we will make a substitution for the variable term in the middle. Let u = x 2. Rewrite the equation in u. Now solve the quadratic. 3u2 βˆ’ 2u βˆ’ 1 = 0 3u2 βˆ’ 2u βˆ’ 1 = 0 (3u + 1)(u βˆ’ 1) = 0 Solve each factor and replace the original term for u. 3u + 1 = 0 3u = βˆ’1 1 _ u = βˆ’ 3 1 _ x2 = βˆ’ 3 x = Β±i √ __ 1 _ 3 __ 1 _ and Β± 1. 3 The solutions are Β± i √ Try It #8 Solve using substitution: x 4 βˆ’ 8x 2 βˆ’ 9 = 0. u βˆ’ 1 = 0 u = 1 x2 = 1 x = Β±1 Example 10 Solving an Equation in Quadratic Form Containing a Binomial Solve the equation in quadratic form: (x + 2)2 + 11(x + 2) βˆ’ 12 = 0. Solution This equation contains a binomial in place of the single variable. The tendency is to expand what is presented. However, recognizing that it fits the criteria for being in quadratic form makes all the difference in the solving process. First, make a substitution, letting u = x + 2. Then rewrite the equation in u. Solve using the zero-factor property and then replace u with the original expression. u2 + 11u βˆ’ 12 = 0 (u + 12)(u βˆ’ 1) = 0 u + 12 = 0 u = βˆ’12 x + 2 = βˆ’12 x = βˆ’14 1 The second factor results in We have two solutions: βˆ’14 and βˆ’1. Try It #9 Solve: (x βˆ’ 5)2 βˆ’ 4(x βˆ’ 5) βˆ’ 21 = 0. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 140 CHAPTER 2 EQUATIONS AND INEQUALITIES Solving Rational Equations Resulting in a Quadratic Earlier, we solved rational equations. Sometimes, solving a rational equation results in a quadratic. When this happens, we continue the solution by simplifying the quadratic equation by one of the methods we have seen. It may turn out that there is no solution. Example 11 Solving a
Rational Equation Leading to a Quadratic Solve the following rational equation: βˆ’4x 8 _. x2 βˆ’ 1 Solution We want all denominators in factored form to find the LCD. Two of the denominators cannot be factored further. However, x2 βˆ’1 = (x + 1)(x βˆ’ 1). Then, the LCD is (x + 1)(x βˆ’ 1). Next, we multiply the whole equation by the LCD. (x + 1)(x βˆ’ 1) ( βˆ’4x 8 __ (x + 1)(x βˆ’ 1) ) (x + 1)(x βˆ’ 1) βˆ’4x(x + 1) + 4(x βˆ’ 1) = βˆ’8 βˆ’4x2 βˆ’ 4x + 4x βˆ’ 4 = βˆ’8 βˆ’4x2 + 4 = 0 βˆ’4(x2 βˆ’ 1) = 0 βˆ’4(x + 1)(x βˆ’ 1) = 0 x = βˆ’1 or x = 1 In this case, either solution produces a zero in the denominator in the original equation. Thus, there is no solution. Try It #10 Solve 3x + 2 ______ x βˆ’ 2 1 __ = + x βˆ’2 ______ x2 βˆ’ 2x. Access these online resources for additional instruction and practice with different types of equations. β€’ Rational Equation with No Solution (http://openstaxcollege.org/l/rateqnosoln) β€’ Solving Equations with Rational Exponents Using Reciprocal Powers (http://openstaxcollege.org/l/ratexprecpexp) β€’ Solving Radical Equations part 1 of 2 (http://openstaxcollege.org/l/radeqsolvepart1) β€’ Solving Radical Equations part 2 of 2 (http://openstaxcollege.org/l/radeqsolvepart2) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.6 SECTION EXERCISES 141 2.6 SECTION EXERCISES VERBAL 1. In a radical equation, what does it mean if a number 2. Explain why possible solutions must be checked in is an extraneous solution? 3 _ 3. Your friend tries to calculate the value βˆ’ 9 and 2 keeps getting an ERROR message. What mistake is he or she probably making? 5. Explain how to change a rational exponent into the correct radical expression. radical equations. 4. Explain why |2x +
5| = βˆ’7 has no solutions. ALGEBRAIC For the following exercises, solve the rational exponent equation. Use factoring where necessary. 2 _ = 16 6. x 3 2 _ = 4 10. (x + 1) 3 3 _ = 27 7 βˆ’ 5x 11 βˆ’ 4x βˆ’ 3x 12. (x βˆ’ 1) 4 For the following exercises, solve the following polynomial equations by grouping and factoring. 13. x 3 + 2x 2 βˆ’ x βˆ’ 2 = 0 16. x 3 + 3x 2 βˆ’ 25x βˆ’ 75 = 0 19. 5x 3 + 45x = 2x 2 + 18 14. 3x 3 βˆ’ 6x 2 βˆ’ 27x + 54 = 0 17. m3 + m2 βˆ’ m βˆ’ 1 = 0 15. 4y 3 βˆ’ 9y = 0 18. 2x 5 βˆ’14x 3 = 0 For the following exercises, solve the radical equation. Be sure to check all solutions to eliminate extraneous solutions. 20. √ β€” 3x βˆ’ 1 βˆ’ 2 = 0 23. √ β€” 3t + 5 = 7 β€” 21. √ x βˆ’ 7 = 5 β€” 24. √ t + 1 + 9 = 7 22 25. √ β€” 12 βˆ’ x = x 26. √ β€” 2x + 3 βˆ’ √ β€” x + 2 = 2 27. √ β€” 3x + 7 + √ β€” x + 2 = 1 28. √ β€” 2x + 3 βˆ’ √ β€” x + 1 = 1 For the following exercises, solve the equation involving absolute value. 29. |3x βˆ’ 4| = 8 33. |2x βˆ’ 1| βˆ’ 7 = βˆ’2 30. |2x βˆ’ 3| = βˆ’2 34. |2x + 1| βˆ’ 2 = βˆ’3 31. |1 βˆ’ 4x| βˆ’ 1 = 5 35. |x + 5| = 0 32. |4x + 1| βˆ’ 3 = 6 36. βˆ’|2x + 1| = βˆ’3 For the following exercises, solve the equation by identifying the quadratic form. Use a substitute variable and find all real solutions by factoring. 37. x 4 βˆ’ 10x 2 + 9 = 0 40. (x + 1)2 βˆ’ 8(x + 1) βˆ’ 9 = 0 38. 4(t βˆ’ 1)2 βˆ’ 9(t βˆ’ 1) = βˆ’2 41. (x βˆ’ 3)2 βˆ’ 4
= 0 39. (x 2 βˆ’ 1)2 + (x 2 βˆ’ 1) βˆ’ 12 = 0 EXTENSIONS For the following exercises, solve for the unknown variable. 42. xβˆ’2 βˆ’ xβˆ’1 βˆ’ 12 = 0 β€” |x|2 = x 43. √ 44. t 25 βˆ’ t 5 + 1 = 0 45. |x 2 + 2x βˆ’ 36| = 12 REAL-WORLD APPLICATIONS For the following exercises, use the model for the period of a pendulum, T, such that T = 2Ο€ √ ___ L _ g, where the length of the pendulum is L and the acceleration due to gravity is g. 46. If the acceleration due to gravity is 9.8 m/s2 and the period equals 1 s, find the length to the nearest cm (100 cm = 1 m). 47. If the gravity is 32 ft/s2 and the period equals 1 s, find the length to the nearest in. (12 in. = 1 ft). Round your answer to the nearest in. For the following exercises, use a model for body surface area, BSA, such that BSA = √ kg and h = height in cm. 48. Find the height of a 72-kg female to the nearest cm whose BSA = 1.8. whose BSA = 2.1. 49. Find the weight of a 177-cm male to the nearest kg _____ wh _ 3600, where w = weight in Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 142 CHAPTER 2 EQUATIONS AND INEQUALITIES LEARNING OBJECTIVES In this section you will: β€’ Use interal notation. β€’ ole linear inequalities. β€’ ole compound linear inequalities of both and and or type. β€’ ole absolute alue inequalities. 2.7 LINEAR INEQUALITIES AND ABSOLUTE VALUE INEQUALITIES It is not easy to make the honor role at most top universities. Suppose students were required to carry a course load of at least 12 credit hours and maintain a grade point average of 3.5 or above. How could these honor roll requirements be expressed mathematically? In this section, we will explore various ways to express different sets of numbers, inequalities, and absolute value inequalities. Figure 1 Using Interval Notation Indicating the solution to an inequality such as x β‰₯ 4 can be achieved in several ways
. We can use a number line as shown in Figure 2. The blue ray begins at x = 4 and, as indicated by the arrowhead, continues to infinity, which illustrates that the solution set includes all real numbers greater than or equal to 4. 5 6 Figure 2 We can use set-builder notation: {x|x β‰₯ 4}, which translates to β€œall real numbers x such that x is greater than or equal to 4.” Notice that braces are used to indicate a set. 9 10 11 8 7 1 2 3 4 0 The third method is interval notation, in which solution sets are indicated with parentheses or brackets. The solutions to x β‰₯ 4 are represented as [4, ∞). This is perhaps the most useful method, as it applies to concepts studied later in this course and to other higher-level math courses. The main concept to remember is that parentheses represent solutions greater or less than the number, and brackets represent solutions that are greater than or equal to or less than or equal to the number. Use parentheses to represent infinity or negative infinity, since positive and negative infinity are not numbers in the usual sense of the word and, therefore, cannot be β€œequaled.” A few examples of an interval, or a set of numbers in which a solution falls, are [βˆ’2, 6), or all numbers between βˆ’2 and 6, including βˆ’2, but not including 6; (βˆ’1, 0), all real numbers between, but not including βˆ’1 and 0; and (βˆ’βˆž, 1], all real numbers less than and including 1. Table 1 outlines the possibilities. Set Indicated Set-Builder Notation Interval Notation All real numbers between a and b, but not including a or b {x| a < x < b} All real numbers greater than a, but not including a All real numbers less than b, but not including b All real numbers greater than a, including a {x| x > a} {x| x < b} {x| x β‰₯ a} (a, b) (a, ∞) (βˆ’βˆž, b) [a, ∞) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.7 LINEAR INEQUALITIES AND ABSOLUTE VALUE INEQUALITIES 143 Set Indicated Set-Builder Notation Interval Notation All real numbers less than b, including b All real numbers between
a and b, including a All real numbers between a and b, including b All real numbers between a and b, including a and b {x| x ≀ b} {x| a ≀ x < b} {x| a < x ≀ b} {x| a ≀ x ≀ b} (βˆ’βˆž, b] [a, b) (a, b] [a, b] All real numbers less than a or greater than b {x| x < a and x > b} (βˆ’βˆž, a) βˆͺ (b, ∞) All real numbers {x| x is all real numbers} (βˆ’βˆž, ∞) Table 1 Example 1 Using Interval Notation to Express All Real Numbers Greater Than or Equal to a Use interval notation to indicate all real numbers greater than or equal to βˆ’2. Solution Use a bracket on the left of βˆ’2 and parentheses after infinity: [βˆ’2, ∞). The bracket indicates that βˆ’2 is included in the set with all real numbers greater than βˆ’2 to infinity. Try It #1 Use interval notation to indicate all real numbers between and including βˆ’3 and 5. Example 2 Using Interval Notation to Express All Real Numbers Less Than or Equal to a or Greater Than or Equal to b Write the interval expressing all real numbers less than or equal to βˆ’1 or greater than or equal to 1. Solution We have to write two intervals for this example. The first interval must indicate all real numbers less than or equal to 1. So, this interval begins at βˆ’ ∞ and ends at βˆ’1, which is written as (βˆ’βˆž, βˆ’1]. The second interval must show all real numbers greater than or equal to 1, which is written as [1, ∞). However, we want to combine these two sets. We accomplish this by inserting the union symbol, βˆͺ, between the two intervals. (βˆ’βˆž, βˆ’1] βˆͺ [1, ∞) Try It #2 Express all real numbers less than βˆ’2 or greater than or equal to 3 in interval notation. Using the Properties of Inequalities When we work with inequalities, we can usually treat them similarly to but not exactly as we treat equalities. We can use the addition property and the multiplication property to help us solve them. The one exception is when we multiply or divide by a negative number; doing so reverses the inequality symbol. properties of inequalities Addition Property If a < b, then a + c < b +
c. Multiplication Property If a < b and c > 0, then ac < bc. If a < b and c < 0, then ac > bc. These properties also apply to a ≀ b, a > b, and a β‰₯ b. Example 3 Demonstrating the Addition Property Illustrate the addition property for inequalities by solving each of the following: a. x βˆ’ 15 < 4 b. 6 β‰₯ x βˆ’ 1 c. x + 7 > 9 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 144 CHAPTER 2 EQUATIONS AND INEQUALITIES Solution The addition property for inequalities states that if an inequality exists, adding or subtracting the same number on both sides does not change the inequality. a. x βˆ’ 15 < 4 x βˆ’ 15 + 15 < 4 + 15 b. x < 19 Try It #3 Solve: 3xβˆ’2 < 1. Add 15 to both sides. Add 1 to both sides. Subtract 7 from both sides. Example 4 Demonstrating the Multiplication Property Illustrate the multiplication property for inequalities by solving each of the following: a. 3x < 6 b. βˆ’2x βˆ’ 1 β‰₯ 5 c. 5 βˆ’ x > 10 Solution a. 3x < 6 1 1 _ _ (3x) < (6) 3 3 x < 2 b. βˆ’2x βˆ’ 1 β‰₯ 5 βˆ’2x β‰₯ 6 1 1 ) ) (βˆ’2x) β‰₯ (6. 5 βˆ’ x > 10 βˆ’x > 5 1 _ Multiply by βˆ’. 2 Reverse the inequality. (βˆ’1)(βˆ’x) > (5)( βˆ’ 1) Multiply by βˆ’ 1. x < βˆ’ 5 Reverse the inequality. Try It #4 Solve: 4x + 7 β‰₯ 2x βˆ’ 3. Solving Inequalities in One Variable Algebraically As the examples have shown, we can perform the same operations on both sides of an inequality, just as we do with equations; we combine like terms and perform operations. To solve, we isolate the variable. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.7 LINEAR INEQUALITIES AND ABSOLUTE VALUE INEQUALITIES 145 Example 5 Solving an Inequality Algebraically Solve the inequality: 13 βˆ’ 7x β‰₯ 10x βˆ’ 4. Solution Solving this inequality is similar
to solving an equation up until the last step. 13 βˆ’ 7x β‰₯ 10x βˆ’ 4 13 βˆ’ 17x β‰₯ βˆ’4 βˆ’17x β‰₯ βˆ’17 x ≀ 1 Move variable terms to one side of the inequality. Isolate the variable term. Dividing both sides by βˆ’17 reverses the inequality. The solution set is given by the interval (βˆ’βˆž, 1], or all real numbers less than and including 1. Try It #5 1 _ Solve the inequality and write the answer using interval notation: βˆ’x + 4 < x + 1. 2 Example 6 Solving an Inequality with Fractions 5 3 2 _ _ _ x β‰₯ βˆ’ Solve the following inequality and write the answer in interval notation: βˆ’ + x. 4 3 8 Solution We begin solving in the same way we do when solving an equation. βˆ’ x βˆ’ 9 _ 12 12 5 17 _ _ x β‰₯ βˆ’ 8 12 5 12 ) ( βˆ’ _ _ x ≀ βˆ’ 17 8 15 _ x ≀ 34 ]. 15_ The solution set is the interval ( βˆ’βˆž, 34 βˆ’ Put variable terms on one side. Write fractions with common denominator. Multiplying by a negative number reverses the inequality. Try It #6 5 8 3 _ _ _ Solve the inequality and write the answer in interval notation: βˆ’ + x ≀ x. 4 3 6 Understanding Compound Inequalities A compound inequality includes two inequalities in one statement. A statement such as 4 < x ≀ 6 means 4 < x and x ≀ 6. There are two ways to solve compound inequalities: separating them into two separate inequalities or leaving the compound inequality intact and performing operations on all three parts at the same time. We will illustrate both methods. Example 7 Solving a Compound Inequality Solve the compound inequality: 3 ≀ 2x + 2 < 6. Solution The first method is to write two separate inequalities: 3 ≀ 2x + 2 and 2x + 2 < 6. We solve them independently. 3 ≀ 2x + 2 1 ≀ 2x 1 _ ≀ x 2 and 2x + 2 < 6 2x < 4 x < 2 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 146 CHAPTER 2 EQUATIONS AND INEQUALITIES Then, we can rewrite the solution as a compound inequality, the same way the problem began. 1_ ≀ x < 2 2 1 In interval notation, the solution is written as [, 2 ). _
2 The second method is to leave the compound inequality intact, and perform solving procedures on the three parts at the same time. 3 ≀ 2x + 2 < 6 1 ≀ 2x < 4 1 _ ≀ x < 2 2 Isolate the variable term, and subtract 2 from all three parts. Divide through all three parts by 2. 1 We get the same solution: [, 2 ). _ 2 Try It #7 Solve the compound inequality: 4 < 2x βˆ’ 8 ≀ 10. Example 8 Solving a Compound Inequality with the Variable in All Three Parts Solve the compound inequality with variables in all three parts: 3 + x > 7x βˆ’ 2 > 5x βˆ’ 10. Solution Let's try the first method. Write two inequalities: and 7x βˆ’ 2 > 5x βˆ’ 10 2x βˆ’ 2 > βˆ’10 2x > βˆ’8 3 + x > 7x βˆ’ 2 3 > 6x βˆ’ 2 5 > 6x 4 < x 6 5 5 ). Notice that when we write the solution in interval or in interval notation ( βˆ’4, _ _ The solution set is βˆ’4 < x < 6 6 notation, the smaller number comes first. We read intervals from left to right, as they appear on a number line. See Figure 3. x > βˆ’4 4 Figure 3 5 6 Try It #8 Solve the compound inequality: 3y < 4 βˆ’ 5y < 5 + 3y. Solving Absolute Value Inequalities As we know, the absolute value of a quantity is a positive number or zero. From the origin, a point located at (βˆ’x, 0) has an absolute value of x, as it is x units away. Consider absolute value as the distance from one point to another point. Regardless of direction, positive or negative, the distance between the two points is represented as a positive number or zero. An absolute value inequality is an equation of the form |A| < B, |A| ≀ B, |A| > B, or |A| β‰₯ B, Where A, and sometimes B, represents an algebraic expression dependent on a variable x. Solving the inequality means finding the set of all x -values that satisfy the problem. Usually this set will be an interval or the union of two intervals and will include a range of values. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.7 LINEAR INEQUALITIES AND ABSOL
UTE VALUE INEQUALITIES 147 There are two basic approaches to solving absolute value inequalities: graphical and algebraic. The advantage of the graphical approach is we can read the solution by interpreting the graphs of two equations. The advantage of the algebraic approach is that solutions are exact, as precise solutions are sometimes difficult to read from a graph. Suppose we want to know all possible returns on an investment if we could earn some amount of money within $200 of $600. We can solve algebraically for the set of x-values such that the distance between x and 600 is less than or equal to 200. We represent the distance between x and 600 as |x βˆ’ 600|, and therefore, |x βˆ’ 600| ≀ 200 or βˆ’200 ≀ x βˆ’ 600 ≀ 200 βˆ’200 + 600 ≀ x βˆ’ 600 + 600 ≀ 200 + 600 400 ≀ x ≀ 800 This means our returns would be between $400 and $800. To solve absolute value inequalities, just as with absolute value equations, we write two inequalities and then solve them independently. absolute value inequalities For an algebraic expression X, and k > 0, an absolute value inequality is an inequality of the form |X| < k is equivalent to βˆ’ k < X < k |X| > k is equivalent to X < βˆ’k or X > k These statements also apply to |X| ≀ k and |X| β‰₯ k. Example 9 Determining a Number within a Prescribed Distance Describe all values x within a distance of 4 from the number 5. Solution We want the distance between x and 5 to be less than or equal to 4. We can draw a number line, such as in Figure 4, to represent the condition to be satisfied. 4 4 5 Figure 4 The distance from x to 5 can be represented using an absolute value symbol, |x βˆ’ 5|. Write the values of x that satisfy the condition as an absolute value inequality. |x βˆ’ 5| ≀ 4 We need to write two inequalities as there are always two solutions to an absolute value equation. x βˆ’ 5 ≀ 4 x ≀ 9 and x βˆ’ 5 β‰₯ βˆ’ 4 x β‰₯ 1 If the solution set is x ≀ 9 and x β‰₯ 1, then the solution set is an interval including all real numbers between and including 1 and 9. So |x βˆ’ 5| ≀ 4 is equivalent to [1, 9] in interval notation. Try It #9 Describe all x-values within a distance of 3 from the number 2. Example 10 Solving
an Absolute Value Inequality Solve |x βˆ’ 1| ≀ 3. Solution |x βˆ’ 1| ≀ 3 βˆ’3 ≀ x βˆ’ 1 ≀ 3 βˆ’2 ≀ x ≀ 4 [βˆ’2, 4] Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 148 CHAPTER 2 EQUATIONS AND INEQUALITIES Example 11 Using a Graphical Approach to Solve Absolute Value Inequalities 1 _ Given the equation y = βˆ’ |4x βˆ’ 5| + 3, determine the x-values for which the y-values are negative. 2 1 _ Solution We are trying to determine where y < 0, which is when βˆ’ |4x βˆ’ 5| + 3 < 0. We begin by isolating the 2 absolute value. 1 _ |4x βˆ’ 5| < βˆ’ 3 βˆ’ 2 |4x βˆ’ 5| > 6 Next, we solve for the equality |4x βˆ’ 5| = 6. Multiply both sides by –2, and reverse the inequality. 4x βˆ’ 5 = 6 or 4x βˆ’ 5 = βˆ’6 4x = 11 x = 11_ 4 4x = βˆ’1 1 _ x = βˆ’ 4 Now, we can examine the graph to observe where the y-values are negative. We observe where the branches are below the x-axis. Notice that it is not important exactly what the graph looks like, as long as we know that it crosses the 11 1 ___ _ horizontal axis at x = βˆ’ and x = 4 4, and that the graph opens downward. See Figure 5. y 5 4 3 2 1 –1 –2 –3 –4 –5 x = βˆ’0.25 –3 –1 –4 –5 Below x-axis –2 21 x = 2.75 x 3 4 5 Below x-axis Figure 5 Try It #10 Solve βˆ’ 2|k βˆ’ 4| ≀ βˆ’ 6. Access these online resources for additional instruction and practice with linear inequalities and absolute value inequalities. β€’ Interval Notation (http://openstaxcollege.org/l/intervalnotn) β€’ How to Solve Linear Inequalities (http://openstaxcollege.org/l/solvelinineq) β€’ How to Solve an Inequality (http://openstaxcollege.org/l/solveineq) β€’ Absolute Value Equations (http://openstaxcollege.org/l/absvaleq) β€’ Compound
Inequalities (http://openstaxcollege.org/l/compndineqs) β€’ Absolute Value Inequalities (http://openstaxcollege.org/l/absvalineqs) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.7 SECTION EXERCISES 149 2.7 SECTION EXERCISES VERBAL 1. 2. βˆ’x > x<βˆ’ x +<x + < 3. 4. x +>x + > 5. y =|x βˆ’| ALGEBRAIC 6. x βˆ’β‰€ 7. x +β‰₯x βˆ’ 8. βˆ’x +>x βˆ’ 9. x +β‰₯x βˆ’ + 10. βˆ’ xβ‰€βˆ’ x 11. βˆ’x βˆ’+>x βˆ’βˆ’x 12. βˆ’x +>βˆ’x + 13. x+ βˆ’ x+ β‰₯ 14. xβˆ’ + x+ ≀ 15. βˆ’<x+≀ 17. y<βˆ’y<+y 19. x+<x+ + +< <+< 21. 23. 25. 27. 29. + ο£Ώ + ο£Ώ +< ο£Ώ ο£Ώ 16. x+>xβˆ’>xβˆ’ 18. xβˆ’<βˆ’x+β‰₯ 20. 22. 24. 26. 28. > ο£Ώ <+ ο£Ώ < < +> ο£Ώ ο£Ώ + ο£Ώ fi 30. |x+|β‰₯βˆ’ 31. |x+|< 32. |xβˆ’|> 33. |x +|+≀ 34. |xβˆ’|+β‰₯ 35. |βˆ’x+|≀ 36. |xβˆ’|<βˆ’ 37. |x βˆ’|>βˆ’ 38. xβˆ’ | |< Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 150 CHAPTER 2 EQUATIONS AND INEQUALITIES x- 39. 40. 41. 42. GRAPHICAL xr fi 43. |xβˆ’|> 44. |x+|β‰₯ 45. |x+|≀ 46. |x βˆ’|< 47. |xβˆ’|< ftyy y- 48. x +<x βˆ’ 49. x βˆ’>
x + 50. x +>x + 51. x βˆ’ x +> 52. x +< x + NUMERIC 53. x|βˆ’< x < 54. x|xβ‰₯ 55. x|x< 56. x| x 57. βˆ’βˆž 58. ∞ 59. βˆ’ 60. βˆ’βˆͺ∞ 61. βˆ’ βˆ’ 62. βˆ’ βˆ’ 63. TECHNOLOGY fts 1 Y2 =MATHNum1:abs( 2 nd CALC 5:intersection1st curve, enter, 2 nd curve, enter, guess, enter x-fi 64. |x +|βˆ’< 65. βˆ’ |x +|< 66. |x +|βˆ’> 67. |x βˆ’|< 68. |x +|β‰₯ EXTENSIONS 69. |x +|=|x +| 70. xβˆ’x> 71. x βˆ’ ≀ x β‰  βˆ’ x + REAL-WORLD APPLICATIONS 72. p=βˆ’x +x βˆ’fi xfi 73. V =T V T 74.. Th C =+x βˆ’ Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 2 REVIEW 151 CHAPTER 2 REVIEW Key Terms absolute value equation an equation in which the variable appears in absolute value bars, typically with two solutions, one accounting for the positive expression and one for the negative expression area in square units, the area formula used in this section is used to find the area of any two-dimensional rectangular region: A = LW Cartesian coordinate system a grid system designed with perpendicular axes invented by RenΓ© Descartes completing the square a process for solving quadratic equations in which terms are added to or subtracted from both sides of the equation in order to make one side a perfect square complex conjugate a complex number containing the same terms as another complex number, but with the opposite operator. Multiplying a complex number by its conjugate yields a real number. complex number the sum of a real number and an imaginary number; the standard form is a + bi, where a is the real part and b is the complex part. complex plane the coordinate plane in which the horizontal axis represents the real component of a complex number, and the vertical axis represents the imaginary component, labeled i. compound inequality a problem or a statement that includes two inequalities conditional equation an equation that is true for some values of the variable discriminant the expression
under the radical in the quadratic formula that indicates the nature of the solutions, real or complex, rational or irrational, single or double roots. distance formula a formula that can be used to find the length of a line segment if the endpoints are known equation in two variables a mathematical statement, typically written in x and y, in which two expressions are equal equations in quadratic form equations with a power other than 2 but with a middle term with an exponent that is one- half the exponent of the leading term extraneous solutions any solutions obtained that are not valid in the original equation graph in two variables the graph of an equation in two variables, which is always shown in two variables in the two- dimensional plane identity equation an equation that is true for all values of the variable imaginary number the square root of βˆ’1: i = √ βˆ’1. β€” inconsistent equation an equation producing a false result intercepts the points at which the graph of an equation crosses the x-axis and the y-axis interval an interval describes a set of numbers within which a solution falls interval notation a mathematical statement that describes a solution set and uses parentheses or brackets to indicate where an interval begins and ends linear equation an algebraic equation in which each term is either a constant or the product of a constant and the first power of a variable linear inequality similar to a linear equation except that the solutions will include sets of numbers midpoint formula a formula to find the point that divides a line segment into two parts of equal length ordered pair a pair of numbers indicating horizontal displacement and vertical displacement from the origin; also known as a coordinate pair, (x, y) origin the point where the two axes cross in the center of the plane, described by the ordered pair (0, 0) perimeter in linear units, the perimeter formula is used to find the linear measurement, or outside length and width, around a two-dimensional regular object; for a rectangle: P = 2L + 2W polynomial equation an equation containing a string of terms including numerical coefficients and variables raised to whole-number exponents Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 152 CHAPTER 2 EQUATIONS AND INEQUALITIES Pythagorean Theorem a theorem that states the relationship among the lengths of the sides of a right triangle, used to solve right triangle problems quadrant one quarter of the coordinate plane, created when the axes divide the plane into four sections quadratic equation an equation containing a
second-degree polynomial; can be solved using multiple methods quadratic formula a formula that will solve all quadratic equations radical equation an equation containing at least one radical term where the variable is part of the radicand rational equation an equation consisting of a fraction of polynomials slope the change in y-values over the change in x-values solution set the set of all solutions to an equation square root property one of the methods used to solve a quadratic equation, in which the x 2 term is isolated so that the square root of both sides of the equation can be taken to solve for x volume in cubic units, the volume measurement includes length, width, and depth: V = LWH x-axis the common name of the horizontal axis on a coordinate plane; a number line increasing from left to right x-coordinate the first coordinate of an ordered pair, representing the horizontal displacement and direction from the origin x-intercept the point where a graph intersects the x-axis; an ordered pair with a y-coordinate of zero y-axis the common name of the vertical axis on a coordinate plane; a number line increasing from bottom to top y-coordinate the second coordinate of an ordered pair, representing the vertical displacement and direction from the origin y-intercept a point where a graph intercepts the y-axis; an ordered pair with an x-coordinate of zero zero-product property the property that formally states that multiplication by zero is zero, so that each factor of a quadratic equation can be set equal to zero to solve equations Key Equations quadratic formula Key Concepts x = β€” βˆ’b Β± √ b2 βˆ’ 4ac ______________ 2a 2.1 The Rectangular Coordinate Systems and Graphs β€’ We can locate, or plot, points in the Cartesian coordinate system using ordered pairs, which are defined as displacement from the x-axis and displacement from the y-axis. See Example 1. β€’ An equation can be graphed in the plane by creating a table of values and plotting points. See Example 2. β€’ Using a graphing calculator or a computer program makes graphing equations faster and more accurate. Equations usually have to be entered in the form y =. See Example 3. β€’ Finding the x- and y-intercepts can define the graph of a line. These are the points where the graph crosses the axes. See Example 4. β€’ The distance formula is derived from the Pythagorean Theorem
and is used to find the length of a line segment. See Example 5 and Example 6. β€’ The midpoint formula provides a method of finding the coordinates of the midpoint dividing the sum of the x-coordinates and the sum of the y-coordinates of the endpoints by 2. See Example 7 and Example 8. 2.2 Linear Equations in One Variable β€’ We can solve linear equations in one variable in the form ax + b = 0 using standard algebraic properties. See Example 1 and Example 2. β€’ A rational expression is a quotient of two polynomials. We use the LCD to clear the fractions from an equation. See Example 3 and Example 4. β€’ All solutions to a rational equation should be verified within the original equation to avoid an undefined term, or zero in the denominator. See Example 5, Example 6, and Example 7. β€’ Given two points, we can find the slope of a line using the slope formula. See Example 8. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 2 REVIEW 153 β€’ We can identify the slope and y-intercept of an equation in slope-intercept form. See Example 9. β€’ We can find the equation of a line given the slope and a point. See Example 10. β€’ We can also find the equation of a line given two points. Find the slope and use the point-slope formula. See Example 11. β€’ The standard form of a line has no fractions. See Example 12. β€’ Horizontal lines have a slope of zero and are defined as y = c, where c is a constant. β€’ Vertical lines have an undefined slope (zero in the denominator), and are defined as x = c, where c is a constant. See Example 13. β€’ Parallel lines have the same slope and different y-intercepts. See Example 14 and Example 15. β€’ Perpendicular lines have slopes that are negative reciprocals of each other unless one is horizontal and the other is vertical. See Example 16. 2.3 Models and Applications β€’ A linear equation can be used to solve for an unknown in a number problem. See Example 1. β€’ Applications can be written as mathematical problems by identifying known quantities and assigning a variable to unknown quantities. See Example 2. β€’ There are many known formulas that can be used to solve applications. Distance problems, for example, are solved using the d = rt formula. See Example 3.
β€’ Many geometry problems are solved using the perimeter formula P = 2L + 2W, the area formula A = LW, or the volume formula V = LWH. See Example 4, Example 5, and Example 6. 2.4 Complex Numbers β€’ The square root of any negative number can be written as a multiple of i. See Example 1. β€’ To plot a complex number, we use two number lines, crossed to form the complex plane. The horizontal axis is the real axis, and the vertical axis is the imaginary axis. See Example 2. β€’ Complex numbers can be added and subtracted by combining the real parts and combining the imaginary parts. See Example 3. β€’ Complex numbers can be multiplied and divided. β—¦ To multiply complex numbers, distribute just as with polynomials. See Example 4 and Example 5. β—¦ To divide complex numbers, multiply both numerator and denominator by the complex conjugate of the denominator to eliminate the complex number from the denominator. See Example 6 and Example 7. β€’ The powers of i are cyclic, repeating every fourth one. See Example 8. 2.5 Quadratic Equations β€’ Many quadratic equations can be solved by factoring when the equation has a leading coefficient of 1 or if the equation is a difference of squares. The zero-product property is then used to find solutions. See Example 1, Example 2, and Example 3. β€’ Many quadratic equations with a leading coefficient other than 1 can be solved by factoring using the grouping method. See Example 4 and Example 5. β€’ Another method for solving quadratics is the square root property. The variable is squared. We isolate the squared term and take the square root of both sides of the equation. The solution will yield a positive and negative solution. See Example 6 and Example 7. β€’ Completing the square is a method of solving quadratic equations when the equation cannot be factored. See Example 8. β€’ A highly dependable method for solving quadratic equations is the quadratic formula, based on the coefficients and the constant term in the equation. See Example 9 and Example 10. β€’ The discriminant is used to indicate the nature of the roots that the quadratic equation will yield: real or complex, rational or irrational, and how many of each. See Example 11. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 154 CHAPTER 2 EQUATIONS AND INEQU
ALITIES β€’ The Pythagorean Theorem, among the most famous theorems in history, is used to solve right-triangle problems and has applications in numerous fields. Solving for the length of one side of a right triangle requires solving a quadratic equation. See Example 12. 2.6 Other Types of Equations β€’ Rational exponents can be rewritten several ways depending on what is most convenient for the problem. To solve, both sides of the equation are raised to a power that will render the exponent on the variable equal to 1. See Example 1, Example 2, and Example 3. β€’ Factoring extends to higher-order polynomials when it involves factoring out the GCF or factoring by grouping. See Example 4 and Example 5. β€’ We can solve radical equations by isolating the radical and raising both sides of the equation to a power that matches the index. See Example 6 and Example 7. β€’ To solve absolute value equations, we need to write two equations, one for the positive value and one for the negative value. See Example 8. β€’ Equations in quadratic form are easy to spot, as the exponent on the first term is double the exponent on the second term and the third term is a constant. We may also see a binomial in place of the single variable. We use substitution to solve. See Example 9 and Example 10. β€’ Solving a rational equation may also lead to a quadratic equation or an equation in quadratic form. See Example 11. 2.7 Linear Inequalities and Absolute Value Inequalities β€’ Interval notation is a method to indicate the solution set to an inequality. Highly applicable in calculus, it is a system of parentheses and brackets that indicate what numbers are included in a set and whether the endpoints are included as well. See Table 1 and Example 1 and Example 2. β€’ Solving inequalities is similar to solving equations. The same algebraic rules apply, except for one: multiplying or dividing by a negative number reverses the inequality. See Example 3, Example 4, Example 5, and Example 6. β€’ Compound inequalities often have three parts and can be rewritten as two independent inequalities. Solutions are given by boundary values, which are indicated as a beginning boundary or an ending boundary in the solutions to the two inequalities. See Example 7 and Example 8. β€’ Absolute value inequalities will produce two solution sets due to the nature of absolute value. We solve by writing two equations: one equal to a positive value and one
equal to a negative value. See Example 9 and Example 10. β€’ Absolute value inequalities can also be solved by graphing. At least we can check the algebraic solutions by graphing, as we cannot depend on a visual for a precise solution. See Example 11. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 2 REVIEW 155 CHAPTER 2 REVIEW EXERCISES THE RECTANGULAR COORDINATE SYSTEMS AND GRAPHS For the following exercises, find the x-intercept and the y-intercept without graphing. 1. 4x βˆ’ 3y = 12 2. 2y βˆ’ 4 = 3x For the following exercises, solve for y in terms of x, putting the equation in slope–intercept form. 3. 5x = 3y βˆ’ 12 4. 2x βˆ’ 5y = 7 For the following exercises, find the distance between the two points. 5. (βˆ’2, 5)(4, βˆ’1) 6. (βˆ’12, βˆ’3)(βˆ’1, 5) 7. Find the distance between the two points (βˆ’71,432) and (511,218) using your calculator, and round your answer to the nearest thousandth. For the following exercises, find the coordinates of the midpoint of the line segment that joins the two given points. 8. (βˆ’1, 5) and (4, 6) 9. (βˆ’13, 5) and (17, 18) For the following exercises, construct a table and graph the equation by plotting at least three points. 1 _ 10. y = x + 4 2 11. 4x βˆ’ 3y = 6 LINEAR EQUATIONS IN ONE VARIABLE For the following exercises, solve for x. 12. 5x + 2 = 7x βˆ’ 8 13. 3(x + 2) βˆ’ 10 = x + 4 14. 7x βˆ’ 3 = 5 15. 12 βˆ’ 5(x + 1) = 2x βˆ’ 5 16. 2x 21 _ 4 For the following exercises, solve for x. State all x-values that are excluded from the solution set. 17. x _ x2 βˆ’ x2 βˆ’ 9 x β‰  3, βˆ’3 3 1 2 _ _ _ = + 18. 4 2 x For the following exercises, find the equation of the line using the point-slope formula. 19. Passes through these two points: (βˆ’2, 1),(4, 2
). 1 _ 20. Passes through the point (βˆ’3, 4) and has a slope of βˆ’. 3 21. Passes through the point (βˆ’3, 4) and is parallel to 22. Passes through these two points: (5, 1),(5, 7). 2 _ the graph y = x + 5. 3 MODELS AND APPLICATIONS For the following exercises, write and solve an equation to answer each question. 23. The number of males in the classroom is five more than three times the number of females. If the total number of students is 73, how many of each gender are in the class? 24. A man has 72 ft of fencing to put around a rectangular garden. If the length is 3 times the width, find the dimensions of his garden. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 156 CHAPTER 2 EQUATIONS AND INEQUALITIES 25. A truck rental is $25 plus $.30/mi. Find out how many miles Ken traveled if his bill was $50.20. COMPLEX NUMBERS For the following exercises, use the quadratic equation to solve. 26. x2 βˆ’ 5x + 9 = 0 27. 2x2 + 3x + 7 = 0 For the following exercises, name the horizontal component and the vertical component. 28. 4 βˆ’ 3i 29. βˆ’2 βˆ’ i For the following exercises, perform the operations indicated. 31. (2 + 3i) βˆ’ (βˆ’5 βˆ’ 8i) 32. 2 √ β€” βˆ’75 + 3 √ β€” 25 34. βˆ’6i(i βˆ’ 5) 37. √ β€” βˆ’2 ( √ β€” βˆ’8 βˆ’ √ β€” 5 ) 35. (3 βˆ’ 5i)2 38. 2 _ 5 βˆ’ 3i 30. (9 βˆ’ i) βˆ’ (4 βˆ’ 7i) 33. √ β€” βˆ’16 + 4 √ β€” βˆ’9 36. √ β€” βˆ’4 Β· √ β€” βˆ’12 39. 3 + 7i ______ i QUADRATIC EQUATIONS For the following exercises, solve the quadratic equation by factoring. 40. 2x2 βˆ’ 7x βˆ’ 4 = 0 41. 3x2 + 18x + 15 = 0 42. 25x2 βˆ’ 9 = 0 43. 7x2 βˆ’ 9x = 0 For the following exercises, solve the quadratic equation by using the square-root
property. 44. x2 = 49 45. (x βˆ’ 4)2 = 36 For the following exercises, solve the quadratic equation by completing the square. 46. x2 + 8x βˆ’ 5 = 0 47. 4x2 + 2x βˆ’ 1 = 0 For the following exercises, solve the quadratic equation by using the quadratic formula. If the solutions are not real, state No real solution. 48. 2x2 βˆ’ 5x + 1 = 0 49. 15x2 βˆ’ x βˆ’ 2 = 0 For the following exercises, solve the quadratic equation by the method of your choice. 50. (x βˆ’ 2)2 = 16 51. x2 = 10x + 3 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 2 REVIEW 157 OTHER TYPES OF EQUATIONS For the following exercises, solve the equations. 3 _ = 27 52. x 2 55. 3x5 βˆ’ 6x3 = 0 58. |3x βˆ’ 7| = 5 1 1 _ _ = 0 βˆ’ 4x 2 4 53. x 56 59. |2x + 3| βˆ’ 5 = 9 54. 4x3 + 8x2 βˆ’ 9x βˆ’ 18 = 0 57. √ β€” 3x + 7 + √ β€” x + 2 = 1 LINEAR INEQUALITIES AND ABSOLUTE VALUE INEQUALITIES For the following exercises, solve the inequality. Write your final answer in interval notation. 60. 5x βˆ’ 8 ≀ 12 61. βˆ’2x + 5 > x βˆ’ 7 63. |3x + 2| + 1 ≀ 9 64. |5x βˆ’ 1| > 14 62. x βˆ’ 1 _____ 3 + x + 2 _____ 5 3__ ≀ 5 65. |x βˆ’ 3| < βˆ’4 For the following exercises, solve the compound inequality. Write your answer in interval notation. 66. βˆ’4 < 3x + 2 ≀ 18 67. 3y < 1 βˆ’ 2y < 5 + y For the following exercises, graph as described. 68. Graph the absolute value function and graph the constant function. Observe the points of intersection and shade the x-axis representing the solution set to the inequality. Show your graph and write your final answer in interval notation. |x + 3| β‰₯ 5 69. Graph both straight lines (left-hand side being y1 and right-hand side being
y2) on the same axes. Find the point of intersection and solve the inequality by observing where it is true comparing the y-values of the lines. See the interval where the inequality is true. x + 3 < 3x βˆ’ 4 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 158 CHAPTER 2 EQUATIONS AND INEQUALITIES CHAPTER 2 PRACTICE TEST 1. Graph the following: 2y = 3x + 4. 2. Find the x- and y-intercepts for the following: 2x βˆ’ 5y = 6. 3. Find the x- and y-intercepts of this equation, and sketch the graph of the line using just the intercepts plotted. 3x βˆ’ 4y = 12 4. Find the exact distance between (5, βˆ’3) and (βˆ’2, 8). Find the coordinates of the midpoint of the line segment joining the two points. 5. Write the interval notation for the set of numbers 6. Solve for x: 5x + 8 = 3x βˆ’ 10. represented by {x|x ≀ 9}. 7. Solve for x: 3(2x βˆ’ 5) βˆ’ 3(x βˆ’ 7) = 2x βˆ’ 9. 9. Solve for x: 5 _____ x + 4 = 4 + 3 _____.. Solve for x: x 2 10. The perimeter of a triangle is 30 in. The longest side is 2 less than 3 times the shortest side and the other side is 2 more than twice the shortest side. Find the length of each side. 11. Solve for x. Write the answer in simplest radical form. 12. Solve: 3x βˆ’ 8 ≀ 4. x2 _ 3 1 _ βˆ’ x = βˆ’ 2 13. Solve: |2x + 3| < 5. 14. Solve: |3x βˆ’ 2| β‰₯ 4. For the following exercises, find the equation of the line with the given information. 15. Passes through the points (βˆ’4, 2) and (5, βˆ’3). 16. Has an undefined slope and passes through the point (4, 3). 17. Passes through the point (2, 1) and is perpendicular 18. Add these complex numbers: (3 βˆ’ 2i) + (4 βˆ’ i). 2 _ x + 3. to y = βˆ’ 5 β€” βˆ’16. 20. Multiply: 5i
(5 βˆ’ 3i). β€” βˆ’4 + 3 √ 19. Simplify: √ 4 βˆ’ i ______. 2 + 3i 21. Divide: 23. Solve: (3x βˆ’ 1)2 βˆ’ 1 = 24. 25. Solve: 4x 2 βˆ’ 4x βˆ’ 1 = 0 27. Solve: 2 + √ β€” 12 βˆ’ 2x = x 22. Solve this quadratic equation and write the two complex roots in a + bi form: x 2 βˆ’ 4x + 7 = 0. 24. Solve: x 2 βˆ’ 6x = 13. 26. Solve 28. Solve: (x βˆ’ 1) 3 For the following exercises, find the real solutions of each equation by factoring. 29. 2x 3 βˆ’ x 2 βˆ’ 8x + 4 = 0 30. (x + 5)2 βˆ’ 3(x + 5) βˆ’ 4 = 0 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. Functions P 1,500 1,000 500 0 Figure 1 Standard and Poor’s Index with dividends reinvested (credit "bull": modification of work by Prayitno Hadinata; credit "graph": modification of work by MeasuringWorth) 1970 1975 1980 1985 1990 1995 2000 2005 2010 3 y CHAPTER OUTLINE 3.1 Functions and Function Notation 3.2 Domain and Range 3.3 Rates of Change and Behavior of Graphs 3.4 Composition of Functions 3.5 Transformation of Functions 3.6 Absolute Value Functions 3.7 Inverse Functions Introduction Toward the end of the twentieth century, the values of stocks of Internet and technology companies rose dramatically. As a result, the Standard and Poor’s stock market average rose as well. Figure 1 tracks the value of that initial investment of just under $100 over the 40 years. It shows that an investment that was worth less than $500 until about 1995 skyrocketed up to about $1,100 by the beginning of 2000. That five-year period became known as the β€œdot-com bubble” because so many Internet startups were formed. As bubbles tend to do, though, the dot-com bubble eventually burst. Many companies grew too fast and then suddenly went out of business. The result caused the sharp decline represented on the graph beginning at the end of 2000. Notice, as we consider this example, that
there is a definite relationship between the year and stock market average. For any year we choose, we can determine the corresponding value of the stock market average. In this chapter, we will explore these kinds of relationships and their properties. 159 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 160 CHAPTER 3 FUNCTIONS LEARNING OBJECTIVES In this section, you will: β€’ etermine whether a relation represents a function. β€’ Find the alue of a function gien its equation or its graph. β€’ etermine whether a function is one-to-one. β€’ Use the ertical line test to identify functions. 3.1 FUNCTIONS AND FUNCTION NOTATION A jetliner changes altitude as its distance from the starting point of a flight increases. The weight of a growing child increases with time. In each case, one quantity depends on another. There is a relationship between the two quantities that we can describe, analyze, and use to make predictions. In this section, we will analyze such relationships. Determining Whether a Relation Represents a Function A relation is a set of ordered pairs. The set consisting of the first components of each ordered pair is called the domain and the set consisting of the second components of each ordered pair is called the range. Consider the following set of ordered pairs. The first numbers in each pair are the first five natural numbers. The second number in each pair is twice that of the first. {(1, 2), (2, 4), (3, 6), (4, 8), (5, 10)} The domain is {1, 2, 3, 4, 5}. The range is {2, 4, 6, 8, 10}. Note that each value in the domain is also known as an input value, or independent variable, and is often labeled with the lowercase letter x. Each value in the range is also known as an output value, or dependent variable, and is often labeled lowercase letter y. A function f is a relation that assigns a single element in the range to each element in the domain. In other words, no x-values are repeated. For our example that relates the first five natural numbers to numbers double their values, this relation is a function because each element in the domain, {1, 2, 3, 4, 5}, is paired with exactly one element in the range, {2, 4, 6, 8, 10
}. Now let’s consider the set of ordered pairs that relates the terms β€œeven” and β€œodd” to the first five natural numbers. It would appear as {(odd, 1), (even, 2), (odd, 3), (even, 4), (odd, 5)} Notice that each element in the domain, {even, odd} is not paired with exactly one element in the range, {1, 2, 3, 4, 5}. For example, the term β€œodd” corresponds to three values from the domain, {1, 3, 5} and the term β€œeven” corresponds to two values from the range, {2, 4}. This violates the definition of a function, so this relation is not a function. Figure 1 compares relations that are functions and not functions. Relation is a Function Outputs Inputs Relation is a Function Inputs Outputs Relation is NOT a Function Inputs p q Outputs x y z (a) (b) (c) Figure 1 ( a ) This relationship is a function because each input is associated with a single output. Note that input q and r both give output n. ( b ) This relationship is also a function. In this case, each input is associated with a single output. ( c ) This relationship is not a function because input q is associated with two different outputs. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.1 FUNCTIONS AND FUNCTION NOTATION 161 function A function is a relation in which each possible input value leads to exactly one output value. We say β€œthe output is a function of the input.” The input values make up the domain, and the output values make up the range. How To… Given a relationship between two quantities, determine whether the relationship is a function. 1. Identify the input values. 2. Identify the output values. 3. If each input value leads to only one output value, classify the relationship as a function. If any input value leads to two or more outputs, do not classify the relationship as a function. Example 1 Determining If Menu Price Lists Are Functions The coffee shop menu, shown in Figure 2 consists of items and their prices. a. Is price a function of the item? b. Is the item a function of the price? Menu Item Price Plain Donut......
........................ 1.49 Jelly Donut.............................. 1.99 Chocolate Donut.......................... 1.99 Figure 2 Solution a. Let’s begin by considering the input as the items on the menu. The output values are then the prices. See Figure 2. Each item on the menu has only one price, so the price is a function of the item. b. Two items on the menu have the same price. If we consider the prices to be the input values and the items to be the output, then the same input value could have more than one output associated with it. See Figure 3. Menu Item Price Plain Donut.............................. 1.49 Jelly Donut.............................. 1.99 Chocolate Donut.......................... Therefore, the item is a not a function of price. Figure 3 Example 2 Determining If Class Grade Rules Are Functions In a particular math class, the overall percent grade corresponds to a grade-point average. Is grade-point average a function of the percent grade? Is the percent grade a function of the grade-point average? Table 1 shows a possible rule for assigning grade points. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 162 CHAPTER 3 FUNCTIONS Percent grade 0-56 57-61 62-66 67-71 72-77 78-86 87-91 92-100 Grade-point average 0.0 1.0 1.5 Table 1 2.0 2.5 3.0 3.5 4.0 Solution For any percent grade earned, there is an associated grade-point average, so the grade-point average is a function of the percent grade. In other words, if we input the percent grade, the output is
a specific grade-point average. In the grading system given, there is a range of percent grades that correspond to the same grade-point average. For example, students who receive a grade-point average of 3.0 could have a variety of percent grades ranging from 78 all the way to 86. Thus, percent grade is not a function of grade-point average Try It #1 Table 2[1] lists the five greatest baseball players of all time in order of rank. Player Rank Babe Ruth Willie Mays Ty Cobb Walter Johnson Hank Aaron Table 2 1 2 3 4 5 a. Is the rank a function of the player name? b. Is the player name a function of the rank? Using Function Notation Once we determine that a relationship is a function, we need to display and define the functional relationships so that we can understand and use them, and sometimes also so that we can program them into graphing calculators and computers. There are various ways of representing functions. A standard function notation is one representation that facilitates working with functions. To represent β€œheight is a function of age,” we start by identifying the descriptive variables h for height and a for age. The letters f, g, and h are often used to represent functions just as we use x, y, and z to represent numbers and A, B, and C to represent sets. h is f of a h = f (a) f (a) We name the function f ; height is a function of age. We use parentheses to indicate the function input. We name the function f ; the expression is read as β€œf of a.” Remember, we can use any letter to name the function; the notation h(a) shows us that h depends on a. The value a must be put into the function h to get a result. The parentheses indicate that age is input into the function; they do not indicate multiplication. We can also give an algebraic expression as the input to a function. For example f (a + b) means β€œfirst add a and b, and the result is the input for the function f. ” The operations must be performed in this order to obtain the correct result. function notation The notation y = f (x) defines a function named f. This is read as β€œy is a function of x.” The letter x represents the input value, or independent variable. The letter y, or f (x), represents the output value, or dependent variable.
1 http://www.baseball-almanac.com/legendary/lisn100.shtml. Accessed 3/24/2014. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.1 FUNCTIONS AND FUNCTION NOTATION 163 Example 3 Using Function Notation for Days in a Month Use function notation to represent a function whose input is the name of a month and output is the number of days in that month. Solution The number of days in a month is a function of the name of the month, so if we name the function f, we write days = f (month) or d = f (m). The name of the month is the input to a β€œrule” that associates a specific number (the output) with each input. 31!f (January) output input rule Figure 4 For example, f (March) = 31, because March has 31 days. The notation d = f (m) reminds us that the number of days, d (the output), is dependent on the name of the month, m(the input). Analysis Note that the inputs to a function do not have to be numbers; function inputs can be names of people, labels of geometric objects, or any other element that determines some kind of output. However, most of the functions we will work with in this book will have numbers as inputs and outputs. Example 4 Interpreting Function Notation A function N = f (y) gives the number of police officers, N, in a town in year y. What does f (2005) = 300 represent? Solution When we read f (2005) = 300, we see that the input year is 2005. The value for the output, the number of police officers (N), is 300. Remember N = f (y). The statement f (2005) = 300 tells us that in the year 2005 there were 300 police officers in the town. Try It #2 Use function notation to express the weight of a pig in pounds as a function of its age in days d. Q & A… Instead of a notation such as y = f (x), could we use the same symbol for the output as for the function, such as y = y (x), meaning β€œy is a function of x?” Yes, this is often done, especially in applied subjects that use higher math, such as physics and engineering. However, in exploring math itself
we like to maintain a distinction between a function such as f, which is a rule or procedure, and the output y we get by applying f to a particular input x. This is why we usually use notation such as y = f (x), P = W(d), and so on. Representing Functions Using Tables A common method of representing functions is in the form of a table. The table rows or columns display the corresponding input and output values. In some cases, these values represent all we know about the relationship; other times, the table provides a few select examples from a more complete relationship. Table 3 lists the input number of each month (January = 1, February = 2, and so on) and the output value of the number of days in that month. This information represents all we know about the months and days for a given year (that is not a leap year). Note that, in this table, we define a days-in-a-month function f where D = f (m) identifies months by an integer rather than by name. Month number, m (input) 1 2 3 4 5 6 Days in month, D (output) 31 28 31 30 31 30 7 31 8 9 31 30 10 31 11 30 12 31 Table 3 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 164 CHAPTER 3 FUNCTIONS Table 4 defines a function Q = g (n). Remember, this notation tells us that g is the name of the function that takes the input n and gives the output Q Table 4 Table 5 below displays the age of children in years and their corresponding heights. This table displays just some of the data available for the heights and ages of children. We can see right away that this table does not represent a function because the same input value, 5 years, has two different output values, 40 in. and 42 in. Age in years, a (input) Height in inches, h (output) 5 40 5 42 6 44 7 47 8 50 9 52 10 54 Table 5 How To… Given a table of input and output values, determine whether the table represents a function. 1. Identify the input and output values. 2. Check to see if each input value is paired with only one output value. If so, the table represents a function. Example 5 Identifying Tables that Represent Functions Which table, Table 6, Table 7, or Table 8, represents a function (if any)? Input Output Input
Output Input Output 2 5 8 1 3 6 βˆ’3 0 4 5 1 5 Table 6 Table 7 0 2 4 1 5 5 Table 8 Solution Table 6 and Table 7 define functions. In both, each input value corresponds to exactly one output value. Table 8 does not define a function because the input value of 5 corresponds to two different output values. When a table represents a function, corresponding input and output values can also be specified using function notation. The function represented by Table 6 can be represented by writing Similarly, the statements f (2) = 1, f (5) = 3, and f (8) = 6 g (βˆ’3) = 5, g (0) = 1, and g (4) = 5 represent the function in table Table 7. Table 8 cannot be expressed in a similar way because it does not represent a function. Try It #3 Does Table 9 represent a function? Input Output 1 2 3 10 100 1000 Table 9 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.1 FUNCTIONS AND FUNCTION NOTATION 165 Finding Input and Output Values of a Function When we know an input value and want to determine the corresponding output value for a function, we evaluate the function. Evaluating will always produce one result because each input value of a function corresponds to exactly one output value. When we know an output value and want to determine the input values that would produce that output value, we set the output equal to the function’s formula and solve for the input. Solving can produce more than one solution because different input values can produce the same output value. Evaluation of Functions in Algebraic Forms When we have a function in formula form, it is usually a simple matter to evaluate the function. For example, the function f (x) = 5 βˆ’ 3x 2 can be evaluated by squaring the input value, multiplying by 3, and then subtracting the product from 5. How To… Given the formula for a function, evaluate. 1. Replace the input variable in the formula with the value provided. 2. Calculate the result. Example 6 Evaluating Functions at Specific Values Evaluate f(x) = x 2 + 3x βˆ’ 4 at: a. 2 b. a c. a + h d. f (a + h) βˆ’ f (a) __ h Solution Replace the x in the function with each specified value. a. Because the
input value is a number, 2, we can use simple algebra to simplify. f (2) = 22 + 3(2. In this case, the input value is a letter so we cannot simplify the answer any further. c. With an input value of a + h, we must use the distributive property. f (a) = a2 + 3a βˆ’ 4 f (a + h) = (a + h)2 + 3(a + h) βˆ’ 4 = a2 + 2ah + h2 + 3a + 3h βˆ’4 d. In this case, we apply the input values to the function more than once, and then perform algebraic operations on the result. We already found that and we know that f (a + h) = a2 + 2ah + h2 + 3a + 3h βˆ’ 4 f(a) = a2 + 3a βˆ’ 4 Now we combine the results and simplify. f (a + h) βˆ’ f(a) __ = h (a2 + 2ah + h2 + 3a + 3h βˆ’ 4) βˆ’ (a2 + 3a βˆ’ 4) ____ h 2ah + h2 + 3h ___________ h h(2a + h + 3) ___________ h Factor out h. = = = 2a + h + 3 Simplify. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 166 CHAPTER 3 FUNCTIONS Example 7 Evaluating Functions Given the function h(p) = p2 + 2p, evaluate h(4). Solution To evaluate h(4), we substitute the value 4 for the input variable p in the given function. h(p) = p2 + 2p h(4) = (4)2 +2 (4) = 16 + 8 = 24 Therefore, for an input of 4, we have an output of 24. Try It #4 Given the function g(m) = √ β€” m βˆ’ 4. Evaluate g(5). Example 8 Solving Functions Given the function h(p) = p2 + 2p, solve for h(p) = 3. Solution h(p) = 3 p2 + 2p = 3 Substitute the original function h(p) = p2 + 2p. p2 + 2p βˆ’ 3 = 0 Subtract 3 from each side. (p + 3)(p βˆ’ 1
) = 0 Factor. If (p + 3)(p βˆ’ 1) = 0, either (p + 3) = 0 or (p βˆ’ 1) = 0 (or both of them equal 0). We will set each factor equal to 0 and solve for p in each case. (p + 3) = 0, p = βˆ’3 (p βˆ’ 1) = 0, p = 1 This gives us two solutions. The output h(p) = 3 when the input is either p = 1 or p = βˆ’3. We can also verify by graphing as in Figure 5. The graph verifies that h(1) = h(βˆ’3) = 3 and h(4) = 24. h(p) 35 30 25 20 15 10 5 1 2 3 4 5 p p –3 –2 h(p) 3 0 0 0 1 3 4 24 Figure 5 Try It #5 Given the function g(m) = √ β€” m βˆ’ 4, solve g(m) = 2. Evaluating Functions Expressed in Formulas Some functions are defined by mathematical rules or procedures expressed in equation form. If it is possible to express the function output with a formula involving the input quantity, then we can define a function in algebraic form. For example, the equation 2n + 6p = 12 expresses a functional relationship between n and p. We can rewrite it to decide if p is a function of n. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.1 FUNCTIONS AND FUNCTION NOTATION 167 How To… Given a function in equation form, write its algebraic formula. 1. Solve the equation to isolate the output variable on one side of the equal sign, with the other side as an expression that involves only the input variable. 2. Use all the usual algebraic methods for solving equations, such as adding or subtracting the same quantity to or from both sides, or multiplying or dividing both sides of the equation by the same quantity. Example 9 Finding an Equation of a Function Express the relationship 2n + 6p = 12 as a function p = f (n), if possible. Solution To express the relationship in this form, we need to be able to write the relationship where p is a function of n, which means writing it as p = [expression involving n]. 2n + 6p = 12 6p = 12 βˆ’ 2n Subtract 2n
from both sides. p = 12 βˆ’ 2n _______ 6 Divide both sides by 6 and simplify. p = 12 __ βˆ’ 6 2n ___ 6 Therefore, p as a function of n is written as 1 __ p = 2 βˆ’ n 3 1 __ p = f (n) = 2 βˆ’ n 3 Example 10 Expressing the Equation of a Circle as a Function Does the equation x 2 + y 2 = 1 represent a function with x as input and y as output? If so, express the relationship as a function y = f (x). Solution First we subtract x 2 from both sides. We now try to solve for y in this equation βˆ’ x2 = + √ β€” 1 βˆ’ x2 and βˆ’ √ β€” 1 βˆ’ x2 We get two outputs corresponding to the same input, so this relationship cannot be represented as a single function y = f (x). If we graph both functions on a graphing calculator, we will get the upper and lower semicircles. Try It #6 If x βˆ’ 8y 3 = 0, express y as a function of x. Q & A… Are there relationships expressed by an equation that do represent a function but that still cannot be represented by an algebraic formula? Yes, this can happen. For example, given the equation x = y + 2y, if we want to express y as a function of x, there is no simple algebraic formula involving only x that equals y. However, each x does determine a unique value for y, and there are mathematical procedures by which y can be found to any desired accuracy. In this case, we say that the equation gives an implicit (implied) rule for y as a function of x, even though the formula cannot be written explicitly. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 168 CHAPTER 3 FUNCTIONS Evaluating a Function Given in Tabular Form As we saw above, we can represent functions in tables. Conversely, we can use information in tables to write functions, and we can evaluate functions using the tables. For example, how well do our pets recall the fond memories we share with them? There is an urban legend that a goldfish has a memory of 3 seconds, but this is just a myth. Goldfish can remember up to 3 months, while the beta fish has a memory of up to 5 months. And while a puppy’s memory span is no longer than 30 seconds, the adult
dog can remember for 5 minutes. This is meager compared to a cat, whose memory span lasts for 16 hours. The function that relates the type of pet to the duration of its memory span is more easily visualized with the use of a table. See Table 10.[2] Pet Puppy Adult dog Cat Goldfish Beta fish Memory span in hours 0.008 0.083 16 2160 3600 Table 10 At times, evaluating a function in table form may be more useful than using equations. Here let us call the function P. The domain of the function is the type of pet and the range is a real number representing the number of hours the pet’s memory span lasts. We can evaluate the function P at the input value of β€œgoldfish.” We would write P(goldfish) = 2160. Notice that, to evaluate the function in table form, we identify the input value and the corresponding output value from the pertinent row of the table. The tabular form for function P seems ideally suited to this function, more so than writing it in paragraph or function form. How To… Given a function represented by a table, identify specific output and input values. 1. Find the given input in the row (or column) of input values. 2. Identify the corresponding output value paired with that input value. 3. Find the given output values in the row (or column) of output values, noting every time that output value appears. 4. Identify the input value(s) corresponding to the given output value. Example 11 Evaluating and Solving a Tabular Function Using Table 11, a. Evaluate g(3) b. Solve g(n) = 6. n g (n Table 11 Solution a. Evaluating g (3) means determining the output value of the function g for the input value of n = 3. The table output value corresponding to n = 3 is 7, so g (3) = 7. b. Solving g (n) = 6 means identifying the input values, n, that produce an output value of 6. Table 11 shows two solutions: 2 and 4. When we input 2 into the function g, our output is 6. When we input 4 into the function g, our output is also 6. 2 http://www.kgbanswers.com/how-long-is-a-dogs-memory-span/4221590. Accessed 3/24/2014. Download the OpenStax text
for free at http://cnx.org/content/col11759/latest. SECTION 3.1 FUNCTIONS AND FUNCTION NOTATION 169 Try It #7 Using Table 11, evaluate g(1). Finding Function Values from a Graph Evaluating a function using a graph also requires finding the corresponding output value for a given input value, only in this case, we find the output value by looking at the graph. Solving a function equation using a graph requires finding all instances of the given output value on the graph and observing the corresponding input value( s ). Example 12 Reading Function Values from a Graph Given the graph in Figure 6, a. Evaluate f (2). b. Solve f (x) = 4. f (x) 7 6 5 4 3 2 1 –1 –1 2 – – 3 –5 –4 –3 –2 21 3 4 5 Figure 6 Solution a. To evaluate f (2), locate the point on the curve where x = 2, then read the y-coordinate of that point. The point has coordinates (2, 1), so f (2) = 1. See Figure 7. f (x) 7 6 5 4 3 2 1 –1 –1 2 – – 3 –5 –4 –3 –2 (2, 1) f (2) = 1 21 3 4 5 Figure 7 b. To solve f (x) = 4, we find the output value 4 on the vertical axis. Moving horizontally along the line y = 4, we locate two points of the curve with output value 4: (βˆ’1, 4) and (3, 4). These points represent the two solutions to f (x) = 4: βˆ’1 or 3. This means f (βˆ’1) = 4 and f (3) = 4, or when the input is βˆ’1 or 3, the output is 4. See Figure 8. f (x) 7 6 5 4 3 2 1 (βˆ’1, 4) (3, 4) 21 3 4 5 –5 –4 –3 –2 –1 –1 2 – – 3 Figure 8 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 170 CHAPTER 3 FUNCTIONS Try It #8 Using Figure 7, solve f (x) = 1. Determining Whether a Function is One-to-One Some functions have a given output value that corresponds to two or more input values. For example, in the
stock chart shown in Figure 1 at the beginning of this chapter, the stock price was $1,000 on five different dates, meaning that there were five different input values that all resulted in the same output value of $1,000. However, some functions have only one input value for each output value, as well as having only one output for each input. We call these functions one-to-one functions. As an example, consider a school that uses only letter grades and decimal equivalents, as listed in Table 12. Letter grade A B C D Grade-point average 4.0 3.0 2.0 1.0 Table 12 This grading system represents a one-to-one function, because each letter input yields one particular grade-point average output and each grade-point average corresponds to one input letter. To visualize this concept, let’s look again at the two simple functions sketched in Figure 1(a) and Figure 1(b). The function in part ( a) shows a relationship that is not a one-to-one function because inputs q and r both give output n. The function in part (b ) shows a relationship that is a one-to-one function because each input is associated with a single output. one-to-one function A one-to-one function is a function in which each output value corresponds to exactly one input value. There are no repeated x- or y-values. Example 13 Determining Whether a Relationship Is a One-to-One Function Is the area of a circle a function of its radius? If yes, is the function one-to-one? Solution A circle of radius r has a unique area measure given by A = Ο€r 2, so for any input, r, there is only one output, A. The area is a function of radius r. If the function is one-to-one, the output value, the area, must correspond to a unique input value, the radius. Any area measure A is given by the formula A = Ο€r2. Because areas and radii are positive numbers, there is exactly one solution: r = √ ___ A __ Ο€ So the area of a circle is a one-to-one function of the circle’s radius. Try It #9 a. Is a balance a function of the bank account number? b. Is a bank account number a function of the balance? c. Is a balance a one-to-one function of
the bank account number? Try It #10 a. If each percent grade earned in a course translates to one letter grade, is the letter grade a function of the percent grade? b. If so, is the function one-to-one? Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.1 FUNCTIONS AND FUNCTION NOTATION 171 Using the Vertical Line Test As we have seen in some examples above, we can represent a function using a graph. Graphs display a great many input-output pairs in a small space. The visual information they provide often makes relationships easier to understand. By convention, graphs are typically constructed with the input values along the horizontal axis and the output values along the vertical axis. The most common graphs name the input value x and the output value y, and we say y is a function of x, or y = f (x) when the function is named f. The graph of the function is the set of all points (x, y) in the plane that satisfies the equation y = f (x). If the function is defined for only a few input values, then the graph of the function consists of only a few points, where the x-coordinate of each point is an input value and the y-coordinate of each point is the corresponding output value. For example, the black dots on the graph in Figure 9 tell us that f (0) = 2 and f (6) = 1. However, the set of all points (x, y) satisfying y = f (x) is a curve. The curve shown includes (0, 2) and (6, 1) because the curve passes through those points. y Figure 9 x The vertical line test can be used to determine whether a graph represents a function. If we can draw any vertical line that intersects a graph more than once, then the graph does not define a function because a function has only one output value for each input value. See Figure 10. Function Not a Function Not a Function Figure 10 How To… Given a graph, use the vertical line test to determine if the graph represents a function. 1. Inspect the graph to see if any vertical line drawn would intersect the curve more than once. 2. If there is any such line, determine that the graph does not represent a function. Download the OpenStax text for free at http://cnx.org/content/col11759/latest
. 172 CHAPTER 3 FUNCTIONS Example 14 Applying the Vertical Line Test Which of the graphs in Figure 11 represent( s ) a function y = f (x)? f (x) (a) y 5 4 3 2 1 f (x) x 4 5 6 7 8 9 10 11 12 x x (b) Figure 11 (c) Solution If any vertical line intersects a graph more than once, the relation represented by the graph is not a function. Notice that any vertical line would pass through only one point of the two graphs shown in parts ( a) and (b) of Figure 11. From this we can conclude that these two graphs represent functions. The third graph does not represent a function because, at most x-values, a vertical line would intersect the graph at more than one point, as shown in Figure 12. Try It #11 Does the graph in Figure 13 represent a function? y Figure 12 y Figure 13 x x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.1 FUNCTIONS AND FUNCTION NOTATION 173 Using the Horizontal Line Test Once we have determined that a graph defines a function, an easy way to determine if it is a one-to-one function is to use the horizontal line test. Draw horizontal lines through the graph. If any horizontal line intersects the graph more than once, then the graph does not represent a one-to-one function. How To… Given a graph of a function, use the horizontal line test to determine if the graph represents a one-to-one function. 1. Inspect the graph to see if any horizontal line drawn would intersect the curve more than once. 2. If there is any such line, determine that the function is not one-to-one. Example 15 Applying the Horizontal Line Test Consider the functions shown in Figure 11(a) and Figure 11(b). Are either of the functions one-to-one? Solution The function in Figure 11(a) is not one-to-one. The horizontal line shown in Figure 14 intersects the graph of the function at two points (and we can even find horizontal lines that intersect it at three points.) f (x) Figure 14 x The function in Figure 11(b) is one-to-one. Any horizontal line will intersect a diagonal line at most once. Try It #12 Is the graph shown here one-to-one? y x
Identifying Basic Toolkit Functions In this text, we will be exploring functionsβ€”the shapes of their graphs, their unique characteristics, their algebraic formulas, and how to solve problems with them. When learning to read, we start with the alphabet. When learning to do arithmetic, we start with numbers. When working with functions, it is similarly helpful to have a base set of building-block elements. We call these our β€œtoolkit functions,” which form a set of basic named functions for which Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 174 CHAPTER 3 FUNCTIONS we know the graph, formula, and special properties. Some of these functions are programmed to individual buttons on many calculators. For these definitions we will use x as the input variable and y = f (x) as the output variable. We will see these toolkit functions, combinations of toolkit functions, their graphs, and their transformations frequently throughout this book. It will be very helpful if we can recognize these toolkit functions and their features quickly by name, formula, graph, and basic table properties. The graphs and sample table values are included with each function shown in Table 13. Name Function Toolkit Functions Graph f (x) Constant f (x) = c, where c is a constant Identity f (x) = x Absolute value f (x) =.#x. Quadratic f (x) = x2 Cubic f (x) = x3 f (x) f (x) f (x) f (x) x x x x x x –2 0 2 x –2 0 2 x –2 0 2 x –2 –1 0 1 2 f(x) 2 2 2 f(x) –2 0 2 f(x) 2 0 2 f(x) 4 1 0 1 4 x –1 –0.5 0 0.5 1 f(x) –1 –0.125 0 0.125 1 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.1 FUNCTIONS AND FUNCTION NOTATION 175 Reciprocal f (x) = 1 __ x Reciprocal squared 1 _ f (x) = x 2 Square root f (x) = √ β€” x Cube root f (x) = 3 √ β€” x f (x) f (x) f (x)
f (x) x –2 –1 –0.5 0.5 1 2 x –2 –1 –0.5 0.5 1 2 x 0 1 4 f(x) –0.5 –1 –2 2 1 0.5 f(x) 0.25 1 4 4 1 0.25 f(x) 0 1 2 x –1 f(x) –1 –0.125 –0.5 0 0.125 1 0 0.5 1 x x x x Table 13 Access the following online resources for additional instruction and practice with functions. β€’ Determine if a Relation is a Function (http://openstaxcollege.org/l/relationfunction) β€’ Vertical Line Test (http://openstaxcollege.org/l/vertlinetest) β€’ Introduction to Functions (http://openstaxcollege.org/l/introtofunction) β€’ Vertical Line Test of Graph (http://openstaxcollege.org/l/vertlinegraph) β€’ One-to-one Functions (http://openstaxcollege.org/l/onetoone) β€’ Graphs as One-to-one Functions (http://openstaxcollege.org/l/graphonetoone) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 176 CHAPTER 3 FUNCTIONS 3.1 SECTION EXERCISES VERBAL 1. What is the difference between a relation and 2. What is the difference between the input and the a function? output of a function? 3. Why does the vertical line test tell us whether the 4. How can you determine if a relation is a one-to-one graph of a relation represents a function? function? 5. Why does the horizontal line test tell us whether the graph of a function is one-to-one? ALGEBRAIC For the following exercises, determine whether the relation represents a function. 6. {(a, b), (c, d), (a, c)} 7. {(a, b),(b, c),(c, c)} For the following exercises, determine whether the relation represents y as a function of x. 9. y = x 2 8. 5x + 2y = 10 10. x = y 2 11. 3x 2 + y = 14 1 __ 14. y = x 3x + 5 ______ 7x βˆ’ 1 17.
y = 20. x = y 3 23. x = Β± √ β€” 1 βˆ’ y 26. y 3 = x 2 12. 2x + y 2 = 6 15. x = 3y + 5 _ 7y βˆ’ 1 18. x 2 + y 2 = 9 21. y = x 3 24. y = Β± √ β€” 1 βˆ’ x 13. y = βˆ’2x 2 + 40x 16. x = √ β€” 1 βˆ’ y 2 19. 2xy =#1 22. y = √ β€” 1 βˆ’ x 2 25. y 2 = x 2 For the following exercises, evaluate the function f at the indicated values f (βˆ’3), f (2), f (βˆ’a), βˆ’f (a), f (a + h). 28. f (x) = βˆ’5x 2 + 2x βˆ’ 1 29. f (x) = √ β€” 2 βˆ’ x + 5 27. f (x) = 2x βˆ’ 5 30. f (x) = 6x βˆ’ 1 ______ 5x + 2 32. Given the function g(x) =#5 βˆ’ x 2, simplify 33. Given the function g(x) =#x 2 + 2x, simplify 31. f (x) = ∣#x βˆ’ 1 ∣ βˆ’ ∣#x + 1 ∣ g(x + h) βˆ’ g(x) __ h g(x) βˆ’ g(a β‰ #0 34. Given the function k(t) = 2t βˆ’ 1: 35. Given the function f (x) =#8 βˆ’ 3x: a. Evaluate k(2). b. Solve k(t) =#7. a. Evaluate f (βˆ’2). b. Solve f (x) = βˆ’1. 36. Given the function p(c) = c 2 + c: 37. Given the function f (x) =#x 2 βˆ’ 3x a. Evaluate p(βˆ’3). b. Solve p(c) = 2. 38. Given the function f (x) =# √ β€” x + 2 : a. Evaluate f (7). b. Solve f (x) =#4 a. Evaluate f (5). b. Solve f (x) =#4 39. Consider the relationship 3r + 2t = 18. a. Write the relationship as a function r = f (t). b. Evaluate f (βˆ’3
). c. Solve f (t) = 2. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.1 SECTION EXERCISES 177 GRAPHICAL For the following exercises, use the vertical line test to determine which graphs show relations that are functions. 40. y 41. y 42. y 43. y 46. y 49. y x x x x 44. y 47. y 50. y x x x x 45. y 48. y 51. y x x x x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 178 CHAPTER 3 FUNCTIONS 52. Given the following graph 53. Given the following graph 54. Given the following graph a. Evaluate f (βˆ’1). b. Solve for f (x) = 3. a. Evaluate f (0). b. Solve for f (x) = βˆ’3. a. Evaluate f (4). b. Solve for f (x) = 1. y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 21 3 4 5 x –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 x –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 1 –1 –2 –3 –4 –5 21 3 4 5 x For the following exercises, determine if the given graph is a one-to-one function. 57. 56. 55. y y 21 3 4 5 x –5 –4 –3 –2 59. 21 3 4 5 x!Ο€ 5 4 3 2 1 –1 –1 –2 –3 –4 –5 y 5 4 3 2 1!1!2!3!4!5 5 4 3 2 1 –1 –1 –2 –3 –4 –5 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 58. –5 –4 –3 –2 NUMERIC 21 3 4 5 x –5 –4 –3 –2 x Ο€ For the following exercises, determine whether the relation represents a function. 60. {(βˆ’1, βˆ’1),(βˆ’
2, βˆ’2),(βˆ’3, βˆ’3)} 61. {(3, 4),(4, 5),(5, 6)} 62. {(2, 5),(7, 11),(15, 8),(7, 9)} For the following exercises, determine if the relation represented in table form represents y as a function of x. 63. x y 5 3 10 8 15 14 64. x y 5 3 10 8 15 8 65. x y 5 3 10 8 10 14 For the following exercises, use the function f represented in Table 14 below. x f (x) 0 74 1 28 2 1 3 53 4 56 Table 14 5 3 6 36 7 45 8 14 9 47 66. Evaluate f (3). 67. Solve f (x) = 1 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.1 SECTION EXERCISES 179 For the following exercises, evaluate the function f at the values f (βˆ’2), f (βˆ’1), f (0), f (1), and f (2). 68. f (x) = 4 βˆ’ 2x 69. f (x) = 8 βˆ’ 3x 70. f (x) = 8x 2 βˆ’ 7x + 3 71. f (x) = 3 + √ β€” x + 3 72. f (x) = x βˆ’ 2 _ x + 3 73. f (x) = 3x For the following exercises, evaluate the expressions, given functions f, g, and h: f (x) = 3x βˆ’ 2 g(x) = 5 βˆ’ x2 h(x) = βˆ’2x2 + 3x βˆ’ 1 74. 3f (1) βˆ’ 4g(βˆ’2) TECHNOLOGY 7 75. f ( ) βˆ’ h(βˆ’2) _ 3 For the following exercises, graph y = x2 on the given viewing window. Determine the corresponding range for each viewing window. Show each graph. 76. [βˆ’0.1, 0.1] 77. [βˆ’10, 10] 78. [βˆ’100, 100] For the following exercises, graph y = x3 on the given viewing window. Determine the corresponding range for each viewing window. Show each graph. 79. [βˆ’0.1, 0.1] 81. [βˆ’100, 100] 80. [βˆ’10, 10] For the following exercises, graph y =
√ viewing window. Show each graph. β€” x on the given viewing window. Determine the corresponding range for each 82. [0, 0.01] 83. [0, 100] 84. [0, 10,000] For the following exercises, graph y = viewing window. Show each graph. 85. [βˆ’0.001, 0.001] β€” 3 √ x on the given viewing window. Determine the corresponding range for each 86. [βˆ’1,000, 1,000] 87. [βˆ’1,000,000, 1,000,000] REAL-WORLD APPLICATIONS 88. The amount of garbage, G, produced by a city with population p is given by G = f (p). G is measured in tons per week, and p is measured in thousands of people. a. The town of Tola has a population of 40,000 and produces 13 tons of garbage each week. Express this information in terms of the function f. b. Explain the meaning of the statement f (5) = 2. 89. The number of cubic yards of dirt, D, needed to cover a garden with area a square feet is given by D = g(a). a. A garden with area 5,000 ft2 requires 50 yd3 of dirt. Express this information in terms of the function g. b. Explain the meaning of the statement g(100) = 1. 90. Let f (t) be the number of ducks in a lake t years after 1990. Explain the meaning of each statement: a. f (5) = 30 b. f (10) = 40 91. Let h(t) be the height above ground, in feet, of a rocket t seconds after launching. Explain the meaning of each statement: a. h(1) = 200 b. h(2) = 350 92. Show that the function f (x) = 3(x βˆ’ 5)2 + 7 is not one-to-one. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 180 CHAPTER 3 FUNCTIONS LEARNING OBJECTIVES In this section, you will: β€’ Find the domain of a function identified by an equation using interal notation. β€’ Gien the graph of a function find the domain and range using interal notation. β€’ Graph piecewise-defined functions. 3.2 DOMAIN AND RANGE If you’re in
the mood for a scary movie, you may want to check out one of the five most popular horror movies of all timeβ€”I am Legend, Hannibal, The Ring, The Grudge, and The Conjuring. Figure 1 shows the amount, in dollars, each of those movies grossed when they were released as well as the ticket sales for horror movies in general by year. Notice that we can use the data to create a function of the amount each movie earned or the total ticket sales for all horror movies by year. In creating various functions using the data, we can identify different independent and dependent variables, and we can analyze the data and the functions to determine the domain and range. In this section, we will investigate methods for determining the domain and range of functions such as these. Market Share of Horror Moveis, by Year Top-Five Grossing Horror Movies for years 2000–201 fl 350 300 250 200 150 100 50 0 8% 7% 6% 5% 4% 3% 2% 1% 0% I am Legend (2007) Hannibal (2001) The Ring (2002) The Grudge (2004) The Conjuring (2013) 2005 Figure 1 Based on data compiled by www.the-numbers.com.[3] 2002 2000 2001 2003 2004 2006 2007 2008 200 9 2010 2011 2012 2013 Finding the Domain of a Function Defined by an Equation In Functions and Function Notation, we were introduced to the concepts of domain and range. In this section, we will practice determining domains and ranges for specific functions. Keep in mind that, in determining domains and ranges, we need to consider what is physically possible or meaningful in real-world examples, such as tickets sales and year in the horror movie example above. We also need to consider what is mathematically permitted. For example, we cannot include any input value that leads us to take an even root of a negative number if the domain and range consist of real numbers. Or in a function expressed as a formula, we cannot include any input value in the domain that would lead us to divide by 0. We can visualize the domain as a β€œholding area” that contains β€œraw materials” for a β€œfunction machine” and the range as another β€œholding area” for the machine’s products. See Figure 2. Domain a b c Function machine Figure 2 Range x y z We can write the domain and range in interval notation, which uses values within brackets to describe a set of numbers. In
interval notation, we use a square bracket [when the set includes the endpoint and a parenthesis (to indicate that the endpoint is either not included or the interval is unbounded. For example, if a person has $100 to spend, he or she would need to express the interval that is more than 0 and less than or equal to 100 and write (0, 100]. We will discuss interval notation in greater detail later. 3 The Numbers: Where Data and the Movie Business Meet. β€œBox Office History for Horror Movies.” http://www.the-numbers.com/market/genre/Horror. Accessed 3/24/2014 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.2 DOMAIN AND RANGE 181 Let’s turn our attention to finding the domain of a function whose equation is provided. Oftentimes, finding the domain of such functions involves remembering three different forms. First, if the function has no denominator or an even root, consider whether the domain could be all real numbers. Second, if there is a denominator in the function’s equation, exclude values in the domain that force the denominator to be zero. Third, if there is an even root, consider excluding values that would make the radicand negative. Before we begin, let us review the conventions of interval notation: β€’ The smallest number from the interval is written first. β€’ The largest number in the interval is written second, following a comma. β€’ β€’ Parentheses, (or), are used to signify that an endpoint value is not included, called exclusive. Brackets, [or], are used to indicate that an endpoint value is included, called inclusive. See Figure 3 for a summary of interval notation. Inequality Interval Notation Graph on Number Line a, ∞) (βˆ’βˆž, a) [a, ∞) (βˆ’βˆž, a] (a, b) [a, b) (a, b] [a, b Figure 3 Description x is greater than a x is less than a x is greater than or equal to a x is less than or equal to a x is strictly between a and b x is between a and b, to include a x is between a and b, to include b x is between a and b, to include a and b Example 1 Finding the Domain of a Function as a Set of Ordered Pairs Find the domain of the following
function: {(2, 10), (3, 10), (4, 20), (5, 30), (6, 40)}. Solution First identify the input values. The input value is the first coordinate in an ordered pair. There are no restrictions, as the ordered pairs are simply listed. The domain is the set of the first coordinates of the ordered pairs. {2, 3, 4, 5, 6} Try It #1 Find the domain of the function: {(βˆ’5, 4), (0, 0), (5, βˆ’4), (10, βˆ’8), (15, βˆ’12)} How To… Given a function written in equation form, find the domain. 1. Identify the input values. 2. Identify any restrictions on the input and exclude those values from the domain. 3. Write the domain in interval form, if possible. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 182 CHAPTER 3 FUNCTIONS Example 2 Finding the Domain of a Function Find the domain of the function f (x) = x2 βˆ’ 1. Solution The input value, shown by the variable x in the equation, is squared and then the result is lowered by one. Any real number may be squared and then be lowered by one, so there are no restrictions on the domain of this function. The domain is the set of real numbers. In interval form, the domain of f is (βˆ’βˆž, ∞). Try It #2 Find the domain of the function: f (x) = 5 βˆ’ x + x 3. How To… Given a function written in an equation form that includes a fraction, find the domain. 1. Identify the input values. 2. Identify any restrictions on the input. If there is a denominator in the function’s formula, set the denominator equal to zero and solve for x. If the function’s formula contains an even root, set the radicand greater than or equal to 0, and then solve. 3. Write the domain in interval form, making sure to exclude any restricted values from the domain. Example 3 Finding the Domain of a Function Involving a Denominator Find the domain of the function f (x) = x + 1 _____. 2 βˆ’ x Solution When there is a denominator, we want to include only values of the input that do not force the denominator to be zero. So, we will set
the denominator equal to 0 and solve for x. 2 βˆ’ x = 0 βˆ’x = βˆ’2 x = 2 Now, we will exclude 2 from the domain. The answers are all real numbers where x < 2 or x > 2 as shown in Figure 4. We can use a symbol known as the union, βˆͺ, to combine the two sets. In interval notation, we write the solution: (βˆ’βˆž, 2) βˆͺ (2, ∞). –3 –2 –1 0 x < 2 or x > 2 1 2 3 ↓ ↓ (βˆ’βˆž, 2) βˆͺ (2, ∞) Figure 4 Try It #3 Find the domain of the function: f (x) = 1 + 4x ______ 2x βˆ’ 1. How To… Given a function written in equation form including an even root, find the domain. 1. Identify the input values. 2. Since there is an even root, exclude any real numbers that result in a negative number in the radicand. Set the radicand greater than or equal to zero and solve for x. 3. The solution(s) are the domain of the function. If possible, write the answer in interval form. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.2 DOMAIN AND RANGE 183 Example 4 Finding the Domain of a Function with an Even Root Find the domain of the function f (x) = √ β€” 7 βˆ’ x. Solution When there is an even root in the formula, we exclude any real numbers that result in a negative number in the radicand. Set the radicand greater than or equal to zero and solve for x. 7 βˆ’ x β‰₯ 0 βˆ’x β‰₯ βˆ’7 x ≀ 7 Now, we will exclude any number greater than 7 from the domain. The answers are all real numbers less than or equal to 7, or (βˆ’βˆž, 7]. Try It #4 Find the domain of the function f (x) = √ β€” 5 + 2x. Q & A… Can there be functions in which the domain and range do not intersect at all? Yes. For example, the function f (x) = βˆ’ # 1 _ β€” x √ negative real numbers as its range. As a more extreme example, a function’s inputs and outputs can be completely different categories (for example, names of weekdays as inputs and numbers as outputs, as
on an attendance chart), in such cases the domain and range have no elements in common. has the set of all positive real numbers as its domain but the set of all Using Notations to Specify Domain and Range In the previous examples, we used inequalities and lists to describe the domain of functions. We can also use inequalities, or other statements that might define sets of values or data, to describe the behavior of the variable in set-builder notation. For example, {x | 10 ≀ x < 30} describes the behavior of x in set-builder notation. The braces { } are read as β€œthe set of,” and the vertical bar | is read as β€œsuch that,” so we would read {x | 10 ≀ x < 30} as β€œthe set of x-values such that 10 is less than or equal to x, and x is less than 30.” Figure 5 compares inequality notation, set-builder notation, and interval notation. 5 5 5 5 5 Inequality Notation 5 < h ≀ 10 Set-builder Notation Interval Notation {h | 5 < h ≀ 10} (5, 10] 5 ≀ h < 10 {h | 5 ≀ h < 10} [5, 10) 5 < h < 10 {h | 5 < h < 10} (5, 10) h < 10 h β‰₯ 10 {h | h < 10} (βˆ’βˆž, 10) {h | h β‰₯ 10} [10, ∞) All real numbers / (βˆ’βˆž, ∞) Figure 5 10 10 10 10 10 10 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 184 CHAPTER 3 FUNCTIONS To combine two intervals using inequality notation or set-builder notation, we use the word β€œor.” As we saw in earlier examples, we use the union symbol, βˆͺ, to combine two unconnected intervals. For example, the union of the sets {2, 3, 5} and {4, 6} is the set {2, 3, 4, 5, 6}. It is the set of all elements that belong to one or the other (or both) of the original two sets. For sets with a finite number of elements like these, the elements do not have to be listed in ascending order of numerical value. If the original two sets have some elements in common, those elements should be listed only once in the union set
. For sets of real numbers on intervals, another example of a union is {x | | x | β‰₯ 3} = (βˆ’βˆž, βˆ’3] βˆͺ [3, ∞) set-builder notation and interval notation Set-builder notation is a method of specifying a set of elements that satisfy a certain condition. It takes the form {x | statement about x} which is read as, β€œthe set of all x such that the statement about x is true.” For example, {x | 4 < x ≀ 12} Interval notation is a way of describing sets that include all real numbers between a lower limit that may or may not be included and an upper limit that may or may not be included. The endpoint values are listed between brackets or parentheses. A square bracket indicates inclusion in the set, and a parenthesis indicates exclusion from the set. For example, (4, 12] How To… Given a line graph, describe the set of values using interval notation. 1. Identify the intervals to be included in the set by determining where the heavy line overlays the real line. 2. At the left end of each interval, use [with each end value to be included in the set (solid dot) or (for each excluded end value (open dot). 3. At the right end of each interval, use] with each end value to be included in the set (filled dot) or) for each excluded end value (open dot). 4. Use the union symbol βˆͺ to combine all intervals into one set. Example 5 Describing Sets on the Real-Number Line Describe the intervals of values shown in Figure 6 using inequality notation, set-builder notation, and interval notation. –2 –1 0 1 2 3 4 5 6 7 Figure 6 Solution To describe the values, x, included in the intervals shown, we would say, β€œx is a real number greater than or equal to 1 and less than or equal to 3, or a real number greater than 5.” Inequality 1 ≀ x ≀ 3 or x > 5 Set-builder notation { x |1 ≀ x ≀ 3 or x > 5 } Interval notation [1, 3] βˆͺ (5, ∞) Remember that, when writing or reading interval notation, using a square bracket means the boundary is included in the set. Using a parenthesis means the boundary is not included in the set. Download the OpenStax text for free at http://cnx.org/
content/col11759/latest. SECTION 3.2 DOMAIN AND RANGE 185 Try It #5 Given this figure, specify the graphed set in a. words b. set-builder notation c. interval notation Finding Domain and Range from Graphs –5 –4 –3 –2 –1 0 1 2 3 4 5 Figure 7 Another way to identify the domain and range of functions is by using graphs. Because the domain refers to the set of possible input values, the domain of a graph consists of all the input values shown on the x-axis. The range is the set of possible output values, which are shown on the y-axis. Keep in mind that if the graph continues beyond the portion of the graph we can see, the domain and range may be greater than the visible values. See Figure 8. Domain y 7 6 5 4 3 2 1 –6 –5 –4 –3 –2 –1 –1 –2 –3 –4 –5 –6 –7 – 8 – 9 Range 21 3 4 5 6 x Figure 8 We can observe that the graph extends horizontally from βˆ’5 to the right without bound, so the domain is [βˆ’5, ∞). The vertical extent of the graph is all range values 5 and below, so the range is (βˆ’βˆž, 5]. Note that the domain and range are always written from smaller to larger values, or from left to right for domain, and from the bottom of the graph to the top of the graph for range. Example 6 Finding Domain and Range from a Graph Find the domain and range of the function f whose graph is shown in Figure 9. –5 –4 –3 –2 –1 f y 1 –1 –2 – 3 – 4 –5 1 2 3 4 5 x Figure 9 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 186 CHAPTER 3 FUNCTIONS Solution We can observe that the horizontal extent of the graph is βˆ’3 to 1, so the domain of f is (βˆ’3, 1]. The vertical extent of the graph is 0 to βˆ’4, so the range is [βˆ’4, 0]. See Figure 10. y Domain 1 –5 –4 –3 –2 –1 f –1 –2 – 3 – 4 –5 1 2 3 4 5 x Range Figure 10 Example 7 Finding Domain and Range from a Graph of Oil Production Find the domain and range of the function f whose graph is shown in Figure 11.
Alaska Crude Oil Production Th 2200 2000 1800 1600 1400 1200 1000 800 600 400 200 0 1975 1980 1985 1990 1995 2000 2005 Figure 11 (credit: modification of work by the U.S. Energy Information Administration) [4] Solution The input quantity along the horizontal axis is β€œyears,” which we represent with the variable t for time. The output quantity is β€œthousands of barrels of oil per day,” which we represent with the variable b for barrels. The graph may continue to the left and right beyond what is viewed, but based on the portion of the graph that is visible, we can determine the domain as 1973 ≀ t ≀ 2008 and the range as approximately 180 ≀ b ≀ 2010. In interval notation, the domain is [1973, 2008], and the range is about [180, 2010]. For the domain and the range, we approximate the smallest and largest values since they do not fall exactly on the grid lines. Try It #6 Given Figure 12, identify the domain and range using interval notation. World Population Increase 100 90 80 70 60 50 40 30 20 10 0 1950 1960 1970 1980 1990 2000 Year Figure 12 Q & A… Can a function’s domain and range be the same? Yes. For example, the domain and range of the cube root function are both the set of all real numbers. 4 http://www.eia.gov/dnav/pet/hist/LeafHandler.ashx?n=PET&s=MCRFPAK2&f=A. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.2 DOMAIN AND RANGE 187 Finding Domains and Ranges of the Toolkit Functions We will now return to our set of toolkit functions to determine the domain and range of each. f (x) f (x) = c Domain: (βˆ’βˆž, ∞) Range: [c, c] x For the constant function f(x) = c, the domain consists of all real numbers; there are no restrictions on the input. The only output value is the constant c, so the range is the set {c} that contains this single element. In interval notation, this is written as [c, c], the interval that both begins and ends with c. Figure 13 f (x) Figure 14 f (x) Figure 15 f (x) Figure 16 f (x) Figure
17 For the identity function f(x) = x, there is no restriction on x. Both the domain and range are the set of all real numbers. Domain: (βˆ’βˆž, ∞) Range: (βˆ’βˆž, ∞) x Domain: (βˆ’βˆž, ∞) Range: [0, ∞) x For the absolute value function f(x) = ∣ x ∣, there is no restriction on x. However, because absolute value is defined as a distance from 0, the output can only be greater than or equal to 0. Domain: (βˆ’βˆž, ∞) Range: [0, ∞) x For the quadratic function f(x) = x 2, the domain is all real numbers since the horizontal extent of the graph is the whole real number line. Because the graph does not include any negative values for the range, the range is only nonnegative real numbers. Domain: (βˆ’βˆž, ∞) Range: (βˆ’βˆž, ∞) x For the cubic function f(x) = x 3, the domain is all real numbers because the horizontal extent of the graph is the whole real number line. The same applies to the vertical extent of the graph, so the domain and range include all real numbers. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 188 CHAPTER 3 FUNCTIONS Domain: (βˆ’βˆž, 0) βˆͺ (0, ∞) Range: (βˆ’βˆž, 0) βˆͺ (0, ∞) x For the reciprocal function f(x) = 1 _ x, we cannot divide by 0, so we must exclude 0 from the domain. Further, 1 divided by any value can never be 0, so the range also will not include 0. In set-builder notation, we could also write {x | x β‰  0}, the set of all real numbers that are not zero. Domain: (βˆ’βˆž, 0) βˆͺ (0, ∞) Range: (0, ∞) x For the reciprocal squared function f(x) = 1 __ x 2, we cannot divide by 0, so we must exclude 0 from the domain. There is also no x that can give an output of 0, so 0 is excluded from the range as well. Note that the output of this function is always positive due to the square in the denominator, so the range includes only positive numbers. Domain
: [0, ∞) Range: [0, ∞) x Domain: (βˆ’βˆž, ∞) Range: (βˆ’βˆž, ∞) x β€” x, we cannot take the For the square root function f(x) = √ square root of a negative real number, so the domain must be 0 or greater. The range also excludes negative numbers because the square root of a positive number x is defined to be positive, even though the square of the negative number βˆ’ √ β€” x also gives us x. β€” 3 √ x the domain and range For the cube root function f(x) = include all real numbers. Note that there is no problem taking a cube root, or any odd-integer root, of a negative number, and the resulting output is negative (it is an odd function). f (x) Figure 18 f (x) Figure 19 f (x) Figure 20 f (x) Figure 21 How To… Given the formula for a function, determine the domain and range. 1. Exclude from the domain any input values that result in division by zero. 2. Exclude from the domain any input values that have nonreal (or undefined) number outputs. 3. Use the valid input values to determine the range of the output values. 4. Look at the function graph and table values to confirm the actual function behavior. Example 8 Finding the Domain and Range Using Toolkit Functions Find the domain and range of f (x) = 2x3 βˆ’ x. Solution There are no restrictions on the domain, as any real number may be cubed and then subtracted from the result. The domain is (βˆ’βˆž, ∞) and the range is also (βˆ’βˆž, ∞). Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.2 DOMAIN AND RANGE 189 Example 9 Finding the Domain and Range Find the domain and range of f (x) = 2 ____ x + 1. Solution We cannot evaluate the function at βˆ’1 because division by zero is undefined. The domain is (βˆ’βˆž, βˆ’1) βˆͺ (βˆ’1, ∞). Because the function is never zero, we exclude 0 from the range. The range is (βˆ’βˆž, 0) βˆͺ (0, ∞). Example 10 Finding the Domain and Range Find the domain and range of f (x) = 2 √ β€” x + 4. Solution We cannot take the
square root of a negative number, so the value inside the radical must be nonnegative. The domain of f (x) is [βˆ’4, ∞). x + 4 β‰₯ 0 when x β‰₯ βˆ’4 We then find the range. We know that f (βˆ’4) = 0, and the function value increases as x increases without any upper limit. We conclude that the range of f is [0, ∞). Analysis Figure 22 represents the function f. f (x) 5 4 3 2 1 f –5 –4 –3 –2 –1 –1 21 3 4 5 x Figure 22 Try It #7 Find the domain and range of f (x) = βˆ’ √ β€” 2 βˆ’ x. Graphing Piecewise-Defined Functions Sometimes, we come across a function that requires more than one formula in order to obtain the given output. For example, in the toolkit functions, we introduced the absolute value function f (x) = |x|. With a domain of all real numbers and a range of values greater than or equal to 0, absolute value can be defined as the magnitude, or modulus, of a real number value regardless of sign. It is the distance from 0 on the number line. All of these definitions require the output to be greater than or equal to 0. If we input 0, or a positive value, the output is the same as the input. f (x) = x if x β‰₯ 0 If we input a negative value, the output is the opposite of the input. f (x) = βˆ’x if x < 0 Because this requires two different processes or pieces, the absolute value function is an example of a piecewise function. A piecewise function is a function in which more than one formula is used to define the output over different pieces of the domain. We use piecewise functions to describe situations in which a rule or relationship changes as the input value crosses certain β€œboundaries.” For example, we often encounter situations in business for which the cost per piece of a certain item is discounted once the number ordered exceeds a certain value. Tax brackets are another real-world example of piecewise functions. For example, consider a simple tax system in which incomes up to $10,000 are taxed at 10%, and any additional income is taxed at 20%. The tax on a total income S would be 0.1S if S ≀ $10,000 and $1000 + 0.2(S βˆ’ $10,000
) if S > $10,000. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 190 CHAPTER 3 FUNCTIONS piecewise function A piecewise function is a function in which more than one formula is used to define the output. Each formula has its own domain, and the domain of the function is the union of all these smaller domains. We notate this idea like this: f (x) = { formula 1 formula 2 formula 3 if x is in domain 1 if x is in domain 2 if x is in domain 3 In piecewise notation, the absolute value function is x βˆ’x { |x| = if x β‰₯ 0 if x < 0 How To… Given a piecewise function, write the formula and identify the domain for each interval. 1. Identify the intervals for which different rules apply. 2. Determine formulas that describe how to calculate an output from an input in each interval. 3. Use braces and if-statements to write the function. Example 11 Writing a Piecewise Function A museum charges $5 per person for a guided tour with a group of 1 to 9 people or a fixed $50 fee for a group of 10 or more people. Write a function relating the number of people, n, to the cost, C. Solution Two di fferent formulas will be needed. For n-values under 10, C = 5n. For values of n that are 10 or greater, C = 50. { C(n) = 5n if 0 < n < 10 50 if n β‰₯ 10 Analysis The function is represented in Figure 23. The graph is a diagonal line from n = 0 to n = 10 and a constant after that. In this example, the two formulas agree at the meeting point where n = 10, but not all piecewise functions have this property. C (n) 50 40 30 20 10 C (n) – 25 – 20 – 15 – 10 – 5 –10 5 10 15 20 25 n Figure 23 Example 12 Working with a Piecewise Function A cell phone company uses the function below to determine the cost, C, in dollars for g gigabytes of data transfer. { C(g) = 25 if 0 < g < 2 25 + 10(g βˆ’ 2) if g β‰₯ 2 Find the cost of using 1.5 gigabytes of data and the cost of using 4 gigabytes of data. Solution To find the cost of using 1.5 gigabytes of data
, C(1.5), we first look to see which part of the domain our input falls in. Because 1.5 is less than 2, we use the first formula. C(1.5) = $25 To fi nd the cost of using 4 gigabytes of data, C(4), we see that our input of 4 is greater than 2, so we use the second formula. C(4) = 25 + 10(4 βˆ’ 2) = $45 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.2 DOMAIN AND RANGE 191 Analysis The function is represented in Figure 24. We can see where the function changes from a constant to a shifted and stretched identity at g = 2. We plot the graphs for the different formulas on a common set of axes, making sure each formula is applied on its proper domain. C(g) 50 40 30 20 10 C (g10 1 2 3 4 5 g Figure 24 How To… Given a piecewise function, sketch a graph. 1. Indicate on the x-axis the boundaries defined by the intervals on each piece of the domain. 2. For each piece of the domain, graph on that interval using the corresponding equation pertaining to that piece. Do not graph two functions over one interval because it would violate the criteria of a function. Example 13 Graphing a Piecewise Function Sketch a graph of the function. f (x) = { x ≀ 1 x2 if 3 x if 1 < x ≀ 2 x > 2 if Solution Each of the component functions is from our library of toolkit functions, so we know their shapes. We can imagine graphing each function and then limiting the graph to the indicated domain. At the endpoints of the domain, we draw open circles to indicate where the endpoint is not included because of a less-than or greater-than inequality; we draw a closed circle where the endpoint is included because of a less-than-or-equal-to or greater-than-or-equal-to inequality. Figure 25 shows the three components of the piecewise function graphed on separate coordinate systems. f (x) 5 4 3 2 1 –1 (ax) 5 4 3 2 1 –1 (b) f (x) 5 4 3 2 1 –1 (c Figure 25 ( a) f (x ) = x 2 if x ≀ 1; ( b) f (x ) = 3
if 1 < x ≀ 2; ( c) f (x ) = x if x > 2 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 192 CHAPTER 3 FUNCTIONS Now that we have sketched each piece individually, we combine them in the same coordinate plane. See Figure 26. f (x) 5 4 3 2 1 – Figure 26 Analysis Note that the graph does pass the vertical line test even at x = 1 and x = 2 because the points (1, 3) and (2, 2) are not part of the graph of the function, though (1, 1) and (2, 3) are. Try It #8 Graph the following piecewise function. f (x) = { x3 βˆ’2 β€” √ x x < βˆ’1 if if βˆ’1 < x < 4 x > 4 if Q & A… Can more than one formula from a piecewise function be applied to a value in the domain? No. Each value corresponds to one equation in a piecewise formula. Access these online resources for additional instruction and practice with domain and range. β€’ Domain and Range of Square Root Functions (http://openstaxcollege.org/l/domainsqroot) β€’ Determining Domain and Range (http://openstaxcollege.org/l/determinedomain) β€’ Find Domain and Range Given the Graph (http://openstaxcollege.org/l/drgraph) β€’ Find Domain and Range Given a Table (http://openstaxcollege.org/l/drtable) β€’ Find Domain and Range Given Points on a Coordinate Plane (http://openstaxcollege.org/l/drcoordinate) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.2 SECTION EXERCISES 193 3.2 SECTION EXERCISES VERBAL 1. Why does the domain differ for different functions? 2. How do we determine the domain of a function defined by an equation? 3. Explain why the domain of f (x) = from the domain of f (x is different 4. When describing sets of numbers using interval notation, when do you use a parenthesis and when do you use a bracket? 5. How do you graph a piecewise function? ALGEBRAIC For the following exercises, find the domain of each function using interval notation. 6.
f (x) = βˆ’2x(x βˆ’ 1)(x βˆ’ 2) 7. f (x) = 5 βˆ’ 2x2 8. f (x) = 3 √ β€” x βˆ’ 2 9. f (x) = 3 βˆ’ √ β€” 6 βˆ’ 2x 10. f (x) = √ β€” 4 βˆ’ 3x 11. f (x) = √ β€” x 2 + 4 12. f (x) = # 3 √ β€” 1 βˆ’ 2x 13. f (x) = # 3 √ β€” x βˆ’ 1 15. f (x) = #3x + 1 ______ 4x + 2 18. f (x) = # 1 ________ x2 βˆ’ x βˆ’ 6 16. f (x) = # β€” √ x + 4 _______ x βˆ’ 4 19. f (x) = # 2x3 βˆ’ 250 __________ x2 βˆ’ 2x βˆ’ 15 14. f (x) = # 9 _____ x βˆ’ 6 17. f (x) = # x βˆ’ 3 __________ x2 + 9x βˆ’ 22 20. f (x) = # 5 _ β€” x βˆ’ 3 √ 21. f (x) = 2x + 1_ 5 βˆ’ x √ β€” 22. f (x √ β€” 23. f (x √ β€” 24. f (x) = #x __ x 25. f (x) = x2 βˆ’ 9x_ x2 βˆ’ 81 26. Find the domain of the function f (x) = √ β€” 2x 3 βˆ’ 50x by: a. using algebra. b. graphing the function in the radicand and determining intervals on the x-axis for which the radicand is nonnegative. GRAPHICAL For the following exercises, write the domain and range of each function using interval notation. 27. y 28. y 29. –10 –8 –6 –4 10 8 6 4 2 –2 –2 –4 –6 –8 –10 42 6 8 10 x –10 –8 –6 –4 10 8 6 4 2 –2 –2 –4 –6 –8 –10 42 6 8 10 x –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 194 30.
33. 36. CHAPTER 3 FUNCTIONS y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 y 5 4 3 2 1 y –1 –1 –2 –3 –4 –5 10 8 6 4 2 –2 –2 –4 –6 –8 –10 –5 –4 –3 –2 –5 –4 –3 –2 –10 –8 –6 –4 31. 21 3 4 5 x –5 –4 –3 –2 34. 21 3 4 5 x –5 –4 –3 –2 37. 42 6 8 10 x –10 –8 –6 –4 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 y 5 4 3 2 1 y –1 –1 –2 –3 –4 –5 10 8 6 4 2 –2 –2 –4 –6 –8 –10 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 32. 21 3 4 5 x –5 –4 –3 –2 35. 21 3 4 5 x –6 –5 –4 –3 –2 1 6 –6, – ) ( ), –6 1 6 ( – 42 6 8 10 x 21, ( 1 6 ) 21 3 4 5 6 x For the following exercises, sketch a graph of the piecewise function. Write the domain in interval notation. x + 1 if x < βˆ’2 { 38. f (x) =# βˆ’2x βˆ’ 3 if x β‰₯ βˆ’2 { 39. f (x) =# 2x βˆ’ 1 if x < 1 if x β‰₯ 1 1 + x { 40. f (x) =# x + 1 if x < 0 x βˆ’ 1 if x > 0 { 41. f (x) =# 3 √ if x < 0 x if x β‰₯ 0 β€” { 42. f (x) =# if x < 0 x2 1 βˆ’ x if x > 0 { 43. f (x) =# if x < 0 x2 x + 2 if x β‰₯ 0 { 44. f (x) =# x + 1 if x < 1 if x β‰₯ 1 x3 { 45. f (x) =# | x | if x < 2 if x β‰₯ 2 1 Download the OpenStax text for free at http://cnx.org/
content/col11759/latest. SECTION 3.2 SECTION EXERCISES 195 NUMERIC For the following exercises, given each function f, evaluate f (βˆ’3), f (βˆ’2), f (βˆ’1), and f (0). { 46. f (x) =# x + 1 if x < βˆ’2 βˆ’2x βˆ’ 3 if x β‰₯ βˆ’2 { 47. f (x) =# 1 if x ≀ βˆ’3 0 if x > βˆ’3 { 48. f (x) =# βˆ’2x 2 + 3 if x ≀ βˆ’1 if x > βˆ’1 5x βˆ’ 7 For the following exercises, given each function f, evaluate f (βˆ’1), f (0), f (2), and f (4). { 49. f (x) =# 7x + 3 if x < 0 7x + 6 if x β‰₯ 0 { 50. f (x) =# if | if x β‰₯ 2 51. f (x) =# { 5x if 3 x 2 if x < 0 if 0 ≀ x ≀ 3 x > 3 For the following exercises, write the domain for the piecewise function in interval notation. { 52. f (x) =# x + 1 if x < βˆ’2 βˆ’2x βˆ’ 3 if x β‰₯ βˆ’2 { 53. f (x) =# x2 βˆ’ 2 if x < 1 βˆ’x2 + 2 if x > 1 { 54. f (x) =# 2x βˆ’ 3 if x < 0 if x β‰₯ 2 βˆ’3x2 TECHNOLOGY 55. Graph y = 1 __ x 2 on the viewing window [βˆ’0.5, βˆ’0.1] and [0.1, 0.5]. Determine the corresponding range for the viewing window. Show the graphs. 1 _ x on the viewing window [βˆ’0.5, βˆ’0.1] and [0.1, 0.5]. Determine the corresponding range for the 56. Graph y = viewing window. Show the graphs. EXTENSION 57. Suppose the range of a function f is [βˆ’5, 8]. What is the range of | f (x) |? 58. Create a function in which the range is all nonnegative real numbers. 59. Create a function in which the domain is x > 2. REAL-WORLD APPLICATIONS 60. The height h of a projectile is a function of the time t it is in the air. The height in feet for t seconds is
given by the function h(t) = βˆ’16t 2 + 96t. What is the domain of the function? What does the domain mean in the context of the problem? 61. The cost in dollars of making x items is given by the function C(x) = 10x + 500. a. The fixed cost is determined when zero items are produced. Find the fixed cost for this item. b. What is the cost of making 25 items? c. Suppose the maximum cost allowed is $1500. What are the domain and range of the cost function, C(x)? Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 196 CHAPTER 3 FUNCTIONS LEARNING OBJECTIVES In this section, you will: β€’ Find the aerage rate of change of a function. β€’ Find the difference quotient of a function. β€’ Use a graph to determine where a function is increasing decreasing or constant. 3.3 RATES OF CHANGE AND BEHAVIOR OF GRAPHS Gasoline costs have experienced some wild fluctuations over the last several decades. Table 1[5] lists the average cost, in dollars, of a gallon of gasoline for the years 2005–2012. The cost of gasoline can be considered as a function of year. y C(y) 2005 2.31 2006 2.62 2007 2.84 2008 3.30 2009 2.41 2010 2.84 2011 3.58 2012 3.68 Table 1 If we were interested only in how the gasoline prices changed between 2005 and 2012, we could compute that the cost per gallon had increased from $2.31 to $3.68, an increase of $1.37. While this is interesting, it might be more useful to look at how much the price changed per year. In this section, we will investigate changes such as these. Finding the Average Rate of Change of a Function The price change per year is a rate of change because it describes how an output quantity changes relative to the change in the input quantity. We can see that the price of gasoline in Table 1 did not change by the same amount each year, so the rate of change was not constant. If we use only the beginning and ending data, we would be finding the average rate of change over the specified period of time. To find the average rate of change, we divide the change in the output value by the change in the input value. Average rate of change = Change in output
__ Change in input βˆ†y_ βˆ†x y2 βˆ’ y1 _ x2 βˆ’ x1 f (x2) βˆ’ f (x1) __ βˆ’ x1 = = = x2 The Greek letter βˆ† (delta) signifies the change in a quantity; we read the ratio as β€œdelta-y over delta-x” or β€œthe change in y divided by the change in x.” Occasionally we write βˆ† f instead of βˆ†y, which still represents the change in the function’s output value resulting from a change to its input value. It does not mean we are changing the function into some other function. In our example, the gasoline price increased by $1.37 from 2005 to 2012. Over 7 years, the average rate of change was βˆ†y _ βˆ†x = $1.37 _ 7 years β‰ˆ 0.196 dollars per year On average, the price of gas increased by about 19.6Β’ each year. Other examples of rates of change include: β€’ A population of rats increasing by 40 rats per week β€’ A car traveling 68 miles per hour (distance traveled changes by 68 miles each hour as time passes) β€’ A car driving 27 miles per gallon (distance traveled changes by 27 miles for each gallon) β€’ The current through an electrical circuit increasing by 0.125 amperes for every volt of increased voltage β€’ The amount of money in a college account decreasing by $4,000 per quarter 5 http://www.eia.gov/totalenergy/data/annual/showtext.cfm?t=ptb0524. Accessed 3/5/2014. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.3 RATES OF CHANGE AND BEHAVIOR OF GRAPHS 197 rate of change A rate of change describes how an output quantity changes relative to the change in the input quantity. The units on a rate of change are β€œoutput units per input units.” The average rate of change between two input values is the total change of the function values (output values) divided by the change in the input values. βˆ†y _ βˆ†x = f (x2) βˆ’ f (x1) __ x2 βˆ’ x1 How To… Given the value of a function at different points, calculate the average rate of change of a function for the interval between two values x1
and x2. 1. Calculate the difference y2 βˆ’ y1 = βˆ†y. 2. Calculate the difference x2 βˆ’ x1 = βˆ†x. βˆ†y _. 3. Find the ratio βˆ†x Example 1 Computing an Average Rate of Change Using the data in Table 1, find the average rate of change of the price of gasoline between 2007 and 2009. Solution In 2007, the price of gasoline was $2.84. In 2009, the cost was $2.41. The average rate of change is βˆ†y ___ βˆ†x = y2 βˆ’ y1 _ x2 βˆ’ x1 $2.41 βˆ’ $2.84 __ 2009 βˆ’ 2007 βˆ’$0.43 ______ 2 years = = Analysis Note that a decrease is expressed by a negative change or β€œnegative increase.” A rate of change is negative when the output decreases as the input increases or when the output increases as the input decreases. = βˆ’$0.22 per year Try It #1 Using the data in Table 1 at the beginning of this section, find the average rate of change between 2005 and 2010. Example 2 Computing Average Rate of Change from a Graph Given the function g (t) shown in Figure 1, find the average rate of change on the interval [βˆ’1, 2]. g(t) –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 t Figure 1 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 198 CHAPTER 3 FUNCTIONS Solution At t = βˆ’1, Figure 2 shows g (βˆ’1) = 4. At t = 2, the graph shows g (2) = 1. g(t) βˆ†g(t) = –3 –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 βˆ†t = 3 t Figure 2 The horizontal change βˆ† t = 3 is shown by the red arrow, and the vertical change βˆ† g(t) = βˆ’3 is shown by the turquoise arrow. The average rate of change is shown by the slope of the red line segment. The output changes by –3 while the input changes by 3, giving an average rate of change of 1 βˆ’ 4 _______ = 2 βˆ’ (βˆ’1) βˆ’3 ___ 3 = βˆ’1 Analysis Note
that the order we choose is very important. If, for example, we use answer. Decide which point will be 1 and which point will be 2, and keep the coordinates fixed as ( x1, y1, we will not get the correct ). ) and ( x2, y2 y2 βˆ’ y1 _ x1 βˆ’ x2 Example 3 Computing Average Rate of Change from a Table After picking up a friend who lives 10 miles away and leaving on a trip, Anna records her distance from home over time. The values are shown in Table 2. Find her average speed over the first 6 hours. t (hours) D(t) (miles) 0 10 1 55 2 90 3 153 4 214 5 240 6 282 7 300 Table 2 Solution Here, the average speed is the average rate of change. She traveled 282 miles in 6 hours. The average speed is 47 miles per hour. 292 βˆ’ 10 _______ = 6 βˆ’ 0 282___ 6 = 47 Analysis Because the speed is not constant, the average speed depends on the interval chosen. For the interval [2, 3], the average speed is 63 miles per hour. Computing Average Rate of Change for a Function Expressed as a Formula Example 4 Compute the average rate of change of f (x) = x 2 βˆ’ #1 __ on the interval [2, 4]. x Solution We can start by computing the function values at each endpoint of the interval. 1 __ f (2) = 22 βˆ’ 2 1 __ f(4) = 42 βˆ’ 4 1 __ = 4 βˆ’ 2 7 __ = 2 1 __ = 16 βˆ’ 4 = 63__ 4 Now we compute the average rate of change. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.3 RATES OF CHANGE AND BEHAVIOR OF GRAPHS 199 Average rate of change = f (4) βˆ’ f (2)_ 4 βˆ’ 2 7 63 __ __ βˆ’ 4 2 _ 4 βˆ’ 2 49 __ 4 _ 2 49__ 8 = = = Try It #2 Find the average rate of change of f (x) = x βˆ’ 2 √ β€” x on the interval [1, 9]. Example 5 Finding the Average Rate of Change of a Force The electrostatic force F, measured in newtons, between two charged particles can be related to the distance between 2 _ the particles d, in centimeters, by the formula F(d) = d 2. Find the average rate
of change of force if the distance between the particles is increased from 2 cm to 6 cm. Solution We are computing the average rate of change of F (d) = 2 __ d 2 on the interval [2, 6]. Average rate of change = = F(6) βˆ’ F(2) __ 6 βˆ’ 2 2 2 _ _ 62 βˆ’ 22 _ 6 βˆ’ 2 2 βˆ’# #2 __ __ 36 4 _ 4 βˆ’ #16 __ 36 _ 4 = βˆ’ #1 __ 9 = = Simplify. Combine numerator terms. Simplify. The average rate of change is βˆ’ #1 __ newton per centimeter. 9 Example 6 Finding an Average Rate of Change as an Expression Find the average rate of change of g(t) = t2 + 3t + 1 on the interval [0, a]. The answer will be an expression involving a in simplest form. Solution We use the average rate of change formula. g(a) βˆ’ g(0)_ a βˆ’ 0 Average rate of change = Evaluate. = (a2 + 3a + 1) βˆ’ (02 + 3(0) + 1) ___ a βˆ’ 0 Simplify. = a2 + 3a + 1 βˆ’ 1 _____________ a a(a + 3) _______ a = a + 3 = Simplify and factor. Divide by the common factor a. This result tells us the average rate of change in terms of a between t = 0 and any other point t = a. For example, on the interval [0, 5], the average rate of change would be 5 + 3 = 8. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 200 CHAPTER 3 FUNCTIONS Try It #3 Find the average rate of change of f (x) = x2 + 2x βˆ’ 8 on the interval [5, a] in simplest forms in terms of a. Using a Graph to Determine Where a Function is Increasing, Decreasing, or Constant As part of exploring how functions change, we can identify intervals over which the function is changing in specific ways. We say that a function is increasing on an interval if the function values increase as the input values increase within that interval. Similarly, a function is decreasing on an interval if the function values decrease as the input values increase over that interval. The average rate of change of an increasing function is positive, and the average rate of change of a decreasing function is negative. Figure 3
shows examples of increasing and decreasing intervals on a function. Increasing –5 –4 –3 –2 f(x) 20 16 12 8 4 –1 –4 –8 –12 –16 –20 Decreasing 21 3 4 5 x Increasing f(b) > f(a) where b > a f(b) < f(a) where b > a f(b) > f(a) where b > a Figure 3 The function f(x ) = x 3 βˆ’ 12x is increasing on (βˆ’βˆž, βˆ’2) βˆͺ (2, ∞) and is decreasing on (βˆ’2, 2). While some functions are increasing (or decreasing) over their entire domain, many others are not. A value of the input where a function changes from increasing to decreasing (as we go from left to right, that is, as the input variable increases) is called a local maximum. If a function has more than one, we say it has local maxima. Similarly, a value of the input where a function changes from decreasing to increasing as the input variable increases is called a local minimum. The plural form is β€œlocal minima.” Together, local maxima and minima are called local extrema, or local extreme values, of the function. (The singular form is β€œextremum.”) Often, the term local is replaced by the term relative. In this text, we will use the term local. Clearly, a function is neither increasing nor decreasing on an interval where it is constant. A function is also neither increasing nor decreasing at extrema. Note that we have to speak of local extrema, because any given local extremum as defined here is not necessarily the highest maximum or lowest minimum in the function’s entire domain. For the function whose graph is shown in Figure 4, the local maximum is 16, and it occurs at x = βˆ’2. The local minimum is βˆ’16 and it occurs at x = 2. Local maximum = 16 occurs at x = –2 f(x) Increasing (–2, 16) 20 16 12 8 4 Decreasing –5 –4 –3 –2 –1 –4 –8 –12 –16 –20 21 3 4 5 x Increasing (2, –16) Local minimum = –16 occurs at x = 2 Figure 4 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.3 RATES
OF CHANGE AND BEHAVIOR OF GRAPHS 201 To locate the local maxima and minima from a graph, we need to observe the graph to determine where the graph attains its highest and lowest points, respectively, within an open interval. Like the summit of a roller coaster, the graph of a function is higher at a local maximum than at nearby points on both sides. The graph will also be lower at a local minimum than at neighboring points. Figure 5 illustrates these ideas for a local maximum. f(x) f(b) Local maximum Increasing function Decreasing function a b c Figure 5 Definition of a local maximum x These observations lead us to a formal definition of local extrema. local minima and local maxima A function f is an increasing function on an open interval if f (b) > f (a) for any two input values a and b in the given interval where b > a. A function f is a decreasing function on an open interval if f (b) < f (a) for any two input values a and b in the given interval where b > a. A function f has a local maximum at x = b if there exists an interval (a, c) with a < b < c such that, for any x in the interval (a, c), f (x) ≀ f (b). Likewise, f has a local minimum at x = b if there exists an interval (a, c) with a < b < c such that, for any x in the interval (a, c), f (x) β‰₯ f (b). Example 7 Finding Increasing and Decreasing Intervals on a Graph Given the function p(t) in Figure 6, identify the intervals on which the function appears to be increasing. p 5 4 3 2 1 –1 1 2 3 4 5 6 t –1 –2 Figure 6 Solution We see that the function is not constant on any interval. The function is increasing where it slants upward as we move to the right and decreasing where it slants downward as we move to the right. The function appears to be increasing from t = 1 to t = 3 and from t = 4 on. In interval notation, we would say the function appears to be increasing on the interval (1, 3) and the interval (4, ∞). Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 202 CHAPTER 3
FUNCTIONS Analysis Notice in this example that we used open intervals (intervals that do not include the endpoints), because the function is neither increasing nor decreasing at t = 1, t = 3, and t = 4. These points are the local extrema (two minima and a maximum). Example 8 Finding Local Extrema from a Graph x 2 __ __ + Graph the function f (x) =. Then use the graph to estimate the local extrema of the function and to determine 3 x the intervals on which the function is increasing. Solution Using technology, we find that the graph of the function looks like that in Figure 7. It appears there is a low point, or local minimum, between x = 2 and x = 3, and a mirror-image high point, or local maximum, somewhere between x = βˆ’3 and x = βˆ’2. f(x) –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 x Figure 7 Analysis Most graphing calculators and graphing utilities can estimate the location of maxima and minima. Figure 8 provides screen images from two different technologies, showing the estimate for the local maximum and minimum. y 6 4 2 0 –2 2.4494898, 1.6329932 2 4 6 x Maximum X = –2.449491 Y = –1.632993 (a) (b) Figure 8 Based on these estimates, the function is increasing on the interval (βˆ’βˆž, βˆ’2.449) and (2.449, ∞). Notice that, while we expect the extrema to be symmetric, the two different technologies agree only up to four decimals due to the differing β€” approximation algorithms used by each. (The exact location of the extrema is at Β± √ 6, but determining this requires calculus.) Try It #4 Graph the function f (x) = x3 βˆ’ 6x2 βˆ’ 15x + 20 to estimate the local extrema of the function. Use these to determine the intervals on which the function is increasing and decreasing. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.3 RATES OF CHANGE AND BEHAVIOR OF GRAPHS 203 Example 9 Finding Local Maxima and Minima from a Graph For the function f whose graph is shown in Figure 9, find all local max
ima and minima. y 10 8 6 4 2 –1 –2 –4 –6 –8 –10 –5 –4 –3 –2 21 3 4 5 x f Figure 9 Solution Observe the graph of f. The graph attains a local maximum at x = 1 because it is the highest point in an open interval around x = 1. The local maximum is the y-coordinate at x = 1, which is 2. The graph attains a local minimum at x = βˆ’1 because it is the lowest point in an open interval around x = βˆ’1. The local minimum is the y-coordinate at x = βˆ’1, which is βˆ’2. Analyzing the Toolkit Functions for Increasing or Decreasing Intervals We will now return to our toolkit functions and discuss their graphical behavior in Figure 10, Figure 11, and Figure 12. Function Increasing/Decreasing Example Constant Function f(x) = c Neither increasing nor decreasing Identity Function f(x) = x Increasing Quadratic Function f(x) = x2 Increasing on (0, ∞) Decreasing on (βˆ’βˆž, 0) Minimum at x = 0 Figure 10 y y y x x x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 204 CHAPTER 3 FUNCTIONS Function Increasing/Decreasing Example Cubic Function f(x) = x3 Increasing Reciprocal f(x) = 1 __ x Decreasing (βˆ’βˆž, 0) βˆͺ (0, ∞) y y y Reciprocal Squared f(x) = 1 _ x2 Increasing on (βˆ’βˆž, 0) Decreasing on (0, ∞) Figure 11 Function Increasing/Decreasing Example y Cube Root β€” x f(x) = 3 √ Increasing y y Square Root f(x) = √ β€” x Increasing on (0, ∞) Absolute Value f(x) = ∣ x ∣ Increasing on (0, ∞) Decreasing on (βˆ’βˆž, 0) Figure 12 x x x x x x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.3 RATES OF CHANGE AND BEHAVIOR OF GRAPHS 205 Use A Graph to Locate the Absolute Maximum and Absolute Minimum There is a difference between locating the highest and lowest points on a graph in a region around an open interval (
locally) and locating the highest and lowest points on the graph for the entire domain. The y-coordinates (output) at the highest and lowest points are called the absolute maximum and absolute minimum, respectively. To locate absolute maxima and minima from a graph, we need to observe the graph to determine where the graph attains it highest and lowest points on the domain of the function. See Figure 13. y 5 4 3 2 1 f Absolute maximum is f (2) = 2 –5 –4 –3 –2 –1 –1 –2 –3 –4 –5 21 3 4 5 x Absolute minimum is f (0) = βˆ’2 Figure 13 Not every function has an absolute maximum or minimum value. The toolkit function f (x) = x3 is one such function. absolute maxima and minima The absolute maximum of f at x = c is f (c) where f (c) β‰₯ f (x) for all x in the domain of f. The absolute minimum of f at x = d is f (d) where f (d) ≀ f (x) for all x in the domain of f. Example 10 Finding Absolute Maxima and Minima from a Graph For the function f shown in Figure 14, find all absolute maxima and minima. y 20 16 12 8 4 –1 –4 –8 –12 –16 –20 –5 –4 –3 –2 f 21 3 4 5 x Figure 14 Solution Observe the graph of f. The graph attains an absolute maximum in two locations, x = βˆ’2 and x = 2, because at these locations, the graph attains its highest point on the domain of the function. The absolute maximum is the y-coordinate at x = βˆ’2 and x = 2, which is 16. The graph attains an absolute minimum at x = 3, because it is the lowest point on the domain of the function’s graph. The absolute minimum is the y-coordinate at x = 3, which is βˆ’10. Access this online resource for additional instruction and practice with rates of change. β€’ Average Rate of Change (http://openstaxcollege.org/l/aroc) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 206 CHAPTER 3 FUNCTIONS 3.3 SECTION EXERCISES VERBAL 1. 3. d diff 2. fab bc fac 4. y=x AL
GEBRAIC fifi bh 5. f x=x 6. hx=βˆ’x βˆ’ 8. g x=x βˆ’βˆ’ 11. __ βˆ’ k t=t+ t 9. y= x 7. 10. qx=xβˆ’ + t βˆ’ t t + pt= βˆ’ 12. f x=x βˆ’b 13. g x=x βˆ’ 14. px=x++h 15. kx=xβˆ’+h 16. f x=x +xx+h 17. gx=xβˆ’xx+h 18. at= +h t+ 19. bx= +h x+ 20. jx=x +h 21. rt=t +h 22. =+ 23. =+ + 26. f(x=xβˆ’xx, x+h 29. = + 24. 27. = = , + + 25. 28. = = + + + 30. = + 31. = + + = = GRAPHICAL fFigure 15 32. x=x= 33. x=x= y – x Figure 15 6 6 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.3 SECTION EXERCISES 207 y – – – – – – – – – – 35. x – – – – y – – – – – – 36 34. 37. 38 – – – – – – – – – – – – Figure 16 – – – 39. 40. f Figure 17. 41. 42 Figure 16 NUMERIC Figure 17 43. Table 3 ab Year Sales (millions of dollars) Table 3 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 208 CHAPTER 3 FUNCTIONS 44. Table 4 ab Year Population (thousands) Table 4 TECHNOLOGY 45. f x=x βˆ’x + 47. 49. β€” t+ gt=t √ mx=x +x βˆ’x βˆ’x+ EXTENSION 51. ThfFigure 18 46. 48. 50. hx=x +x +x +x βˆ’ kt=t βˆ’t nx=x βˆ’x +x βˆ’x+ a. b. c. d. = = Figure 18 52. f x= c x f c βˆ’ " __ 53. f(x)= "
__.b x fb __ βˆ’ REAL-WORLD APPLICATIONS 54. 56. dt=ttdt t=t= 55. o fi 57. ThFigure 19 Time (days) Figure 19 t t=t= Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.4 COMPOSITION OF FUNCTIONS 209 LEARNING OBJECTIVES In this section, you will: β€’ Combine functions using algebraic operations and find their domains. β€’ Create a new function by composition of functions and find its domain. β€’ aluate composite functions. β€’ ecompose a composite function into its component functions. 3.4 COMPOSITION OF FUNCTIONS Suppose we want to calculate how much it costs to heat a house on a particular day of the year. The cost to heat a house will depend on the average daily temperature, and in turn, the average daily temperature depends on the particular day of the year. Notice how we have just defined two relationships: The cost depends on the temperature, and the temperature depends on the day. Using descriptive variables, we can notate these two functions. The function C(T) gives the cost C of heating a house for a given average daily temperature in T degrees Celsius. The function T(d) gives the average daily temperature on day d of the year. For any given day, Cost = C(T(d)) means that the cost depends on the temperature, which in turns depends on the day of the year. Thus, we can evaluate the cost function at the temperature T(d). For example, we could evaluate T(5) to determine the average daily temperature on the 5th day of the year. Then, we could evaluate the cost function at that temperature. We would write C(T(5)). Cost for the temperature C(T(5)) Temperature on day 5 By combining these two relationships into one function, we have performed function composition, which is the focus of this section. Combining Functions Using Algebraic Operations Function composition is only one way to combine existing functions. Another way is to carry out the usual algebraic operations on functions, such as addition, subtraction, multiplication and division. We do this by performing the operations with the function outputs, defining the result as the output of our new function. Suppose we need to add two columns of numbers that represent a husband and wife’s separate annual incomes over a period of years, with the result being their