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dollars, to produce x bottles of 100% All-Natural Certified Free-Trade Organic Sasquatch Tonic is C(x) = 10x + 100, x β‰₯ 0 and the price-demand function, in dollars per bottle, is p(x) = 35 βˆ’ x, 0 ≀ x ≀ 35. 12. The cost, in cents, to produce x cups of Mountain Thunder Lemonade at Junior’s Lemonade Stand is C(x) = 18x + ...
0 ≀ t ≀ 12 What is the warmest temperature of the day? When does this happen? 18. Suppose C(x) = x2 βˆ’ 10x + 27 represents the costs, in hundreds, to produce x thousand pens. How many pens should be produced to minimize the cost? What is this minimum cost? 19. Skippy wishes to plant a vegetable garden along one side of...
no air resistance or forces other than the Earth’s gravity, the height above the ground at time t of a falling object is given by s(t) = βˆ’4.9t2 + v0t + s0 where s is in meters, t is in seconds, v0 is the object’s initial velocity in meters per second and s0 is its initial position in meters. (a) What is the applied do...
. x2 βˆ’ 10y2 = 0 for x 32. y2 βˆ’ 4y = x2 βˆ’ 4 for x 33. x2 βˆ’ mx = 1 for x 34. y2 βˆ’ 3y = 4x for y 35. y2 βˆ’ 4y = x2 βˆ’ 4 for y 36. βˆ’gt2 + v0t + s0 = 0 for t (Assume g = 0.) 11The weight of the bridge deck forces the bridge cable into a parabola and a free hanging cable such as a power line does not form a parabola. We shall ...
οΏ½, βˆ’1] Decreasing on [βˆ’1, ∞) Vertex (βˆ’1, 4) is a maximum Axis of symmetry x = βˆ’1 y 10 2 βˆ’1 1 2 x βˆ’4 βˆ’3 βˆ’2 βˆ’1 βˆ’1 y x βˆ’2 βˆ’3 βˆ’4 βˆ’5 βˆ’6 βˆ’7 βˆ’8 y 2 1 βˆ’2 βˆ’1 βˆ’1 1 2 3 4 x βˆ’2 βˆ’3 βˆ’4 βˆ’5 βˆ’6 βˆ’7 βˆ’8 βˆ’9 y 4 3 2 1 βˆ’1 βˆ’2 βˆ’3 βˆ’4 1 x βˆ’3 βˆ’2 βˆ’1 204 Linear and Quadratic Functions 5. f (x) = 2x2 βˆ’ 4x βˆ’ 1 = 2(x βˆ’ 1)2 βˆ’ 3 √ 6 √ 6 2+ 2, 0 2βˆ’ 2, 0 ...
: (βˆ’βˆž, ∞) Range: βˆ’βˆž, 73 12 Increasing on βˆ’βˆž, 5 6 Decreasing on 5 6, ∞ Vertex 5 6, 73 12 Axis of symmetry x = 5 6 is a maximum 100 x βˆ’ 1 = x βˆ’ 1 200 1βˆ’ 1+ 40001 2 βˆ’ 40001 40000 √ 40001 and 200 √ 9. f (x) = x2 βˆ’ 1 x-intercepts 200 y-intercept (0, βˆ’1) Domain: (βˆ’βˆž, ∞) Range: βˆ’ 40001 40000, ∞ Decreasing on βˆ’βˆž, 1 200 200, ∞ ...
12 cups of lemonade need to be made and sold to maximize profit. The maximum profit is 192Β’ or $1.92. The price per cup should be set at 54Β’ per cup to maximize profit. The break even points are x = 4 and x = 20, so to make a profit, between 4 and 20 cups of lemonade need to be made and sold. P (x) = βˆ’0.5x2 + 9x βˆ’ 36, for...
hammer reaches a maximum height of approximately 13.62 feet. The hammer is in the air approximately 1.61 seconds. 2.3 Quadratic Functions 207 25. (a) The applied domain is [0, ∞). (d) The height function is this case is s(t) = βˆ’4.9t2 + 15t. The vertex of this parabola is approximately (1.53, 11.48) so the maximum heig...
with Absolute Value and Quadratic Functions In this section, not only do we develop techniques for solving various classes of inequalities analytically, we also look at them graphically. The first example motivates the core ideas. Example 2.4.1. Let f (x) = 2x βˆ’ 1 and g(x) = 5. 1. Solve f (x) = g(x). 2. Solve f (x) < g...
is the solution to f (x) = g(x). Sure enough, f (3) = 5 and g(3) = 5 so that the point (3, 5) is on both graphs. In other words, the graphs of f and g intersect at (3, 5). In part 2, we set f (x) < g(x) and solved to find x < 3. For x < 3, the point (x, f (x)) is below (x, g(x)) since the y values on the graph of f are...
x) 1. Solve f (x) = g(x). 2. Solve f (x) < g(x). 3. Solve f (x) β‰₯ g(x). Solution. 1. To solve f (x) = g(x), we look for where the graphs of f and g intersect. These appear to be at the points (βˆ’1, 2) and (1, 2), so our solutions to f (x) = g(x) are x = βˆ’1 and x = 1. 2. To solve f (x) < g(x), we look for where the graph...
no solution. For c β‰₯ 0, |x| > c is equivalent to x < βˆ’c or x > c. For c β‰₯ 0, |x| β‰₯ c is equivalent to x ≀ βˆ’c or x β‰₯ c. For c < 0, |x| > c and |x| β‰₯ c are true for all real numbers. As with Theorem 2.1 in Section 2.2, we could argue Theorem 2.4 using cases. However, in light of what we have developed in this section, w...
= 4, so we have graphical confirmation of our analytic solution. 2. To solve 4 βˆ’ 3|2x + 1| > βˆ’2 analytically, we first isolate the absolute value before applying Theorem 2.4. To that end, we get βˆ’3|2x + 1| > βˆ’6 or |2x + 1| < 2. Rewriting, we now have. Graphically we βˆ’2 < 2x + 1 < 2 so that βˆ’ 3 2 and 1 see that the graph...
Section 1.1.1. 2.4 Inequalities with Absolute Value and Quadratic Functions 213 x βˆ’ 1| y = 5 y = 2 βˆ’8 βˆ’7 βˆ’6 βˆ’5 βˆ’4 βˆ’3 βˆ’2 βˆ’. We need to exercise some special caution when solving |x + 1| β‰₯ x+4 2. As we saw in Example 2.2.1 in Section 2.2, when variables are both inside and outside of the absolute value, it’s usually bes...
Quadratic Functions the values of x for which the graph of y = f (x) = x2 βˆ’ x βˆ’ 6 (the parabola) is below the graph of y = g(x) = 0 (the x-axis). We’ve provided the graph again for reference. y 6 5 4 3 2 1 βˆ’3βˆ’2βˆ’1 βˆ’1 1 2 3 x βˆ’2 βˆ’3 βˆ’4 βˆ’5 βˆ’6 y = x2 βˆ’ x βˆ’ 6 We can see that the graph of f does dip below the x-axis between ...
ratic Functions 215 Steps for Solving a Quadratic Inequality 1. Rewrite the inequality, if necessary, as a quadratic function f (x) on one side of the in- equality and 0 on the other. 2. Find the zeros of f and place them on the number line with the number 0 above them. 3. Choose a real number, called a test value, in ...
1. We know 2x2 + x βˆ’ 3 = 0 when x = βˆ’ 3 ; f (0) = βˆ’3 < 0, so (βˆ’) goes above the interval βˆ’ 3, we choose3 x = βˆ’2; for βˆ’ 3 2 and x = 1, so our final answer is βˆ’ 3, βˆ’ 3 2, 1. To verify our solution graphically, we refer to the original inequality, 2x2 ≀ 3 βˆ’ x. We let g(x) = 2x2 and h(x) = 3 βˆ’ x. We are looking for the x v...
x2 βˆ’ 2x > 1 graphically, we set g(x) = x2 βˆ’ 2x and h(x) = 1. We are looking for the x values where the graph of g is above the graph of h. As before we present the graphs on the right and the sign chart on the left. 2 β‰ˆ βˆ’0.4 and 1 + 2 βˆͺ 1 + √ √ √ √ (+) βˆ’1 0 √ 1 βˆ’ (βˆ’) 2 0 0 √ 1 + (+) 3 βˆ’2 βˆ’1 1 2 3 x y = x2 βˆ’ 2x 3. To s...
y 4 3 2 1 y = x2 + 1 y = 2x βˆ’1 1 x 4. To solve our last inequality, 2x βˆ’ x2 β‰₯ |x βˆ’ 1| βˆ’ 1, we re-write the absolute value using cases. For x < 1, |x βˆ’ 1| = βˆ’(x βˆ’ 1) = 1 βˆ’ x, so we get 2x βˆ’ x2 β‰₯ 1 βˆ’ x βˆ’ 1, or x2 βˆ’ 3x ≀ 0. Finding the zeros of f (x) = x2 βˆ’ 3x, we get x = 0 and x = 3. However, we are only concerned with ...
where the graph of h is above the the graph of i (the solution of h(x) > i(x)) as well as the x-coordinates of the intersection points of both graphs (where h(x) = i(x)). The combined sign chart is given on the left and the graphs are on the right. (+) 0 (βˆ’) 0 (+) 0 2 βˆ’1 0 3 y y = 2x βˆ’ x2 1 βˆ’1 1 2 3 x y = |x βˆ’ 1| βˆ’ 1 ...
οΏ½ a ≀ b. To use this property, we proceed as follows βˆ’0.25 ≀ x2 βˆ’ 576 ≀ 0.25 x2 575.75 ≀ √ x2 575.75 ≀ |x| 575.75 ≀ ≀ 576.25 √ ≀ √ ≀ √ √ √ (add 576 across the inequalities.) (take square roots.) √ x2 = |x|) ( 575.75 βˆͺ √ √ √ 575.75 ≀ |x| ≀ 576.25 576.25 575.75 ≀ |x| to be βˆ’βˆž, βˆ’ 576.25. To solve 576.25, √ √ √ 575.75] βˆͺ [...
the points on the graph. Dotting the graph of y = |x| as we have done before to indicate that the points on the graph itself are not in the relation, we get the shaded region below on the left. 2. For a point to be in S, its y-coordinate must be less than or equal to the y-coordinate on the parabola y = 2 βˆ’ x2. This i...
βˆ’ 1 > 0 27. 2 ≀ |x2 βˆ’ 9| < 9 29. x2 + x + 1 β‰₯ 0 31. x|x + 5| β‰₯ βˆ’6 12. x + 5 < |x + 5| 14. |2x + 1| ≀ 6 βˆ’ x 16. |3 βˆ’ x| β‰₯ x βˆ’ 5 18. 16x2 + 8x + 1 > 0 20. 9x2 + 16 β‰₯ 24x 22. x2 + 1 < 0 24. x > x2 26. 5x + 4 ≀ 3x2 28. x2 ≀ |4x βˆ’ 3| 30. x2 β‰₯ |x| 32. x|x βˆ’ 3| < 2 33. The profit, in dollars, made by selling x bottles of 100%...
οΏ½F? (Refer to Exercise 35 in Section 2.1 for assistance if needed.) In Exercises 39 - 42, write and solve an inequality involving absolute values for the given statement. 39. Find all real numbers x so that x is within 4 units of 2. 40. Find all real numbers x so that 3x is within 2 units of βˆ’1. 41. Find all real numbe...
Solution. 15. 1, 5 3 17. (βˆ’βˆž, βˆ’3] βˆͺ [1, ∞) 19. No solution 21. {2} 23. βˆ’ 1 3, 4 25. βˆ’βˆž,, ∞ 27. √ √ 2, βˆ’ 11 βˆͺ √ βˆ’ 7, 0 √ 7 0, βˆͺ βˆͺ √ √ 11, 3 2 βˆ’3 2. βˆ’βˆž, βˆ’ 12 7 βˆͺ 8 7, ∞ 4. (βˆ’βˆž, 1] βˆͺ [3, ∞) 6. (βˆ’βˆž, ∞) 8. [3, 4) βˆͺ (5, 6] 10. (βˆ’βˆž, βˆ’4) βˆͺ 2 3, ∞ 12. (βˆ’βˆž, βˆ’5) 14. βˆ’7, 5 3 16. (βˆ’βˆž, ∞) 18. βˆ’βˆž, βˆ’ 1 4 βˆͺ βˆ’ 1 4, ∞ 20. (βˆ’βˆž, ∞) 22. No...
22 AM to 6 PM. 2.4 Inequalities with Absolute Value and Quadratic Functions 223 36. h(t) β‰₯ 250 on [10 βˆ’ 5 √ √ 2, 10 + 5 2] β‰ˆ [2.93, 17.07]. This means the rocket is at least 250 feet off the ground between 2.93 and 17.07 seconds after lift off. 37. s(t) = βˆ’4.9t2 + 30t + 2. s(t) > 35 on (approximately) (1.44, 4.68). This ...
order to generate linear and quadratic models. Our goal is to give the reader an understanding of the basic processes involved, but we are quick to refer the reader to a more advanced course1 for a complete exposition of this material. Suppose we collected three data points: {(1, 2), (3, 1), (4, 3)}. By plotting these...
and perform the Linear Regression feature and we get = 9 1and authors with more expertise in this area, 2Like Calculus and Linear Algebra 226 Linear and Quadratic Functions The calculator tells us that the line of best fit is y = ax + b where the slope is a β‰ˆ 0.214 and the y-coordinate of the y-intercept is b β‰ˆ 1.428. ...
can tell both from the correlation coefficient as well as the graph that the regression line is a good fit to the data. 3. The slope of the regression line is a β‰ˆ 1.287. To interpret this, recall that the slope is the rate of change of the y-coordinates with respect to the x-coordinates. Since the y-coordinates represent...
of best fit in the same way we found a line of best fit. The process is called quadratic regression and its goal is to minimize the least square error of the data with their corresponding points on the parabola. The calculator has a built in feature for this as well which yields 2.5 Regression 229 The coefficient of deter...
data for Lorain County, Ohio is: Year Population 1970 256843 1980 274909 1990 271126 2000 284664 (a) Find the least squares regression line for these data and comment on the goodness of fit. Interpret the slope of the line of best fit. (b) Use the regression line to predict the population of Lorain County in 2010. (The ...
64 53.57 62.62 71.93 81.69 90.43 41 356 731 1051 1347 1631 1966 2310 2670 3030 3371 Find the least squares line for this data. Is it a good fit? What does the slope of the line represent? Do you and your classmates believe this model would have held for ten years had I not crashed the car on the Turnpike a few years ago...
and com- ment on its goodness of fit. (d) Compare and contrast the predictions the three models make for my weight on January 1, 2010 (Day #366). Can any of these models be used to make a prediction of my weight 20 years from now? Explain your answer. 232 Linear and Quadratic Functions (e) Why is this a Civics lesson i...
this exercise I used the sunrise/sunset times in Fairbanks, Alaska for 2009 to give you a chart of the number of hours of daylight they get on the 21st of each month. We’ll let x = 1 represent January 21, 2009, x = 2 represent February 21, 2009, and so on. Month Number Hours of Daylight 1 2 3 4 5 6 7 8 9 10 11 12 5.8 ...
=.9997 so this is an excellent fit to the data. The slope 36.8 represents miles per gallon. 5. (a) The line for the Thursday data is y = βˆ’.12x + 237.69. We have r = βˆ’.9568 and r2 =.9155 so this is a really good fit. (b) The line for the Saturday data is y = βˆ’0.000693x + 235.94. We have r = βˆ’0.008986 and r2 = 0.0000807 w...
when x = 13. So January 21, 2010 will be β€œextra dark”? Obviously a parabola pointing down isn’t telling us the whole story. Chapter 3 Polynomial Functions 3.1 Graphs of Polynomials Three of the families of functions studied thus far – constant, linear and quadratic – belong to a much larger group of functions called p...
2 + 4 5. h(x) = |x| 6. z(x) = 0 1Enjoy this while it lasts. Before we’re through with the book, you’ll have been exposed to the most terrible of algebraic beasts. We will tame them all, in time. 236 Solution. Polynomial Functions 1. We note directly that the domain of g(x) = x3+4 x is x = 0. By definition, a polynomial ...
goodness polynomial. Definition 3.2. Suppose f is a polynomial function. Given f (x) = anxn + anβˆ’1xnβˆ’1 +... + a2x2 + a1x + a0 with an = 0, we say – The natural number n is called the degree of the polynomial f. – The term anxn is called the leading term of the polynomial f. – The real number an is called the leading coe...
.1.2. Find the degree, leading term, leading coefficient and constant term of the following polynomial functions. 1. f (x) = 4x5 βˆ’ 3x2 + 2x βˆ’ 5 2. g(x) = 12x + x3 3. h(x) = 4 βˆ’ x 5 Solution. 4. p(x) = (2x βˆ’ 1)3(x βˆ’ 2)(3x + 2) 1. There are no surprises with f (x) = 4x5 βˆ’ 3x2 + 2x βˆ’ 5. It is written in the form of Definitio...
is 5 and the leading coefficient is 24. As for the constant term, we can perform a similar trick. The constant term is obtained by multiplying the constant terms from each of the factors (βˆ’1)3(βˆ’2)(2) = 4. Our next example shows how polynomials of higher degree arise β€˜naturally’4 in even the most basic geometric applicat...
x > 0 or x < 5. Hence, our applied domain is 0 < x < 5. 2. Using a graphing calculator, we see that the graph of y = V (x) has a relative maximum. For 0 < x < 5, this is also the absolute maximum. Using the β€˜Maximum’ feature of the calculator, we get x β‰ˆ 1.81, y β‰ˆ 96.77. This yields a height of x β‰ˆ 1.81 inches, a width...
(written x β†’ βˆ’βˆž) and, on the flip side, as x becomes large without bound11 (written x β†’ ∞). For example, given f (x) = x2, as x β†’ βˆ’βˆž, we imagine substituting x = βˆ’100, x = βˆ’1000, etc., into f to get f (βˆ’100) = 10000, f (βˆ’1000) = 1000000, and so on. Thus the function values are becoming larger and larger positive number...
case when n = 1, since the graph of f (x) = x is a line and doesn’t fit the general pattern of higher-degree odd polynomials.) Below we have graphed y = x3, y = x5, and y = x7. The β€˜ο¬‚attening’ and β€˜steepening’ that we saw with the even powers presents itself here as well, and, it should come as no surprise that all of ...
the left fails to be continuous where it has a β€˜break’ or β€˜hole’ in the graph; everywhere else, the function is continuous. The function is continuous at the β€˜corner’ and the β€˜cusp’, but we consider these β€˜sharp turns’, so these are places where the function fails to be smooth. Apart from these four places, the functi...
Polynomial Functions Example 3.1.4. Use the Intermediate Value Theorem to establish that Solution. Consider the polynomial function f (x) = x2 βˆ’ 2. Then f (1) = βˆ’1 and f (3) = 7. Since f (1) and f (3) have different signs, the Intermediate Value Theorem guarantees us a real number c between 1 and 3 with f (c) = 0. If c...
is (+), its graph is above the x-axis; wherever f is (βˆ’), its graph is below the x-axis. The x-intercepts of the graph of f are (βˆ’2, 0), (0, 0) and (3, 0). Knowing f is smooth and continuous allows us to sketch its graph. y (+) 0 (βˆ’) βˆ’2 0 (+) 0 (+) 0 3 βˆ’3 βˆ’1 1 4 x A sketch of y = f (x) A couple of notes about the Exam...
indicates. 4x3 become closer and closer to x βˆ’1000 βˆ’100 βˆ’10 10 100 1000 1 4x2 5 4x3 0.00000025 βˆ’0.00000000125 βˆ’0.00000125 βˆ’0.00125 0.00125 0.00000125 0.00000000125 0.000025 0.0025 0.0025 0.000025 0.00000025 In other words, as x β†’ ±∞, f (x) β‰ˆ 4x3 (1 βˆ’ 0 + 0) = 4x3, which is the leading term of f. The formal proof of Th...
points, as it does at x = βˆ’2 and x = 0, or does it simply touch and rebound like it does at x = 3. From the sign diagram, the graph of f will cross the x-axis whenever the signs on either side of the zero switch (like they do at x = βˆ’2 and x = 0); it will touch when the signs are the same on either side of the zero (a...
factor the same on either side of 3. If we look back at the exponents on the factors (x + 2) and x3, we see that they are both odd, so as we substitute values to the left and right of the corresponding zeros, the signs of the corresponding factors change which results in the sign of the function value changing. This i...
ish below the x-axis, in Quadrant IV. Next, we find the zeros of f. Fortunately for us, f is factored.15 Setting each factor equal to zero gives is x = 1 2 we note that it corresponds to the factor (2x βˆ’ 1). This isn’t strictly in the form required in Definition 3.3. If we factor out the 2, however, we get (2x βˆ’ 1) = 2 x...
)(t2 + t + 4) 9. f (x) = βˆ’2x3(x + 1)(x + 2)2 10. G(t) = 4(t βˆ’ 2)2 t + 1 2 In Exercises 11 - 20, find the real zeros of the given polynomial and their corresponding multiplicities. Use this information along with a sign chart to provide a rough sketch of the graph of the polynomial. Compare your answer with the result fr...
of the following intervals: [βˆ’4, βˆ’3], [0, 1] and [2, 3]. 28. Rework Example 3.1.3 assuming the box is to be made from an 8.5 inch by 11 inch sheet of paper. Using scissors and tape, construct the box. Are you surprised?16 16Consider decorating the box and presenting it to your instructor. If done well enough, maybe yo...
x as drawn below. The length is the longest side and is clearly labeled. The girth is the distance around the box in the other two dimensions so in our case it is the sum of the four sides of the square, 4x. (a) Assuming that we’ll be mailing a box via Parcel Post where Length + Girth = 130 inches, express the length ...
W ) 1.012 1.063 2.199 1.496 3.275 1.610 4.676 1.613 6.805 1.505 9.975 1.314 (a) Make a scatter diagram of the data using the Resistance as the independent variable and Power as the dependent variable. (b) Use your calculator to find quadratic (2nd degree), cubic (3rd degree) and quartic (4th degree) regression models fo...
1 and one of higher odd multiplicity. Look again closely at the graphs of a(x) = x(x + 2)2 and F (x) = x3(x + 2)2 from Exercise 3.1.1. Discuss with your classmates how the graphs are fundamentally different at the origin. It might help to use a graphing calculator to zoom in on the origin to see the different crossing b...
r) β†’ βˆ’βˆž As r β†’ ∞, q(r) β†’ βˆ’βˆž √ 5. f (x) = 3x17 + 22.5x10 βˆ’ Ο€x7 + 1 3 √ 3x17 √ Degree 17 Leading term Leading coefficient Constant term 1 3 As x β†’ βˆ’βˆž, f (x) β†’ βˆ’βˆž As x β†’ ∞, f (x) β†’ ∞ 3 Polynomial Functions 2. g(x) = 3x5 βˆ’ 2x2 + x + 1 Degree 5 Leading term 3x5 Leading coefficient 3 Constant term 1 As x β†’ βˆ’βˆž, g(x) β†’ βˆ’βˆž As x β†’ ∞...
β†’ βˆ’βˆž 10. G(t) = 4(t βˆ’ 2)2 t + 1 2 Degree 3 Leading term 4t3 Leading coefficient 4 Constant term 8 As t β†’ βˆ’βˆž, G(t) β†’ βˆ’βˆž As t β†’ ∞, G(t) β†’ ∞ 11. a(x) = x(x + 2)2 x = 0 multiplicity 1 x = βˆ’2 multiplicity 2 12. g(x) = x(x + 2)3 x = 0 multiplicity 1 x = βˆ’2 multiplicity 3 y y βˆ’2 βˆ’1 x βˆ’2 βˆ’1 x 13. f (x) = βˆ’2(x βˆ’ 2)2(x + 1) x = 2...
3 + 1 domain: (βˆ’βˆž, ∞) range: (βˆ’βˆž, ∞) 22. g(x) = (x + 2)4 + 1 domain: (βˆ’βˆž, ∞) range: [1, ∞) y x 12 11 10 1 βˆ’2 βˆ’3 βˆ’4 βˆ’5 βˆ’6 βˆ’7 βˆ’8 βˆ’9 βˆ’10 βˆ’4 βˆ’3 βˆ’2 βˆ’1 23. g(x) = 2 βˆ’ 3(x βˆ’ 1)4 domain: (βˆ’βˆž, ∞) range: (βˆ’βˆž, 2] y 1 2 x 2 1 βˆ’1 βˆ’2 βˆ’3 βˆ’4 βˆ’5 βˆ’6 βˆ’7 βˆ’8 βˆ’9 βˆ’10 βˆ’11 βˆ’12 βˆ’13 y 21 20 19 18 17 16 15 14 13 12 11 10 4 βˆ’3 βˆ’2 βˆ’1 x 24. g(x) = βˆ’...
34 inches, and depth β‰ˆ 7.84 inches. The maximum volume is β‰ˆ 66.15 cubic inches. 29. The calculator gives the location of the absolute maximum (rounded to three decimal places) as x β‰ˆ 6.305 and y β‰ˆ 1115.417. Since x represents the number of TVs sold in hundreds, x = 6.305 corresponds to 630.5 TVs. Since we can’t sell ha...
volume of 20342.59in.3. (c) If we start with Length + Girth = 108 then the length is 108 βˆ’ 4x and the volume is V (x) = βˆ’4x3 + 108x2. Graphing y = V (x) on [0, 27] Γ— [0, 11700] shows a maximum at (18.00, 11664.00) so the dimensions of the box with maximum volume are 18.00in. Γ— 18.00in. Γ— 36in. for a volume of 11664.00...
(x = 12). The quartic regression model is p4(x) = 0.0144x4 βˆ’0.3507x3 +2.259x2 βˆ’1.571x+5.513. It has R2 = 0.98594 which is good. The graph of y = p4(x) in the viewing window [βˆ’1, 15] Γ— [0, 35] along with the scatter plot is shown below on the right. Notice that p4(15) is above 24 making this a bad model as well for fut...
x) = 0 results in the polynomial equation x3 + 4x2 βˆ’ 5x βˆ’ 14 = 0. Despite all of the factoring techniques we learned1 in Intermediate Algebra, this equation foils2 us at every turn. If we graph f using the graphing calculator, we get The graph suggests that the function has three zeros, one of which is x = 2. It’s easy...
x) and p(x) are nonzero polynomials where the degree of p is greater than or equal to the degree of d. There exist two unique polynomials, q(x) and r(x), such that p(x) = d(x) q(x) + r(x), where either r(x) = 0 or the degree of r is strictly less than the degree of d. As you may recall, all of the polynomials in Theore...
x) for some polynomial q. Hence, p(c) = (c βˆ’ c) q(c) = 0, so c is a zero of p. Conversely, if c is a zero of p, then p(c) = 0. In this case, The Remainder Theorem tells us the remainder when p(x) is divided by (x βˆ’ c), namely p(c), is 0, which means (x βˆ’ c) is a factor of p. What we have established is the fundamental ...
5x βˆ’14 2x2 12x 14 6x2 0 7x Note that by arranging things in this manner, each term in the last row is obtained by adding the two terms above it. Notice also that the quotient polynomial can be obtained by dividing each of the first three terms in the last row by x and adding the results. If you take the time to work ba...
, take the 2 from the divisor and multiply by the 1 that was β€˜brought down’ to get 2. Write this underneath the 4, then add to get 6. 2 1 4 βˆ’5 βˆ’14 ↓ 2 1 2 1 4 βˆ’5 βˆ’14 ↓ 2 1 6 Now take the 2 from the divisor times the 6 to get 12, and add it to the βˆ’5 to get 7. 2 1 4 βˆ’5 βˆ’14 ↓ 2 12 1 6 2 1 4 βˆ’5 βˆ’14 ↓ 2 12 7 1 6 Finally, t...
the synthetic division tableau, we need to enter 0 for the coefficient of x in the dividend. Doing so gives 3 5 βˆ’2 ↓ 5 0 15 39 13 39 1 117 118 Since the dividend was a third degree polynomial, the quotient is a quadratic polynomial with coefficients 5, 13 and 39. Our quotient is q(x) = 5x2 + 13x + 39 and the remainder is ...
polynomial is βˆ’6x βˆ’ 13 and the remainder is βˆ’35? The answer is the word β€˜unique’ in Theorem 3.4. The theorem states that there is only one way to decompose βˆ’12x2 βˆ’ 8x + 4 into a multiple of (2x βˆ’ 3) plus a constant term. Since we have found such a way, we can be sure it is the only way. (βˆ’6x βˆ’ 13) βˆ’ 35 The next exampl...
. Since this doesn’t factor nicely, we use the quadratic formula to find that the remaining zeros are x = βˆ’1Β±. 2 √ 7 3.2 The Factor Theorem and the Remainder Theorem 263 In Section 3.1, we discussed the notion of the multiplicity of a zero. Roughly speaking, a zero with multiplicity 2 can be divided twice into a polynom...
guarantees x βˆ’ 3 and βˆ’ 12 βˆ’3 3 0 βˆ’4 βˆ’11 βˆ’2 βˆ’12 2 βˆ’1 βˆ’6 6 0 βˆ’4 4 2 0 βˆ’12 12 3 0 3 3 0 This gives us 4x4 βˆ’ 4x3 βˆ’ 11x2 + 12x βˆ’ 3 = x βˆ’ 1 2 with the constant in front √ 4), or, when written p(x 264 Polynomial Functions We have shown that p is a product of its leading coefficient times linear factors of the form (x βˆ’ c) wher...
the form p(x) = d(x)q(x) + r(x). 1. 4x2 + 3x βˆ’ 1 Γ· (x βˆ’ 3) 2. 2x3 βˆ’ x + 1 Γ· x2 + x + 1 3. 5x4 βˆ’ 3x3 + 2x2 βˆ’ 1 Γ· x2 + 4 4. βˆ’x5 + 7x3 βˆ’ x Γ· x3 βˆ’ x2 + 1 5. 9x3 + 5 Γ· (2x βˆ’ 3) 6. 4x2 βˆ’ x βˆ’ 23 Γ· x2 βˆ’ 1 In Exercises 7 - 20 use synthetic division to perform the indicated division. Write the polynomial in the form p(x) = d(x)...
3 24. p(x) = x3 + 2x2 + 3x + 4, c = βˆ’1 25. p(x) = 3x3 βˆ’ 6x2 + 4x βˆ’ 8, c = 2 26. p(x) = 8x3 + 12x2 + 6x + 1, c = βˆ’ 1 2 27. p(x) = x4 βˆ’ 2x2 + 4, c = 3 2 29. p(x) = x4 + x3 βˆ’ 6x2 βˆ’ 7x βˆ’ 7, c = βˆ’ √ 28. p(x) = 6x4 βˆ’ x2 + 2, c = βˆ’ 2 3 7 30. p(x) = x2 βˆ’ 4x + 1, c = 2 βˆ’ √ 3 266 Polynomial Functions In Exercises 31 - 40, you ar...
zeros of p are c = 1 and c = 3 c = 3 is a zero of multiplicity 2. The leading term of p(x) is βˆ’5x3 The solutions to p(x) = 0 are x = Β±3 and x = 6 The leading term of p(x) is 7x4 The point (βˆ’3, 0) is a local minimum on the graph of y = p(x). The solutions to p(x) = 0 are x = Β±3, x = βˆ’2, and x = 4. The leading term of p...
+ 3 = (x + 3) (4x βˆ’ 17) + 54 11. x3 + 8 = (x + 2) x2 βˆ’ 2x + 4 + 0 12. 4x3 + 2x βˆ’ 3 = (x βˆ’ 3) 4x2 + 12x + 38 + 111 13. 18x2 βˆ’ 15x βˆ’ 25 = x βˆ’ 5 3 (4x + 2) + 0 14. 4x2 βˆ’ 1 = x βˆ’ 1 2 15. 2x3 + x2 + 2x + 1 = x + 1 2 16. 3x3 βˆ’ x + 4 = x βˆ’ 2 3 17. 2x3 βˆ’ 3x + 1 = x βˆ’ 1 2 18. 4x4 βˆ’ 12x3 + 13x2 βˆ’ 12x + 9 = x βˆ’ 3 2 19. x4 βˆ’ 6x2 ...
οΏ½ 7 31. x3 βˆ’ 6x2 + 11x βˆ’ 6 = (x βˆ’ 1)(x βˆ’ 2)(x βˆ’ 3) 32. x3 βˆ’ 24x2 + 192x βˆ’ 512 = (x βˆ’ 8)3 33. 3x3 + 4x2 βˆ’ x βˆ’ 2 = 3 x βˆ’ 2 3 34. 2x3 βˆ’ 3x2 βˆ’ 11x + 6 = 2 x βˆ’ 1 2 (x + 1)2 (x + 2)(x βˆ’ 3) √ √ 35. x3 + 2x2 βˆ’ 3x βˆ’ 6 = (x + 2)(x + 36. 2x3 βˆ’ x2 βˆ’ 10x + 5 = 2 x βˆ’ 1 2 (x + 3)(x βˆ’ √ 5)(x βˆ’ 3) √ 5) 37. 4x4 βˆ’ 28x3 + 61x2 βˆ’ 42x + 9 =...
of a polynomial function. This section presents results which will help us determine good candidates to test using synthetic division. There are two approaches to the topic of finding the real zeros of a polynomial. The first approach (which is gaining popularity) is to use a little bit of Mathematics followed by a good...
omial, the next theorem gives us a list of possible real zeros. Theorem 3.9. Rational Zeros Theorem: Suppose f (x) = anxn + anβˆ’1xnβˆ’1 +... + a1x + a0 is a polynomial of degree n with n β‰₯ 1, and a0, a1,... an are integers. If r is a rational zero of f, then r is of the form Β± p q, where p is a factor of the constant term...
+... + a1pqnβˆ’1 + a0qn = 0 a0qn = βˆ’anpn βˆ’ anβˆ’1pnβˆ’1q βˆ’... βˆ’ a1pqnβˆ’1 we can play the same game and conclude a0 is a multiple of p, and we have the result. Example 3.3.2. Let f (x) = 2x4 + 4x3 βˆ’ x2 βˆ’ 6x βˆ’ 3. Use the Rational Zeros Theorem to list all of the possible rational zeros of f. Solution. To generate a complete li...
. In Example 3.3.2, we learned that any rational zero of f must be in the list Β± 1 2, Β± 3. From the graph, it looks as if we can rule out any of the positive rational zeros, since the graph seems to cross the x-axis at a value just a little greater than 1. On the negative side, βˆ’1 looks good, so we try that for our syn...
βˆ’1.22 then turns back upwards to touch the xβˆ’axis at x = βˆ’1. This tells us that, despite what the calculator showed us the first time, there is a relative maximum occurring at x = βˆ’1 and not a β€˜ο¬‚attened crossing’ as we originally 272 Polynomial Functions believed. After resizing the window, we see not only the relative...
atic in form’. In other words, we have three terms: x4, x2 and 12, and the exponent on the first term, x4, is exactly twice that of the second term, x2. We may rewrite this as x22 + x2 βˆ’ 12 = 0. To better see the forest for the trees, we momentarily replace x2 with the variable u. In terms of u, our equation becomes u2 ...
) = 2x4 + 4x3 βˆ’ x2 βˆ’ 6x βˆ’ 3 without using the calculator. In this subsection, we present some more advanced mathematical tools (theorems) to help us. Our first result is due to RenΒ΄e Descartes. Theorem 3.10. Descartes’ Rule of Signs: Suppose f (x) is the formula for a polynomial function written with descending powers o...
negative real zeros always starts with the number of sign changes and decreases by an even number. For example, if f (x) has 7 sign changes, then, counting multplicities, f has either 7, 5, 3 or 1 positive real zero. This implies that the graph of y = f (x) crosses the positive x-axis at least once. If f (βˆ’x) results ...
NOTE: If the number 0 occurs in the final line of the division tableau in either of the above cases, it can be treated as (+) or (βˆ’) as needed. The Upper and Lower Bounds Theorem works because of Theorem 3.4. For the upper bound part of the theorem, suppose c > 0 is divided into f and the resulting line in the division...
real zeros lie in the interval [βˆ’4, 4] and that our possible rational zeros are Β± 1 2 and Β± 3. Descartes’ Rule of Signs guarantees us at least one negative real zero and exactly one positive real zero, counting multiplicity. We try our positive rational zeros, starting with the smallest, 1 2. Since the remainder isn’t...
negative real zeros, counting multiplicity, so we try βˆ’1 again, and it works once more. At this point, we have taken f, a fourth degree polynomial, and performed two successful divisions. Our quotient polynomial is quadratic, so we look at it to find the remaining zeros. 276 βˆ’ 1 2 4 βˆ’1 βˆ’6 5 4 2 ↓ βˆ’1 βˆ’ 3 2 3 βˆ’ 5 2 βˆ’ 19 ...
4This is apparently a bad thing. 5We don’t use this word lightly; it can be proven that the zeros of some polynomials cannot be expressed using the usual algebraic symbols. 6See this page. 3.3 Real Zeros of Polynomials 277 Test gives us Β±1 as rational zeros to try but neither of these work since f (1) = f (βˆ’1) = βˆ’1. I...
οΏ½οΏ½nding zeros plays in solving equations and inequalities. Example 3.3.7. 1. Find all of the real solutions to the equation 2x5 + 6x3 + 3 = 3x4 + 8x2. 2. Solve the inequality 2x5 + 6x3 + 3 ≀ 3x4 + 8x2. 3. Interpret your answer to part 2 graphically, and verify using a graphing calculator. Solution. 1. Finding the real ...
3x4 + 8x2. We recall that the solution to f (x) ≀ g(x) is the set of x values for which the graph of f is below the graph of g (where f (x) < g(x)) along with the x values where the two graphs intersect (f (x) = g(x)). Graphing f and g on the calculator produces the picture on the lower left. (The end behavior should ...
1 = 0. We find three real zeros, x = 1 βˆ’ 2 = βˆ’0.414..., x = 1 + 2 = 2.414..., and x = 5, of which only the last two fall in the applied domain of [0, 10.07]. We choose x = 0, x = 3 and x = 10.07 as our test values and plug them into the function P (x) = βˆ’5x3 + 35x2 βˆ’ 45x βˆ’ 25 (not f (x) = x3 βˆ’ 7x2 + 9x βˆ’ 5) to get the ...
οΏ½nd an interval containing all of the real zeros. Use the Rational Zeros Theorem to make a list of possible rational zeros. Use Descartes’ Rule of Signs to list the possible number of positive and negative real zeros, counting multiplicities. 1. f (x) = x3 βˆ’ 2x2 βˆ’ 5x + 6 2. f (x) = x4 + 2x3 βˆ’ 12x2 βˆ’ 40x βˆ’ 32 3. f (x) =...
f (x) = 6x4 βˆ’ 5x3 βˆ’ 9x2 23. f (x) = x4 + 2x2 βˆ’ 15 24. f (x) = x4 βˆ’ 9x2 + 14 25. f (x) = 3x4 βˆ’ 14x2 βˆ’ 5 26. f (x) = 2x4 βˆ’ 7x2 + 6 27. f (x) = x6 βˆ’ 3x3 βˆ’ 10 28. f (x) = 2x6 βˆ’ 9x3 + 10 29. f (x) = x5 βˆ’ 2x4 βˆ’ 4x + 8 30. f (x) = 2x5 + 3x4 βˆ’ 18x βˆ’ 27 3.3 Real Zeros of Polynomials 281 In Exercises 31 - 33, use your calculato...
βˆ’ 49x + 20 42. x4 + 2x2 = 15 44. 2x5 + 3x4 = 18x + 27 In Exercises 45 - 54, solve the polynomial inequality and state your answer using interval notation. 45. βˆ’2x3 + 19x2 βˆ’ 49x + 20 > 0 46. x4 βˆ’ 9x2 ≀ 4x βˆ’ 12 47. (x βˆ’ 1)2 β‰₯ 4 49. x4 ≀ 16 + 4x βˆ’ x3 48. 4x3 β‰₯ 3x + 1 50. 3x2 + 2x < x4 51. x3 + 2x2 2 53. 2x4 > 5x2 + 3 < x...
the important features of the graph. 58. With the help of your classmates, create a list of five polynomials with different degrees whose real zeros cannot be found using any of the techniques in this section. 3.3 Real Zeros of Polynomials 283 3.3.4 Answers 1. For f (x) = x3 βˆ’ 2x2 βˆ’ 5x + 6 All of the real zeros lie in t...
3 + 5x2 + 34x βˆ’ 10 All of the real zeros lie in the interval [βˆ’3, 3] Possible rational zeros are Β± 1 17, Β± 2 There are 2 or 0 positive real zeros; there is 1 negative real zero 17, Β±1, Β±2, Β±5, Β±10 17, Β± 10 17, Β± 5 284 Polynomial Functions 8. For f (x) = 36x4 βˆ’ 12x3 βˆ’ 11x2 + 2x + 1 All of the real zeros lie in the inter...
βˆ’ 49x + 20 x = 1 2, x = 4, x = 5 (each has mult. 1) 17. f (x) = βˆ’17x3 + 5x2 + 34x βˆ’ 10 √ x = 5 17, x = Β± 2 (each has mult. 1) 18. f (x) = 36x4 βˆ’ 12x3 βˆ’ 11x2 + 2x + 1 2 (mult. 2), x = βˆ’ 1 3 (mult. 2) x = 1 19. f (x) = 3x3 + 3x2 βˆ’ 11x βˆ’ 10 x = βˆ’2, x = 3Β± 69 (each has mult. 1) √ 6 3.3 Real Zeros of Polynomials 285 20. f ...
+ 3x4 βˆ’ 18x βˆ’ 27 √ each has mult. 1) 31. f (x) = x5 βˆ’ 60x3 βˆ’ 80x2 + 960x + 2304 x = βˆ’4 (mult. 3), x = 6 (mult. 2) 32. f (x) = 25x5 βˆ’ 105x4 + 174x3 βˆ’ 142x2 + 57x βˆ’ 9 x = 3 5 (mult. 2), x = 1 (mult. 3) 33. f (x) = 90x4 βˆ’ 399x3 + 622x2 βˆ’ 399x + 90 each has mult. 1) 34. We choose q(x) = 72x3 βˆ’ 6x2 βˆ’ 7x + 1 = 72 Β· f (x). C...
, 5 βˆ’ 5] lies in the applied domain, however. In the context of the problem, this says for the volume of the box to be at least 80 cubic inches, the square removed from each corner needs to have a side length of at least 1 inch, but no more than 5 βˆ’ 5 β‰ˆ 2.76 inches. √ 56. C(x) ≀ 5000 on (approximately) (βˆ’βˆž, 82.18]. The...
’ of a negative real number. In property 2, it is important to remember the restriction on c. For example, it is perfectly acceptable to say √ √ √ βˆ’4 = i 4 = i(2) = 2i. However, βˆ’(βˆ’4) = i βˆ’4, otherwise, we’d get √ 2 = 4 = βˆ’(βˆ’4) = i √ βˆ’4 = i(2i) = 2i2 = 2(βˆ’1) = βˆ’2, which is unacceptable.3 We are now in the position to d...
])(x βˆ’ [1 βˆ’ 2i]) 1. As mentioned earlier, we treat expressions involving i as we would any other radical. We combine like terms to get (1 βˆ’ 2i) βˆ’ (3 + 4i) = 1 βˆ’ 2i βˆ’ 3 βˆ’ 4i = βˆ’2 βˆ’ 6i. 2. Using the distributive property, we get (1 βˆ’ 2i)(3 + 4i) = (1)(3) + (1)(4i) βˆ’ (2i)(3) βˆ’ (2i)(4i) = 3 + 4i βˆ’ 6i βˆ’ 8i2. Since i2 = βˆ’1, ...