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Answers 1. y + 1 = 3(x βˆ’ 3) y = 3x βˆ’ 10 3. y + 1 = βˆ’(x + 7) y = βˆ’x βˆ’ 8 169 2. y βˆ’ 8 = βˆ’2(x + 5) y = βˆ’2x βˆ’ 2 4x + 2) 5x βˆ’ 10) 7. y βˆ’ 117 = 0 y = 117 √ 9. y βˆ’ 2 3 = βˆ’5(x βˆ’ √ y = βˆ’5x + 7 3 11. y = βˆ’ 5 3 x 13. y = 8 5 x βˆ’ 8 15. y = 5 17. y = βˆ’ 5 4 x + 11 8 19. y = βˆ’x 21. f (x) = 2x βˆ’ 1 slope: m = 2 y-intercept: (0, βˆ’1) 2,...
βˆ’2 1 2 x βˆ’2 βˆ’1 1 2 x βˆ’1 27. (βˆ’1, βˆ’1) and 11 5, 27 5 28. d(t) = 3t, t β‰₯ 0. 29. E(t) = 360t, t β‰₯ 0. 30. C(x) = 45x + 20, x β‰₯ 0. 31. C(t) = 80t + 50, 0 ≀ t ≀ 8. 32. W (x) = 200 +.05x, x β‰₯ 0 She must make $5500 in weekly sales. 33. C(p) = 0.035p + 1.5 The slope 0.035 means it costs 3.5Β’ per page. C(0) = 1.5 means there is...
P (c) = 43. (a) (b) (c) 6p + 1.5 5.5p if if 1 ≀ p ≀ 5 p β‰₯ 6 15n 12.5n if if 1 ≀ n ≀ 9 n β‰₯ 10 10 10 + 0.15(m βˆ’ 500) 0 ≀ m ≀ 500 if if m > 500 0.12c 12 + 0.1(c βˆ’ 100) if if 1 ≀ c ≀ 100 c > 100 D(d) = ο£±   βˆ’ 1 8 2 d + 31 2 2 if if if 0 ≀ d ≀ 15 15 ≀ d ≀ 27 27 ≀ d ≀ 37 D(s) = ο£±   2 1 2 s βˆ’ 3 8 if if if 0 ≀ s ≀ 10 1...
), it is 56β—¦F. T (8) = 64, so at 2 PM (8 hours after 6 AM), it is 64β—¦F. T (12) = 56, so at 6 PM (12 hours after 6 AM), it is 56β—¦F. (b) The average rate of change is T (8)βˆ’T (4) 8βˆ’4 = 2. Between 10 AM and 2 PM, the temperature increases, on average, at a rate of 2β—¦F per hour. (c) The average rate of change is T (12)βˆ’T (...
positive counterparts while it leaves positive numbers alone. This last description is the one we shall adopt, and is summarized in the following definition. 25 = 5 and | βˆ’ 5| = (βˆ’5)2 = √ Definition 2.4. The absolute value of a real number x, denoted |x|, is given by |x| = βˆ’x, x, if x < 0 if x β‰₯ 0 In Definition 2.4, we d...
to show that this equation is true for all real numbers a and b. If a and b are both positive, then so is ab. Hence, |a| = a, |b| = b and |ab| = ab. Hence, the equation |ab| = |a||b| is the same as ab = ab which is true. If both a and b are negative, then ab is positive. Hence, |a| = βˆ’a, |b| = βˆ’b and |ab| = ab. The eq...
3. 3|2x + 1| βˆ’ 5 = 0 4. 4 βˆ’ |5x + 3| = 5 5. |x| = x2 βˆ’ 6 6. |x βˆ’ 2| + 1 = x Solution. 1. The equation |3x βˆ’ 1| = 6 is of the form |x| = c for c > 0, so by the Equality Properties, |3x βˆ’ 1| = 6 is equivalent to 3x βˆ’ 1 = 6 or 3x βˆ’ 1 = βˆ’6. Solving the former, we arrive at x = 7 3, and solving the latter, we get x = βˆ’ 5 3...
risk of introducing extraneous solutions, or worse, losing solutions. For this reason, we break equations like this into cases by rewriting the term in absolute values, |x|, using Definition 2.4. For x < 0, |x| = βˆ’x, so for x < 0, the equation |x| = x2 βˆ’ 6 is equivalent to βˆ’x = x2 βˆ’ 6. Rearranging this gives us x2 + x ...
x βˆ’ 1 becomes x βˆ’ 2 = x βˆ’ 1. Here, the equation reduces to βˆ’2 = βˆ’1, which signifies we have no solutions here. Hence, our only solution is x = 3 2. Next, we turn our attention to graphing absolute value functions. Our strategy in the next example is to make liberal use of Definition 2.4 along with what we know about gra...
|, x β‰₯ 0 Notice that we have an β€˜open circle’ at (0, 0) in the graph when x < 0. As we have seen before, this is due to the fact that the points on y = βˆ’x approach (0, 0) as the x-values approach 0. Since x is required to be strictly less than zero on this stretch, the open circle is drawn at the origin. However, notic...
we get g(x) = βˆ’x + 3, x βˆ’ 3, if x < 3 if x β‰₯ 3 As before, the open circle we introduce at (3, 0) from the graph of y = βˆ’x + 3 is filled by the point (3, 0) from the line y = x βˆ’ 3. We determine the domain as (βˆ’βˆž, ∞) and the range as [0, ∞). The function g is increasing on [3, ∞) and decreasing on (βˆ’βˆž, 3]. The relative ...
= 0. Isolating the absolute 3 or x = βˆ’1, so 3, 0 and (βˆ’1, 0). Substituting x = 0 gives y = i(0) = 4βˆ’2|3(0)+1| = 2, value term gives |3x + 1| = 2, so either 3x + 1 = 2 or 3x + 1 = βˆ’2. We get x = 1 our x-intercepts are 1 for a y-intercept of (0, 2). Rewriting the formula for i(x) without absolute values gives i(x) = 4 βˆ’...
We begin by graphing f (x) = |x| and labeling three points, (βˆ’1, 1), (0, 0) and (1, 1). y 4 3 2 1 (1, 1) (βˆ’1, 1) βˆ’3 βˆ’2 βˆ’1 (0, 0) 1 2 3 x f (x) = |x| 1. Since g(x) = |x βˆ’ 3| = f (x βˆ’ 3), Theorem 1.7 tells us to add 3 to each of the x-values of the points on the graph of y = f (x) to obtain the graph of y = g(x). This s...
subtract 3 from each y-coordinate y 1 βˆ’1 βˆ’3 βˆ’2 βˆ’1 1 2 3 x (βˆ’1, βˆ’2) βˆ’2 (1, βˆ’2) (0, βˆ’3) βˆ’3 βˆ’4 h(x) = f (x) βˆ’ 3 = |x| βˆ’ 3 180 Linear and Quadratic Functions 3. We re-write i(x) = 4 βˆ’ 2|3x + 1| = 4 βˆ’ 2f (3x + 1) = βˆ’2f (3x + 1) + 4 and apply Theorem 1.7. First, we take care of the changes on the β€˜inside’ of the absolute va...
in Section 1.7 can be used to graph an entire family of absolute value functions, not all functions involving absolute values posses the characteristic β€˜βˆ¨β€™ shape. As the next example illustrates, often there is no substitute for appealing directly to the definition. Example 2.2.4. Graph each of the following functions....
0, ∞). The local minimum value of f is the absolute minimum value of f, namely βˆ’1; the local maximum and absolute maximum values for f also coincide βˆ’ they both are 1. Every point on the graph of f is simultaneously a relative maximum and a relative minimum. (Can you remember why in light of Definition 1.11? This was ex...
x β‰₯ 3 To solve g(x) = 0, we see that the only piece which contains a variable is g(x) = 2x for βˆ’2 ≀ x < 3. Solving 2x = 0 gives x = 0. Since x = 0 is in the interval [βˆ’2, 3), we keep this solution and have (0, 0) as our only x-intercept. Accordingly, the y-intercept is also (0, 0). To graph g, we start with x < βˆ’2 and...
6 4. 4 βˆ’ |x| = 3 7. 5 βˆ’ |x| 2 = 1 2. |3x βˆ’ 1| = 10 3. |4 βˆ’ x| = 7 5. 2|5x + 1| βˆ’ 3 = 0 6. |7x βˆ’ 1| + 2 = 0 8. 2 3 |5 βˆ’ 2x| βˆ’ 1 2 = 5 9. |x| = x + 3 10. |2x βˆ’ 1| = x + 1 11. 4 βˆ’ |x| = 2x + 1 12. |x βˆ’ 4| = x βˆ’ 5 13. |x| = x2 14. |x| = 12 βˆ’ x2 15. |x2 βˆ’ 1| = 3 Prove that if |f (x)| = |g(x)| then either f (x) = g(x) or f ...
) = |x + 2| βˆ’ |x| 33. f (x) = |x + 4| + |x βˆ’ 2| 34. With the help of your classmates, find an absolute value function whose graph is given below. y 4 3 2 1 βˆ’8βˆ’7βˆ’6βˆ’5βˆ’4βˆ’3βˆ’2βˆ’ 35. With help from your classmates, prove the second, third and fifth parts of Theorem 2.1. 36. Prove The Triangle Inequality: For all real numbers a ...
βˆ’3 βˆ’2 βˆ’4 βˆ’3 βˆ’2 βˆ’1 1 2 3 4 x 2.2 Absolute Value Functions 185 24. f (x) = |4x| f (0) = 0 x-intercept (0, 0) y-intercept (0, 0) Domain (βˆ’βˆž, ∞) Range [0, ∞) Decreasing on (βˆ’βˆž, 0] Increasing on [0, ∞) Relative and absolute minimum at (0, 0) No relative or absolute maximum 25. f (x) = βˆ’3|x| f (0) = 0 x-intercept (0, 0) y-i...
x) = |2 βˆ’ x| 2 βˆ’ x No zeros No x-intercept y-intercept (0, 1) Domain (βˆ’βˆž, 2) βˆͺ (2, ∞) Range {βˆ’1, 1} Constant on (βˆ’βˆž, 2) Constant on (2, ∞) Absolute minimum at every point (x, βˆ’1) 30. Re-write f (x) = x + |x| βˆ’ 3 as x < 0 x β‰₯ 0 βˆ’3 2x βˆ’ 3 f (x) = if if = 0 f 3 2 x-intercept 3 2, 0 y-intercept (0, βˆ’3) Domain (βˆ’βˆž, ∞) Range...
οΏ½οΏ½ f (βˆ’1) = 0 x-intercept (βˆ’1, 0) y-intercept (0, 2) Domain (βˆ’βˆž, ∞) Range [βˆ’2, 2] Increasing on [βˆ’2, 0] Constant on (βˆ’βˆž, βˆ’2] Constant on [0, ∞) Absolute minimum at every point (x, βˆ’2) where x ≀ βˆ’2 33. Re-write f (x) = |x + 4| + |x βˆ’ 2| as βˆ’2x βˆ’ 2 6 2x + 2 x < βˆ’4 if if βˆ’4 ≀ x < 2 x β‰₯ 2 if f (x) = ο£±   No zeros No x-i...
ratic function is (βˆ’βˆž, ∞). The most basic quadratic function is f (x) = x2, whose graph appears below. Its shape should look familiar from Intermediate Algebra – it is called a parabola. The point (0, 0) is called the vertex of the parabola. In this case, the vertex is a relative minimum and is also the where the absol...
1) y (2, 4) (βˆ’4, 1) 1 (0, 1) (1, 1) (βˆ’3, βˆ’2) (βˆ’1, βˆ’2) βˆ’4 βˆ’3 βˆ’2 βˆ’1 x βˆ’1 βˆ’2 βˆ’1 (0, 0) 1 2 x βˆ’3 (βˆ’2, βˆ’3) f (x) = x2 βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’βˆ’β†’ g(x) = f (x + 2) βˆ’ 3 = (x + 2)2 βˆ’ 3 From the graph, we see that the vertex has moved from (0, 0) on the graph of y = f (x) to (βˆ’2, βˆ’3) on the graph of y = g(x). This sets [βˆ’3, ∞) as the range ...
this is a vertical stretch by a factor of 2, followed by a reflection about the x-axis, followed by a vertical shift up 1 unit. This moves (1, 4) to (1, βˆ’7), (2, 1) to (2, βˆ’1), (3, 0) to (3, 1), (4, 1) to (4, βˆ’1) and (5, 4) to (5, βˆ’7). y (3, 1) 2 1 (2, βˆ’1) 3 4 5 (4, βˆ’1) x 1 βˆ’1 βˆ’2 βˆ’3 βˆ’4 βˆ’5 βˆ’6 (2, 4) (1, 1) (1, βˆ’7) (5, βˆ’...
convert both g(x) and h(x) into that form by expanding and collecting like terms. Doing so, we find g(x) = (x + 2)2 βˆ’ 3 = x2 + 4x + 1 and h(x) = βˆ’2(x βˆ’ 3)2 + 1 = βˆ’2x2 + 12x βˆ’ 17. While these β€˜simplified’ formulas for g(x) and h(x) satisfy Definition 2.5, they do not lend themselves to graphing easily. For that reason, th...
)2 + 1, no rewrite is needed. We can directly identify h = 3 and k = 1 and, sure enough, we found the vertex of the graph of y = h(x) to be (3, 1). To see why the formula in Theorem 2.2 produces the vertex, consider the graph of the equation y = a(x βˆ’ h)2 + k. When we substitute x = h, we get y = k, so (h, k) is on the...
) = g2(x) + k = a(x βˆ’ h)2 + k which effects a vertical shift up or down k units5 resulting in the vertex moving from (h, 0) to (h, k). In addition to verifying Theorem 2.2, the arguments in the two preceding paragraphs have also shown us the role of the number a in the graphs of quadratic functions. The graph of y = a(x...
οΏ½οΏ½cient of x, in this case βˆ’4, and take half of it to get 1 2 (βˆ’4) = βˆ’2. This tells us that our target perfect square quantity is (x βˆ’ 2)2. To get an expression equivalent to (x βˆ’ 2)2, we need to add (βˆ’2)2 = 4 to the x2 βˆ’ 4x to create a perfect square trinomial, but to keep the balance, we must also subtract it. We col...
without the need to plot additional points. We see that the range of f is [βˆ’1, ∞) and we are done. 2. To get started, we rewrite g(x) = 6 βˆ’ x βˆ’ x2 = βˆ’x2 βˆ’ x + 6 and note that the coefficient of x2 is βˆ’1, not 1. This means our first step is to factor out the (βˆ’1) from both the x2 and x terms. We then follow the completing...
) = 6 βˆ’ x βˆ’ x2 With Example 2.3.2 fresh in our minds, we are now in a position to show that every quadratic function can be written in standard form. We begin with f (x) = ax2 + bx + c, assume a = 0, and complete the square in complete generality. f (x) = ax2 + bx + c x + c (Factor out coefficient of x2 from x2 and x.) x...
οΏ½οΏ½cient of x2 in the general form of the quadratic function. In other words, it is the coefficient of x2 alone which determines this behavior – a result that is generalized in Section 3.1. The second treasure is a re-discovery of the quadratic formula. Equation 2.5. The Quadratic Formula: If a, b and c are real numbers w...
solutions we get from a quadratic equation. These cases, and their relation to the discriminant, are summarized below. Theorem 2.3. Discriminant Trichotomy: Let a, b and c be real numbers with a = 0. If b2 βˆ’ 4ac < 0, the equation ax2 + bx + c = 0 has no real solutions. If b2 βˆ’ 4ac = 0, the equation ax2 + bx + c = 0 ha...
(x) = R(x) βˆ’ C(x) = βˆ’1.5x2 + 250x βˆ’ (80x + 150) = βˆ’1.5x2 + 170x βˆ’ 150. Since the revenue function is valid when 0 ≀ x ≀ 166, P is also restricted to these values. 2. To find the x-intercepts, we set P (x) = 0 and solve βˆ’1.5x2 + 170x βˆ’ 150 = 0. The mere thought of trying to factor the left hand side of this equation cou...
οΏ½οΏ½erently than the y-axis. This results in the left-hand x-intercept and the y-intercept being uncomfortably close to each other and to the origin in the picture.) = 14000 3 y 4000 3000 2000 1000 10 20 30 40 50 60 70 80 90 100 110 120 x 3. The zeros of P are the solutions to P (x) = 0, which we have found to be approxi...
The time is finally right for him to pursue his dream of farming alpaca. He wishes to build a rectangular pasture, and estimates that he has enough money for 200 linear feet of fencing material. If he makes the pasture adjacent to a stream (so no fencing is required on that side), what are the dimensions of the pasture...
nd w < 100. Hence, the function we wish to maximize is A(w) = βˆ’2w2 +200w for 0 < w < 100. Since A is a quadratic function (of w), we know that the graph of y = A(w) is a parabola. Since the coefficient of w2 is βˆ’2, we know that this parabola opens downwards. This means that there is a maximum value to be found, and we kn...
cepts, we solve x2 βˆ’ x βˆ’ 6 = 0. Factoring gives (x βˆ’ 3)(x + 2) = 0 so x = βˆ’2 or x = 3. Hence, (βˆ’2, 0) and (3, 0) are x-intercepts. The y-intercept (0, βˆ’6) is found by setting x = 0. To plot the vertex, we find x = βˆ’ b 2, and y = 1 4 = βˆ’6.25. Plotting, we get the parabola seen below on the left. To obtain 2 points on the...
(x) and calling to mind Theorem 1.4 from Section 1.7. f (x) = βˆ’g(x), g(x), if g(x) < 0 if g(x) β‰₯ 0 The function f is defined so that when g(x) is negative (i.e., when its graph is below the x-axis), the graph of f is its refection across the x-axis. This is a general template to graph functions of the form f (x) = |g(x...
οΏ½t. Find and interpret break-even points. 10. The cost, in dollars, to produce x β€œI’d rather be a Sasquatch” T-Shirts is C(x) = 2x + 26, x β‰₯ 0 and the price-demand function, in dollars per shirt, is p(x) = 30 βˆ’ 2x, 0 ≀ x ≀ 15. 11. The cost, in dollars, to produce x bottles of 100% All-Natural Certified Free-Trade Organi...
F (t) = βˆ’0.0076t2 + 0.45t + 16, 0 ≀ t ≀ 28, where t is the number of years since 1980. Find and interpret the coordinates of the vertex of the graph of y = F (t). 17. The temperature T, in degrees Fahrenheit, t hours after 6 AM is given by: T (t) = βˆ’ 1 2 t2 + 8t + 32, 0 ≀ t ≀ 12 What is the warmest temperature of the ...
Games. In one event, the hammer throw, the height h in feet of the hammer above the ground t seconds after Jason lets it go is modeled by h(t) = βˆ’16t2 + 22.08t + 6. What is the hammer’s maximum height? What is the hammer’s total time in the air? Round your answers to two decimal places. 202 Linear and Quadratic Functi...
to find the point on L closest to (0, 0). 30. With the help of your classmates, show that if a quadratic function f (x) = ax2 + bx + c has two real zeros then the x-coordinate of the vertex is the midpoint of the zeros. In Exercises 31 - 36, solve the quadratic equation for the indicated variable. 31. x2 βˆ’ 10y2 = 0 for...
minimum Axis of symmetry x = 1 4. f (x) = βˆ’2(x + 1)2 + 4 = βˆ’2x2 βˆ’ 4x + 2 √ 2, 0) 2, 0) and (βˆ’1 + √ x-intercepts (βˆ’1 βˆ’ y-intercept (0, 2) Domain: (βˆ’βˆž, ∞) Range: (βˆ’βˆž, 4] Increasing on (βˆ’βˆž, 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’...
1 2 Vertex βˆ’ 1 is a minimum 2, 3 4 Axis of symmetry 2 βˆ’1 1 x 2.3 Quadratic Functions 205 8. f (x) = βˆ’3x2 + 5x + 4 = βˆ’3 x βˆ’ 5 6 5βˆ’ 73, 0 5+ and, 0 73 √ √ 2 + 73 12 6 x-intercepts 6 y-intercept (0, 4) Domain: (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 i...
, so to make a profit, between 5 and 20 bottles of tonic need to be made and sold. 12You’ll need to use your calculator to zoom in far enough to see that the vertex is not the y-intercept. 206 12. 13. 14. Linear and Quadratic Functions P (x) = βˆ’3x2 + 72x βˆ’ 240, for 0 ≀ x ≀ 30 12 cups of lemonade need to be made and sold...
be produced for a cost of $200. 19. 8 feet by 16 feet; maximum area is 128 square feet. 20. 50 feet by 50 feet; maximum area is 2500 feet; he can raise 100 average alpacas. 21. The largest rectangle has area 12.25 square inches. 22. 2 seconds. 23. The rocket reaches its maximum height of 500 feet 10 seconds after lift...
) is βˆ’ 2 5, 1 5 31. x = Β±y √ 10 34. y = √ 3 Β± 16x + 9 2 32. x = Β±(y βˆ’ 2) 35. y = 2 Β± x 33. x = 36. t = √ m Β± m2 + 4 2 v0 Β± v2 0 + 4gs0 2g 208 Linear and Quadratic Functions 2.4 Inequalities with Absolute Value and Quadratic Functions In this section, not only do we develop techniques for solving various classes of ineq...
is (x, f (x)), and a generic 2.4 Inequalities with Absolute Value and Quadratic Functions 209 point on the graph of y = g(x) is (x, g(x)). When we seek solutions to f (x) = g(x), we are looking for x values whose y values on the graphs of f and g are the same. In part 1, we found x = 3 is the solution to f (x) = g(x)....
asks for solutions to equations and inequalities. 210 Linear and Quadratic Functions Example 2.4.2. The graphs of f and g are below. (The graph of y = g(x) is bolded.) Use these graphs to answer the following questions. y 4 3 2 1 (βˆ’1, 2) y = g(x) (1, 2) βˆ’2 βˆ’1 1 2 x βˆ’1 y = f (x) 1. Solve f (x) = g(x). 2. Solve f (x) < ...
Theorem 2.4. Inequalities Involving the Absolute Value: Let c be a real number. For c > 0, |x| < c is equivalent to βˆ’c < x < c. For c > 0, |x| ≀ c is equivalent to βˆ’c ≀ x ≀ c. For c ≀ 0, |x| < c has no solution, and for c < 0, |x| ≀ c has no solution. For c β‰₯ 0, |x| > c is equivalent to x < βˆ’c or x > c. For c β‰₯ 0, |x|...
4, ∞). Graphically, we have 212 Linear and Quadratic Functions x βˆ’ 1| βˆ’4 βˆ’3 βˆ’2 βˆ’1 1 2 3 4 5 x We see that the graph of y = |x βˆ’ 1| is above the horizontal line y = 3 for x < βˆ’2 and x > 4 hence this is where |x βˆ’ 1| > 3. The two graphs intersect when x = βˆ’2 and x = 4, so we have graphical confirmation of our analytic sol...
οΏ½ (3, 6]. Graphically, we see that the graph of y = |x βˆ’ 1| is β€˜between’ the horizontal lines y = 2 and y = 5 for x values between βˆ’4 and βˆ’1 as well as those between 3 and 6. Including the x values where y = |x βˆ’ 1| and y = 5 intersect, we get 1See Definition 1.2 in Section 1.1.1. 2.4 Inequalities with Absolute Value an...
re-visit this problem using some of the techniques developed in this section not only to reinforce our solution in Section 2.3, but to also help formulate a general analytic procedure for solving all quadratic inequalities. If we consider f (x) = x2 βˆ’ x βˆ’ 6 and g(x) = 0, then solving x2 βˆ’ x βˆ’ 6 < 0 corresponds graphic...
be above the x-axis at one point and below the x-axis at another point without crossing the x-axis. This allows us to determine the sign of all of the function values on a given interval by testing the function at just one value in the interval. This gives us the following. 2We will give this property a name in Chapte...
is placed above (1, ∞). Since we are solving 2x2 + x βˆ’ 3 ≀ 0, we look for solutions to 2x2 + x βˆ’ 3 < 0 as well as solutions for 2x2 + x βˆ’ 3 = 0. For 2x2 + x βˆ’ 3 < 0, we need the intervals which we have a (βˆ’). Checking the sign diagram, we see this is βˆ’ 3 2, 1. We know 2x2 + x βˆ’ 3 = 0 when x = βˆ’ 3 ; f (0) = βˆ’3 < 0, so ...
and x = 3 as our test values and find f (βˆ’1) = 2, which is (+); f (0) = βˆ’1 which is (βˆ’); and f (3) = 2 which is (+) again. Our solution to x2 βˆ’ 2x βˆ’ 1 > 0 is where 2, ∞. To check the inequality we have (+), so, in interval notation βˆ’βˆž, 1 βˆ’ x2 βˆ’ 2x > 1 graphically, we set g(x) = x2 βˆ’ 2x and h(x) = 1. We are looking for ...
), but the parabola is always above the line otherwise.4 4In this case, we say the line y = 2x is tangent to y = x2 + 1 at (1, 2). Finding tangent lines to arbitrary functions is a fundamental problem solved, in general, with Calculus. 2.4 Inequalities with Absolute Value and Quadratic Functions 217 0 1 (+) 0 (+) 2 y 4...
(+) βˆ’1 (βˆ’) 1 2 1 (βˆ’) 0 2 (+) 3 1 Combining these into one sign diagram, we have that our solution is [0, 2]. Graphically, to check 2x βˆ’ x2 β‰₯ |x βˆ’ 1| βˆ’ 1, we set h(x) = 2x βˆ’ x2 and i(x) = |x βˆ’ 1| βˆ’ 1 and look for the x values where the graph of h is above the the graph of i (the solution of h(x) > i(x)) as well as the ...
25. One way to proceed at this point is to solve the two inequalities βˆ’0.25 ≀ x2 βˆ’ 576 and x2 βˆ’ 576 ≀ 0.25 individually using sign diagrams and then taking the intersection of the solution sets. While this way will (eventually) lead to the correct answer, we take this opportunity to showcase the increasing property of ...
5The underlying concept of Calculus can be phrased in terms of tolerances, so this is well worth your attention. 2.4 Inequalities with Absolute Value and Quadratic Functions 219 Solution. 1. The relation R consists of all points (x, y) whose y-coordinate is greater than |x|. If we graph y = |x|, then we want all of th...
|2x + 1| < 0 11. |1 βˆ’ 2x| β‰₯ x + 5 13. x β‰₯ |x + 1| 15. x + |2x βˆ’ 3| < 2 17. x2 + 2x βˆ’ 3 β‰₯ 0 19. x2 + 9 < 6x 21. x2 + 4 ≀ 4x 23. 3x2 ≀ 11x + 4 25. 2x2 βˆ’ 4x 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‰₯ 2...
used to shoot a marble straight up into the air from 2 meters above the ground with an initial velocity of 30 meters per second, for what values of time t will the marble be over 35 meters above the ground? (Refer to Exercise 25 in Section 2.3 for assistance if needed.) Round your answers to two decimal places. 38. Wh...
. 6Understanding this type of inequality is really important in Calculus. 222 Linear and Quadratic Functions 2.4.2 Answers 1. 1 3, 3 3. (βˆ’3, 2) 5. No solution 7. (βˆ’3, 2] βˆͺ [6, 11) 9. βˆ’ 12 7, βˆ’ 6 5 11. βˆ’βˆž, βˆ’ 4 3 βˆͺ [6, ∞) 13. No 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 11, 8 + 2 11) β‰ˆ (1.37, 14.63), which corresponds to between 7:22 AM (1.37 hours after 6 AM) to 8:38 PM (14.63 hours after 6 AM.) However, since the model is valid only for t, 0 ≀ t ≀ 12, we restrict our answer and find it is warmer than 42β—¦ Fahrenheit from 7:22 AM to 6 PM. 2.4 Inequalities with Absolute Value and Qua...
1 βˆ’2 βˆ’3 βˆ’4 2.5 Regression 2.5 Regression 225 We have seen examples already in the text where linear and quadratic functions are used to model a wide variety of real world phenomena ranging from production costs to the height of a projectile above the ground. In this section, we use some basic tools from statistical ana...
E. This line is called the least squares regression line, or sometimes the β€˜line of best fit’. The formula for the line of best fit requires notation we won’t present until Chapter 9.1, so we will revisit it then. The graphing calculator can come to our assistance here, since it has a built-in feature to compute the reg...
of fit. 3. Interpret the slope of the line of best fit. 4. Use the regression line to predict the annual US energy consumption in the year 2013. 5. Use the regression line to predict when the annual consumption will reach 120 Quads. Solution. 1. Entering the data into the calculator gives The data certainly appears to b...
is to convert the times into the 24 hour clock time so that 1 PM is 13, 2 PM is 14, etc.. If we enter these data into the graphing calculator and plot the points we get While the beginning of the data looks linear, the temperature begins to fall in the afternoon hours. This sort of behavior reminds us of parabolas, an...
on the goodness of fit.7 Interpret the slope of the line of best fit. (b) Use the regression line to predict the population of Lake County in 2010. (The recorded figure from the 2010 census is 230,041) (c) Use the regression line to predict when the population of Lake County will reach 250,000. 2. According to this websi...
row is the odometer reading when I refilled the gas tank. So, for example, the fourth entry is the point (28.25, 1051) which says that I had used a total of 28.25 gallons of gasoline when the odometer read 1051 miles. Gasoline Used (Gallons) Odometer (Miles) 0 9.26 19.03 28.25 36.45 44.64 53.57 62.62 71.93 81.69 90.43 ...
.2 236.2 236.2 235.2 233.2 236.8 238.2 (a) Find the least squares line for the Thursday data and comment on its goodness of fit. (b) Find the least squares line for the Saturday data and comment on its goodness of fit. (c) Use Quadratic Regression to find a parabola which models the Saturday data and com- ment on its good...
15 2423316 13968290 80399780 Use Quadratic Regression to find a parabola which models this data and comment on its goodness of fit. (Spoiler Alert: Does anyone know what type of function we need here?) (b) This next data set comes from the U.S. Naval Observatory. That site has loads of awesome stuff on it, but for this ex...
266 indicates the country’s energy production is increasing at a rate of 0.266 Quad per year. (d) According to the model, the production in 2010 will be 74.8 Quad. (e) According to the model, the production will reach 100 Quad in the year 2105. 4. The line is y = 36.8x + 16.39. We have r =.99987 and r2 =.9997 so this i...
= 0 but we know N (0) = 2. 234 Linear and Quadratic Functions (b) The quadratic model for the hours of daylight in Fairbanks, Alaska is y =.51x2 +6.23xβˆ’.36. Even with R2 =.92295 we should be wary of making predictions beyond the data. Case in point, the model gives βˆ’4.84 hours of daylight when x = 13. So January 21, 2...
natural number lets us focus on well-behaved algebraic animals.1 Example 3.1.1. Determine if the following functions are polynomials. Explain your reasoning. 1. g(x) = 4 + x3 x √ 4. f (x) = 3 x 2. p(x) = 4x + x3 x 3. q(x) = 4x + x3 x2 + 4 5. h(x) = |x| 6. z(x) = 0 1Enjoy this while it lasts. Before we’re through with ...
as a piecewise function involving polynomials. As we shall see in this section, graphs of polynomials possess a quality2 that the graph of h does not. 6. There’s nothing in Definition 3.1 which prevents all the coefficients an, etc., from being 0. Hence, z(x) = 0, is an honest-to-goodness polynomial. Definition 3.2. Suppo...
οΏ½t include the constant functions, so we might as well treat them as outsiders from the start. One good thing that comes from Definition 3.2 is that we can now think of linear functions as degree 1 (or β€˜ο¬rst degree’) polynomial functions and quadratic functions as degree 2 (or β€˜second degree’) polynomial functions. Exam...
term is to multiply the terms with the highest power of x from each factor together - in other words, the leading term of p(x) is the product of the leading terms of the factors of p(x). Hence, the leading term of p is (2x)3(x)(3x) = 24x5. This means that the degree of p is 5 and the leading coefficient is 24. As for th...
x)(12 βˆ’ 2x) = 4x3 βˆ’ 44x2 + 120x To find a suitable applied domain, we note that to make a box at all we need x > 0. Also the shorter of the two dimensions of the cardboard is 10 inches, and since we are removing 2x inches from this dimension, we also require 10 βˆ’ 2x > 0 or x < 5. Hence, our applied domain is 0 < x < 5. ...
equally important feature of these functions which we can be seen graphically – their end behavior. The end behavior of a function is a way to describe what is happening to the function values (the y-values) as the x-values approach the β€˜ends’ of the x-axis.9 That is, what happens to y as x becomes small without bound...
for a < 0, as x β†’ βˆ’βˆž, f (x) β†’ βˆ’βˆž and as x β†’ ∞, f (x) β†’ βˆ’βˆž Graphically: a > 0 a < 0 We now turn our attention to functions of the form f (x) = xn where n β‰₯ 3 is an odd natural number. (We ignore the 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 polynomia...
functions have no β€˜sharp turns’. It turns out that these traits are preserved when functions are added together, so general polynomial functions inherit these qualities. Below we find the graph of a function which is neither smooth nor continuous, and to its right we have a graph of a polynomial, for comparison. The fu...
uses the Intermediate Value Theorem to establish a fact that that most students take for granted. Many students, and sadly some instructors, will find it silly. 14In fact, if you take Calculus, you’ll find that smooth functions are automatically continuous, so that saying β€˜polynomials are continuous and smooth’ is redun...
), (βˆ’2, 0), (0, 3) and (3, ∞). We select the test values x = βˆ’3, x = βˆ’1, x = 1 and x = 4. We find f (βˆ’3) is (+), f (βˆ’1) is (βˆ’) and f (1) is (+) as is f (4). Wherever f 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...
concerned with x’s far down the x-axis, we are far away from x = 0 so can rewrite f (x) for these values of x as f (x) = 4x3 1 βˆ’ 1 4x2 + 5 4x3 4x2 and 5 1 As x becomes unbounded (in either direction), the terms 0, as the table below indicates. 4x3 become closer and closer to x βˆ’1000 βˆ’100 βˆ’10 10 100 1000 1 4x2 5 4x3 0....
(+) as x β†’ ±∞, and as a result, we no longer need to evaluate f at the test values x = βˆ’3 and x = 4. Is there a way to eliminate the need to evaluate f at the other test values? What we would really need to know is how the function behaves near its zeros does it cross through the x-axis at these points, as it does at ...
f (4), we substitute to get (+4)3(+2)2(+5) which is (+)(+)(+) or (+). The sign didn’t change for the middle factor (x βˆ’ 3)2. Even though this is the factor which corresponds to the zero x = 3, the fact that the quantity is squared kept the sign of the middle factor the same on either side of 3. If we look back at the ...
Solution. The end behavior of the graph of f will match that of its leading term. To find the leading term, we multiply by the leading terms of each factor to get (βˆ’3)(2x)(x)2 = βˆ’6x3. This tells us that the graph will start above the x-axis, in Quadrant II, and finish below the x-axis, in Quadrant IV. Next, we find the z...
x2 + x + 1 4. Z(b) = 42b βˆ’ b3 6. s(t) = βˆ’4.9t2 + v0t + s0 7. P (x) = (x βˆ’ 1)(x βˆ’ 2)(x βˆ’ 3)(x βˆ’ 4) 8. p(t) = βˆ’t2(3 βˆ’ 5t)(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 al...
= βˆ’x5 βˆ’ 3 25. f (x) = x5, g(x) = (x + 1)5 + 10 26. f (x) = x6, g(x) = 8 βˆ’ x6 27. Use the Intermediate Value Theorem to prove that f (x) = x3 βˆ’ 9x + 5 has a real zero in each 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 shee...
by producing and selling x PortaBoy game systems. 33. According to US Postal regulations, a rectangular shipping box must satisfy the inequality β€œLength + Girth ≀ 130 inches” for Parcel Post and β€œLength + Girth ≀ 108 inches” for other services. Let’s assume we have a closed rectangular box with a square face of side l...
the elusive creatures? 35. An electric circuit is built with a variable resistor installed. For each of the following resistance values (measured in kilo-ohms, kΩ), the corresponding power to the load (measured in milliwatts, mW ) is given in the table below. 17 Resistance: (kΩ) Power: (mW ) 1.012 1.063 2.199 1.496 3....
the end behavior of a linear function behaves like every other polynomial of odd degree, so what doesn’t f (x) = x do that g(x) = x3 does? It’s the β€˜ο¬‚attening’ for values of x near zero. It is this local behavior that will distinguish between a zero of multiplicity 1 and one of higher odd multiplicity. Look again clos...
cient βˆ’3 Constant term 4 As x β†’ βˆ’βˆž, f (x) β†’ βˆ’βˆž As x β†’ ∞, f (x) β†’ βˆ’βˆž 3. q(r) = 1 βˆ’ 16r4 Degree 4 Leading term βˆ’16r4 Leading coefficient βˆ’16 Constant term 1 As r β†’ βˆ’βˆž, q(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†’...
οΏ½ 3.1 Graphs of Polynomials 251 9. f (x) = βˆ’2x3(x + 1)(x + 2)2 Degree 6 Leading term βˆ’2x6 Leading coefficient βˆ’2 Constant term 0 As x β†’ βˆ’βˆž, f (x) β†’ βˆ’βˆž As x β†’ ∞, f (x) 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)...
Z(b) = b(42 βˆ’ b2) √ 42 multiplicity 1 b = βˆ’ b = 0 multiplicity 1 b = 42 multiplicity 1 √ y βˆ’6βˆ’5βˆ’4βˆ’3βˆ’2βˆ’1 1 2 3 4 5 6 b 3.1 Graphs of Polynomials 253 21. g(x) = (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...
V (x) = x(8.5 βˆ’ 2x)(11 βˆ’ 2x) = 4x3 βˆ’ 39x2 + 93.5x, 0 < x < 4.25. Volume is maximized when x β‰ˆ 1.58, so the dimensions of the box with maximum volume are: height β‰ˆ 1.58 inches, width β‰ˆ 5.34 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...
4x3 + 130x2. (b) Graphing y = V (x) on [0, 33] Γ— [0, 21000] shows a maximum at (21.67, 20342.59) so the dimensions of the box with maximum volume are 21.67in. Γ— 21.67in. Γ— 43.32in. for a 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....
3(1.96) β‰ˆ 10 and p3(10.05) β‰ˆ 10 so it seems reasonable to say that we’ll have at least 14 hours of darkness from December 21, 2008 (x = 0) to February 21, 2009 (x = 2) and then again from October 21,2009 (x = 10) to December 21, 2009 (x = 12). The quartic regression model is p4(x) = 0.0144x4 βˆ’0.3507x3 +2.259x2 βˆ’1.571x+...