1 Chapter P: Preparing for Calculus P.1: Functions and Their Graphs Concepts and Vocabulary 1. If f is a function defined by y = f(x), then x is called the independent variable and y is the dependent variable. 2. True . The independent variable is sometimes referred to as the argument of the function. 3. False . If no domain is specified for a function f, then the domain of f is taken to be the largest set of real numbers for which the value f(x) is defined and is a real number. 3(x2−1) 4. False . The domain of the function f(x)= is the set {x|x(cid:54)=1}. x−1 5. False . A function can have at most one y-intercept; otherwise, there would be more than one y-value corresponding to x=0. 6. A set of points in the xy-plane is the graph of a function if and only if every vertical line intersects the graph in at most one point. 7. If the point (5,−3) is on the graph of f, then f( 5 )= −3 . 8. Let f(x) = ax2 +4. For the point (−1,2) to be on the graph of f, we must have f(−1) = 2. But f(−1)=a(−1)2+4=a+4, so we must have a+4=2, or a=−2 . 9. A function f is (a) increasing on an interval I if, for any choice of x and x in I, with x < x , 1 2 1 2 then f(x )<f(x ). 1 2 10. Afunctionis (a) even ifforeverynumberxinitsdomain,thenumber−xisalsointhedomainand f(−x)=f(x). A function f is (b) odd if for every number x in its domain, the number −x is also in the domain and f(−x)=−f(x). 11. False . Evenfunctionshavegraphsthataresymmetricwithrespecttothey-axis. Oddfunctionshave graphs that are symmetric with respect to the origin. 12. The average rate of change of f(x)=2x3−3 from 0 to 2 is f(2)−f(0) 2(2)3−3−(2(0)3−3) 13−(−3) 16 = = = = 8 . 2−0 2 2 2 Practice Problems 13. Let f(x)=3x2+2x−4. Then (a) f(0)=3(0)2+2(0)−4= −4 . (b) f(−x)=3(−x)2+2(−x)−4= 3x2−2x−4 . (c) −f(x)=−(3x2+2x−4)= −3x2−2x+4 . (d) f(x+1)=3(x+1)2+2(x+1)−4=3(x2+2x+1)+2x+2−4= 3x2+8x+1 . 2 (e) f(x+h)=3(x+h)2+2(x+h)−4=3(x2+2xh+h2)+2x+2h−4= 3x2+6xh+3h2+2x+2h−4 . x 14. Let f(x)= . Then x2+1 0 (a) f(0)= = 0 . 02+1 −x x (b) f(−x)= = − . (−x)2+1 x2+1 x (c) −f(x)= − . x2+1 x+1 x+1 (d) f(x+1)= = . (x+1)2+1 x2+2x+2 x+h x+h (e) f(x+h)= = . (x+h)2+1 x2+2xh+h2+1 15. Let f(x)=|x|+4. Then (a) f(0)=|0|+4= 4 . (b) f(−x)=|−x|+4= |x|+4 . (c) −f(x)=−(|x|+4)= −|x|−4 . (d) f(x+1)= |x+1|+4 . (e) f(x+h)= |x+h|+4 . √ 16. Let f(x)= 3−x. Then √ √ (a) f(0)= 3−0= 3 . (cid:112) √ (b) f(−x)= 3−(−x)= 3+x . √ (c) −f(x)= − 3−x . (cid:112) √ (d) f(x+1)= 3−(x+1)= 2−x . √ (cid:112) (e) f(x+h)= 3−(x+h)= 3−x−h . 17. Becausef(x)=x3−1isdefinedforanyrealnumberx,thedomainoff isthesetof all real numbers , or in interval notation (−∞,∞) . x 18. Because x2 +1 is never equal to zero for any real number x, f(x) = is defined for any real x2+1 numberx. Thedomainoff isthereforethesetof all real numbers ,orinintervalnotation (−∞,∞) . 19. Because the square root of a negative number is not a real number, the value of t2 − 9 must be nonnegative. The solution of the inequality t2−9≥0 is {t|t≤−3}∪{t|t≥3}, so the domain of v is the set of real numbers {t|t≤−3}∪{t|t≥3} , or in interval notation (−∞,−3]∪[3,∞) . 20. Because the expression x−1 appears under thesquare rootand in thedenominator, the value of x−1 must be positive; that is, x−1>0. The solution of this inequality is {x|x>1}, so the domain of g is the set of real numbers {x|x>1} , or in interval notation (1,∞) . 3 21. Because division by zero is not defined, x3−4x=x(x2−4)=x(x−2)(x+2) cannot be zero; that is, x(cid:54)=0,x(cid:54)=2,andx(cid:54)=−2. Thus,thedomainofhisthesetofrealnumbers {x|x(cid:54)=−2,x(cid:54)=0,x(cid:54)=2} . 22. Because the square root of a negative number is not a real number, the value of t + 1 must be nonnegative. The solution of the inequality t+1 ≥ 0 is {t|t ≥ −1}. Moreover, because division by zero is not defined, t−5 cannot be 0; that is, t cannot be equal to 5. The intersection of the sets {t|t≥−1} and {t|t(cid:54)=5} is the set {t|−1≤t<5}∪{t|t>5}. The domain of s is therefore the set of real numbers {t|−1≤t<5}∪{t|t>5} , or in interval notation [−1,5)∪(5,∞) . 23. Let f(x)=−3x+1. Then f(x+h)−f(x) −3(x+h)+1−(−3x+1) −3x−3h+1+3x−1 −3h = = = = −3 . h h h h 1 24. Let f(x)= . Then x+3 f(x+h)−f(x) 1 − 1 (x+h+3)(x+3) = x+h+3 x+3 · h h (x+h+3)(x+3) x+3−(x+h+3) −h 1 = = = − . h(x+h+3)(x+3) h(x+h+3)(x+3) (x+h+3)(x+3) √ 25. Let f(x)= x+7. Then √ √ √ √ f(x+h)−f(x) x+h+7− x+7 x+h+7+ x+7 = · √ √ h h x+h+7+ x+7 x+h+7−(x+7) h 1 = √ √ = √ √ = √ √ . h( x+h+7+ x+7) h( x+h+7+ x+7) x+h+7+ x+7 2 26. Let f(x)= √ . Then x+7 √ √ f(x+h)−f(x) √ 2 − √2 x+h+7 x+7 = x+h+7 x+7 · √ √ h h x+h+7 x+7 √ √ √ √ 2 x+7−2 x+h+7 x+7+ x+h+7 = √ √ · √ √ h x+h+7 x+7 x+7+ x+h+7 2(x+7)−2(x+h+7) = √ √ √ √ h x+h+7 x+7( x+7+ x+h+7) −2h = √ √ √ √ h x+h+7 x+7( x+7+ x+h+7) 2 = −√ √ √ √ . x+h+7 x+7( x+7+ x+h+7) 27. Let f(x)=x2+2x. Then f(x+h)−f(x) (x+h)2+2(x+h)−(x2+2x) x2+2xh+h2+2x+2h−x2−2x = = h h h 2xh+h2+2h h(2x+h+2) = = = 2x+h+2 . h h 4 28. Let f(x)=(2x+3)2. Then f(x+h)−f(x) (2(x+h)+3)2−(2x+3)2 4(x+h)2+12(x+h)+9−(4x2+12x+9) = = h h h 4x2+8xh+4h2+12x+12h+9−4x2−12x−9 = h 8xh+4h2+12h h(8x+4h+12) = = = 8x+4h+12 . h h 29. This graph fails the vertical line test (for example, the vertical line x = 2 intersects the graph in two points), so it does not represent a function . 30. This graph passes the vertical line test, and so it does represent a function . (a) From the graph, the domain of the function is the set of all real numbers or the interval (−∞,∞) , and the range of the function is the set of all positive real numbers or the inter- val (0,∞) . (b) The graph has no x-intercepts, and the y-intercept is 1. Thus, the only intercept is (0,1) . (c) The graph is not symmetric with respect to the x-axis, the y-axis, or the origin. 31. This graph passes the vertical line test, and so it does represent a function . (a) From the graph, the domain of the function is the set {x|−π ≤x≤π} or the closed interval [−π,π] ,andtherangeofthefunctionistheset {y|−1≤y ≤1} ortheclosedinterval [−1,1] . π (cid:16) π (cid:17) (b) Thegraphhasx-interceptsof± anday-interceptof1,sotheinterceptsare ± ,0 and (0,1) . 2 2 (c) The graph is symmetric with respect to the y-axis but not symmetric with respect to the x-axis or the origin. 32. This graph fails the vertical line test (for example, the vertical line x = 2 intersects the graph in two points), so it does not represent a function . 33. (a) f(−1)=−1+3=2; f(0)=0+3=3; f(1)=5; f(8)=−8+2=−6 (b) The graph of f consists of three pieces corresponding to each equation in the definition. On the interval [−2,1), the graph is the line y = x+3, and on the interval (1,∞), the graph is the line y =−x+2. The graph also contains the point (1,5). 5 4 3 2 1 -2 -1 1 2 3 -1 5 (c) The individual components of f have domains of {x|−2 ≤ x < 1}, {x|x = 1}, and {x|x > 1}. Thedomainoff istheunionofthesethreesets;thatis, {x|x≥−2} insetnotationor [−2,∞) in interval notation. Moreover, the individual components of f have ranges of {y|1 ≤ y < 4}, {y|y =5},and{y|y <1}. Therangeoff istheunionofthesethreesets;thatis, {y|y =5 or y <4} in set notation or (−∞,4)∪{5} in interval notation. The x-intercept is 2, and the y-intercept is 3, so the intercepts are (2,0) and (0,3) . 34. (a) f(−1)=2(−1)+5=3; f(0)=−3; f(1)=−5(1)=−5; f(8)=−5(8)=−40 (b) The graph of f consists of three pieces corresponding to each equation in the definition. On the interval [−3,0), the graph is the line y =2x+5, and on the interval (0,∞), the graph is the line y =−5x. The graph also contains the point (0,−3). 5 -3 -2 -1 1 2 3 -5 -10 -15 (c) The individual components of f have domains of {x|−3 ≤ x < 0}, {x|x = 0}, and {x|x > 0}. Thedomainoff istheunionofthesethreesets;thatis, {x|x≥−3} insetnotationor [−3,∞) in interval notation. Moreover, the individual components of f have ranges of {y|−1 ≤ y < 5}, {y|y =−3}, and {y|y <0}. The range of f is the union of these three sets; that is, {y|y <5} in 5 set notation or (−∞,5) in interval notation. The x-intercept is − , and the y-intercept is −3, 2 (cid:18) (cid:19) 5 so the intercepts are − ,0 and (0,−3) . 2 35. (a) f(−1)=1+(−1)=0; f(0)=02 =0; f(1)=12 =1; f(8)=82 =64 (b) The graph of f consists of two pieces corresponding to each equation in the definition. On the interval (−∞,0), the graph is the line y = 1+x, and on the interval [0,∞), the graph is the parabola y =x2. 6 8 6 4 2 -5 -4 -3 -2 -1 1 2 3 -2 -4 (c) The individual components of f have domains of {x|x < 0} and {x|x ≥ 0}. The domain of f is the union of these two sets; that is, all real numbers or the interval (−∞,∞) . Moreover, the individual components of f have ranges of {y|y <1} and {y|y ≥0}. The range of f is the union of these two sets; that is, all real numbers or the interval (−∞,∞) . The x-intercepts are −1 and 0, and the y-intercept is 0, so the intercepts are (−1,0) and (0,0) . 1 √ √ √ 36. (a) f(−1)= =−1; f(0)= 30=0; f(1)= 31=1; f(8)= 38=2 −1 (b) The graph of f consists of two pieces corresponding to each equation in the definition. On the √ interval (−∞,0), the graph is y = 1, and on the interval [0,∞), the graph is y = 3x. x 4 2 -4 -2 2 4 6 8 10 -2 -4 -6 -8 (c) The individual components of f have domains of {x|x < 0} and {x|x ≥ 0}. The domain of f is the union of these two sets; that is, all real numbers or the interval (−∞,∞) . Moreover, the individual components of f have ranges of {y|y <0} and {y|y ≥0}. The range of f is the union of these two sets; that is, all real numbers or the interval (−∞,∞) . The x-intercept is 0, and the y-intercept is 0, so the only intercept is (0,0) . 37. Because the graph of f includes the point (0,3), f(0)=3 ; because the graph also includes the point (−6,−3), f(−6)=−3 . 38. Because the graph of f lies above the x-axis at x=3, f(3) is positive . 7 39. Because the graph of f lies below the x-axis at x=−4, f(−4) is negative . 40. Because the graph of f includes the points (−3,0), (6,0), and (10,0), f(x) = 0 for x=−3, x=6, and x=10 . 41. Because the graph of f lies above the x-axis for −3 < x < 6 and for 10 < x ≤ 11, f(x) > 0 for −3<x<6 and for 10<x≤11 . In interval notation, this can be written as (−3,6)∪(10,11] . 42. The points on the graph of f have x-coordinates between −6 and 11 inclusive. The domain of f is therefore the set of real numbers {x|−6≤x≤11} or the closed interval [−6,11] . 43. Thepointsonthegraphoff havey-coordinatesbetween−3and3inclusive. Therangeoff istherefore the set of real numbers {y|−3≤y ≤3} or the closed interval [−3,3] . 44. Becausethegraphoff includesthepoints(−3,0),(6,0),and(10,0),thex-interceptsare −3, 6, and 10 . 45. Because the graph of f includes the point (0,3), the y-intercept is 3 . 1 46. The graph of the horizontal line y = will intersect the graph of f three times . 2 47. The graph of the vertical line x=5 will intersect the graph of f once . 48. Because the graph of f includes the point (x,3) for 0≤x≤4, f(x)=3 for 0≤x≤4 . 49. Because the graph of f includes the points (−5,−2) and (8,−2), f(x) = −2 for x=−5 and for x=8 . 50. The function f is increasing on the intervals (−6,0) and (8,11) . 51. The function f is decreasing on the interval (4,8) . 52. The function f is constant on the interval (0,4) . 53. Thefunctionf isnonincreasing(thatis, thefunctionisconstantordecreasing)ontheinterval (0,8) . 54. The function f is nondecreasing (that is, the function is constant or increasing) on the intervals (−6,4) and (8,11) . 55. Because division by zero is not defined, x−6 cannot be equal to 0. The domain of g is therefore the set {x|x(cid:54)=6} . 3+2 5 56. Note that g(3)= =− . Because g(3)(cid:54)=14, the point (3,14) is not on the graph of g. 3−6 3 4+2 57. If x=4, then g(x)=g(4)= =−3 ; therefore, the point (4,−3) is on the graph of g. 4−6 x+2 58. If g(x)=2, then =2. Multiplying both sides of this equation by x−6 yields x−6 x+2=2(x−6)=2x−12, so that x=14 . Accordingly, the point (14,2) is on the graph of g. 8 59. The x-intercepts occur when g(x) = 0. The value of g can only be zero when its numerator is zero; therefore, g(x) = 0 only when x+2 = 0. Thus, the graph of g has only one x-intercept, and this intercept is −2 . 0+2 1 1 60. Because g(0)= =− , the y-intercept is − . 0−6 3 3 61. The domain of h is {x|x (cid:54)= ±1}, so for every number x in the domain of h, the number −x is also in the domain. Moreover, −x x h(−x)= =− =−h(x), (−x)2−1 x2−1 so that the function h is odd . Because h is an odd function, its graph is symmetric with respect to the origin . 62. The domain of f is all real numbers, so for every number x in the domain of f, the number −x is also in the domain. Moreover, (cid:112) (cid:112) f(−x)= 3 3(−x)2+1= 3 3x2+1=f(x), so that the function f is even . Because f is an even function, its graph is symmetric with respect to the y-axis . 63. The domain of G is {x|x≥0}. The number x=1 is in the domain of G, but x=−1 is not; therefore, the function G is neither even nor odd . Because G is neither an even nor an odd function, its graph is not symmetric with respect to either the y-axis or the origin . 64. The domain of F is {x|x (cid:54)= 0}, so for every number x in the domain of F, the number −x is also in the domain. Moreover, 2(−x) 2x F(−x)= =− =−F(x), |−x| |x| so that the function F is odd . Because F is an odd function, its graph is symmetric with respect to the origin . 65. Let f(x)=−2x2+4. (a) The average rate of change of f from 1 to 2 is f(2)−f(1) −4−2 −6 = = = −6 . 2−1 2−1 1 (b) The average rate of change of f from 1 to 3 is f(3)−f(1) −14−2 −16 = = = −8 . 3−1 3−1 2 (c) The average rate of change of f from 1 to 4 is f(4)−f(1) −28−2 −30 = = = −10 . 4−1 4−1 3 (d) The average rate of change of f from 1 to x for x(cid:54)=1 is f(x)−f(1) −2x2+4−2 2(1−x2) 2(1−x)(1+x) = = = = −2(1+x) . x−1 x−1 x−1 x−1 9 66. Let s(t)=20−0.8t2. (a) The average rate of change of s from 1 to 4 is s(4)−s(1) 7.2−19.2 −12 = = = −4 . 4−1 4−1 3 (b) The average rate of change of s from 1 to 3 is s(3)−s(1) 12.8−19.2 −6.4 = = = −3.2 . 3−1 3−1 2 (c) The average rate of change of s from 1 to 2 is s(2)−f(1) 16.8−19.2 −2.4 = = = −2.4 . 2−1 2−1 1 (d) The average rate of change of s from 1 to t for t(cid:54)=1 is s(t)−s(1) 20−0.8t2−19.2 0.8(1−t2) 0.8(1−t)(1+t) = = = = −0.8(1+t) . t−1 t−1 t−1 t−1 67. For −1 ≤ x < 0, the graph is a line through the points (−1,1) and (0,0). The slope of this line is m = −1, and the y-intercept is 0; therefore, the equation for this component of the function is y =−x. For0≤x≤2, thegraphisalinethroughthepoints(0,0)and(2,1). Theslopeofthislineis m= 1, and the y-intercept is 0; therefore, the equation for this component of the function is y = 1x. 2 2 Combining these two equations, the definition for this piecewise function is (cid:26) −x, if −1≤x<0 f(x)= 1x, if 0≤x≤2. 2 The domain of f is the set {x|−1≤x≤2} or the closed interval [−1,2] , and the range is the set {y|0≤y ≤1} or the closed interval [0,1] . 68. For x ≤ 0, the graph is a line with slope 1 through the origin, and for 0 < x ≤ 2, the graph is a horizontal line at height y =1. The definition for this piecewise function is therefore (cid:26) x, if x≤0 f(x)= 1, if 0<x≤2. The domain of f is {x|x≤2} in set notation or (−∞,2] in interval notation, and the range is {y|y =1 or y ≤0} in set notation or (−∞,0]∪{1} in interval notation. 69. For x<1, the graph is a horizontal line at height y =−1; at x=1, the value of the function is 0; and for 1 < x ≤ 2, the graph is a line with slope −1 and x-intercept 2. The definition for this piecewise function is therefore  −1, if x<1  f(x)= 0, if x=1  2−x, if 1<x≤2. The domain of f is {x|x≤2} in set notation or (−∞,2] in interval notation, and the range is {y|y =−1 or 0≤y <1} in set notation or {−1}∪[0,1) in interval notation. 10 70. For x < −1, the graph is a horizontal line at height y = 2; at x = −1, the value of the function is 0; and for −1 < x ≤ 1, the graph is a line with slope −2 and y-intercept −1. The definition for this piecewise function is therefore  2, if x<−1  f(x)= 0, if x=−1  −2x−1, if −1<x≤1. The domain of f is {x|x≤1} in set notation or (−∞,1] in interval notation, and the range is {y|y =2 or −3≤y <1} in set notation of [−3,1)∪{2} in interval notation. 71. The definition for this piecewise function is  −x3, if −2<x<1  f(x)= 0, if x=1  x2, if 1<x≤3. The domain of f is the set {x|−2<x≤3} or the interval (−2,3] , and the range is the set {y|−1<y ≤9} or the interval (−1,9] . 72. The definition for this piecewise function is (cid:26) x2−1, if −3≤x≤0 f(x)= x3, if 0<x≤2. The domain of f is the set {x|−3≤x≤2} or the closed interval [−3,2] , and the range is the set {y|−1≤y ≤8} or the closed interval [−1,8] . 73. (a) The cost of manufacturing 100 road bikes is C(100)=0.004(100)3−0.6(100)2+250(100)+100,500=123,500; the cost of manufacturing 101 road bikes is C(101)=0.004(101)3−0.6(101)2+250(101)+100,500≈123,750.60. The average rate of change of C from 100 to 101 road bikes is therefore C(101)−C(100) 123,750.60−123,500 ≈ = $250.60 per bike . 101−100 1 (b) The cost of manufacturing 500 road bikes is C(500)=0.004(500)3−0.6(500)2+250(500)+100,500=575,500; the cost of manufacturing 501 road bikes is C(501)=0.004(501)3−0.6(501)2+250(501)+100,500≈578,155.40. The average rate of change of C from 500 to 501 road bikes is therefore C(501)−C(500) 578,155.40−575,500 ≈ = $2655.40 per bike . 501−500 1 (c) Answers will vary. One possible interpretation is that the unit cost per road bike increases as the number of road bikes manufactured increases.