Solutions for Graphical Approach To Algebra And Trigonometry 7th Us Edition by Hornsby

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Chapter 2: Analysis of Graphs and Functions

2.1: Graphs of Basic Functions and Relations; Symmetry

1. (,). 

2. (,); [0,)

3. (0,0)

4. [0,); [0,)

5. increases

6. (,0]; [0,)

7. x-axis

8. even

9. odd

10. y-axis; origin

11. The domain can be all real numbers; therefore, the function is continuous for the interval (,)  .

12. The domain can be all real numbers; therefore, the function is continuous for the interval (,)  .

13. The domain can only be values where 0; x  therefore, the function is continuous for the interval [0,). 

14. The domain can only be values where 0; x  therefore, the function is continuous for the interval (,0]. 

15. The domain can be all real numbers except 3; therefore, the function is continuous for the interval (,3)(3,). 

16. The domain can be all real numbers except 1; therefore, the function is continuous for the interval (,1)(1,). 

17. (a) The function is increasing for the interval  3, 

(b) The function is decreasing for the interval  ,3

(c) The function is never constant; therefore, none.

(d) The domain can be all real numbers; therefore, the interval (,). 

(e) The range can only be values where 0; y  therefore, the interval [0,). 

18. (a) The function is increasing for the interval  4, 

(b) The function is decreasing for the interval  ,1

(c) The function is constant for the interval  1,4

(d) The domain can be all real numbers; therefore, the interval (,). 

(e) The range can only be values where 3; y  therefore, the interval [3,). 

19. (a) The function is increasing for the interval  ,1

(b) The function is decreasing for the interval  4, 

(c) The function is constant for the interval  1,4

(d) The domain can be all real numbers; therefore, the interval (,). 

(e) The range can only be values where 3; y  therefore, the interval (,3]. 

20. (a) The function is never increasing; therefore, none.

(b) The function is always decreasing; therefore, the interval (,). 

(c) The function is never constant; therefore, none.

(d) The domain can be all real numbers; therefore, the interval (,). 

(e) The range can be all real numbers; therefore, the interval (,). 

21. (a) The function is never increasing; therefore, none

(b) The function is decreasing for the intervals  ,2 and  3, 

(c) The function is constant for the interval (2,3).

(d) The domain can be all real numbers; therefore, the interval (,). 

(e) The range can only be values where 1.5 y  or 2; y  therefore, the interval (,1.5][2,). 

22. (a) The function is increasing for the interval (3,). 

(b) The function is decreasing for the interval (,3). 

(c) The function is constant for the interval  3,3

(d) The domain can be all real numbers except 3; therefore, the interval (,3)(3,). 

(e) The range can only be values where 1; y  therefore, the interval (1,). 

23. Graph 5 ().fxx  See Figure 23. As x increases for the interval (,),  y increases; therefore, the function is increasing.

24. Graph 3 ().fxx  See Figure 24. As x increases for the interval (,),  y decreases; therefore, the function is decreasing.

25. Graph 4 ().fxx  See Figure 25. As x increases for the interval  ,0 y decreases; therefore, the function is decreasing on  ,0

26. Graph 4 ().fxx  See Figure 26. As x increases for the interval  0,  , y increases; therefore, the function is increasing on   0,  [-10,10] by [-10,10] [-10,10] by [-10,10] [-10,10] by [-10,10] [-10,10] by [-10,10] Xscl = 1 Yscl = 1 Xscl = 1 Yscl = 1 Xscl= 1 Yscl = 1 Xscl = 1 Yscl= 1

Figure 23

Figure 24

Figure 25

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Figure 26

27. Graph ()||.fxx  See Figure 27. As x increases for the interval ,0, y increases; therefore, the function is increasing on ,0. 

28. Graph ()||.fxx  See Figure 28. As x increases for the interval  0,  , y decreases; therefore, the function is decreasing on  0,  .

29. Graph 3 (). fxx  See Figure 29. As x increases for the interval (,),  y decreases; therefore, the function is decreasing.

30. Graph (). fxx  See Figure 30. As x increases for the interval  0,  y decreases; therefore, the function is decreasing.

[-10,10] by [-10,10] [-10,10] by [-10,10] [-10,10] by [-10,10] [-10,10] by [-10,10]

Xscl = 1 Yscl = 1 Xscl = 1 Yscl = 1 Xscl= 1 Yscl = 1 Xscl = 1 Yscl= 1

Figure 27

Figure 28 Figure 29 Figure 30

31. Graph 3 ()1.fxx  See Figure 31. As x increases for the interval (,),  y decreases; therefore, the function is decreasing.

32. Graph 2 ()2. fxxx  See Figure 32. As x increases for the interval  1,  y increases; therefore, the function is increasing on  1,  .

33. Graph 2 ()2.fxx  See Figure 33. As x increases for the interval  ,0 y increases; therefore, the function is increasing on  ,0

34. Graph ()|1|.fxx See Figure 34. As x increases for the interval  ,1 y decreases; therefore, the function is decreasing on  ,1 . [-10,10] by [-10,10] [-10,10] by [-10,10] [-10,10] by [-10,10] [-10,10] by [-10,10]

Xscl = 1 Yscl = 1 Xscl = 1 Yscl = 1 Xscl= 1 Yscl = 1 Xscl = 1 Yscl= 1

Figure 31

Figure 32

Figure 33

Figure 34

35. (a) No (b) Yes (c) No

36. (a) Yes (b) No (c) No

37. (a) Yes (b) No (c) No

38. (a) No (b) No (c) Yes

39. (a) Yes (b) Yes (c) Yes

40. (a) Yes (b) Yes (c) Yes

41. (a) No (b) No (c) Yes

42. (a) No (b) Yes (c) No

43. Since the graph is symmetric with respect to the y-axis the function is even.

44. Since the graph is symmetric with respect to the origin the function is odd.

45. Since the graph is symmetric with respect to the origin the function is odd.

46. Since the graph is symmetric with respect to the y-axis the function is even.

47. Since the graph is not symmetric with respect to the y-axis or orign, the function is neither even nor odd.

48. Since the graph is not symmetric with respect to the y-axis or orign, the function is neither even nor odd.

49. (a) Since ()(),fxfx  this is an even function and is symmetric with respect to the y-axis.

See Figure 49a. (b) Since ()(), fxfx  this is an odd function and is symmetric with respect to the origin.

See Figure 49b.

Figure 49a

Figure 49b

50. (a) Since this is an odd function and is symmetric with respect to the origin. See Figure 50a. (b) Since this is an even function and is symmetric with respect to the y-axis. See Figure 50b

Figure 50a

Figure 50b

51. If f is an even function then ()() fxfx  or opposite domains have the same range. See Figure 51

52. If g is an odd function then ()() gxgx  or opposite domains have the opposite range. See Figure 52

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Figure 51

Figure 52

53. This is an even function since opposite domains have the same range.

54. This is an even function since opposite domains have the same range.

55. This is an odd function since opposite domains have the opposite range.

56. This is an odd function since opposite domains have the opposite range.

57. This is neither even nor odd since the opposite domains are neither the opposite or same range.

58. This is neither even nor odd since the opposite domains are neither the opposite or same range.

59. If 42 ()76, fxxx then 4242 ()()7()6()76. fxxxfxxx 

Since ()(),fxfx  the function is even.

60. If 62 ()28, fxxx  then 6262 ()2()8()()28. fxxxfxxx 

Since ()(),fxfx  the function is even.

61. If 3 ()3, fxxx  then 33 ()3()()()3 fxxxfxxx 

and 33 ()(3)()3. fxxxfxxx   Since ()(), fxfx  the function is odd.

62. If 53 ()23, fxxxx  then 5353 ()()2()3()()23 fxxxxfxxxx    and 5353 ()(23)()23. fxxxxfxxxx    Since ()(), fxfx  the function is odd.

63. If 64 ()45 fxxx then 6464 ()()4()5()45. fxxxfxxx    Since ()(),fxfx  the function is even.

64. If ()8,fx  then ()8.fx Since ()(),fxfx  the function is even.

65. If 53 ()37, fxxxx  then 5353 ()3()()7()()37 fxxxxfxxxx    and 5353 ()(37)()37. fxxxxfxxxx    Since ()(), fxfx  the function is odd.

66. If 3 ()4, fxxx  then 33 ()()4()()4 fxxxfxxx 

and 33 ()(4)()4. fxxxfxxx

Since ()(), fxfx  the function is odd.

67. If ()|5|,fxx  then ()|5()|()|5|. fxxfxx  

Since ()(),fxfx  the function is even.

68. If 2 ()1,fxx then 22 ()()1()1. fxxfxx  Since ()(),fxfx  the function is even.

81. If 3 ()3,fxxx then 33 ()()()3()3 fxxxfxxx    and 33 ()(3)()3. fxxxfxxx   Since ()()(), fxfxfx  the function is not symmetric with respect to the y-axis or the origin.

82. If 4 ()52, fxxx then 44 ()()5()2()52 fxxxfxxx    and 44 ()(52)()52. fxxxfxxx    Since ()()(), fxfxfx  the function is not symmetric with respect to the y-axis or the origin. Graph 4 ()52; fxxx the graph supports no symmetry with respect to the y-axis or the origin.

83. If 63 ()4, fxxx  then 6363 ()()4()()4 fxxxfxxx    and 6363 ()(4)()4. fxxxfxxx

   Since ()()(), fxfxfx  the function is not symmetric with respect to the y-axis or the origin. Graph 63 ()4; fxxx  the graph supports no symmetry with respect to the y-axis or the origin.

84. If 3 ()3, fxxx  then 33 ()()3()()3 fxxxfxxx    and 33 ()(3)()3. fxxxfxxx   Since ()(), fxfx  the function is symmetric with respect to the origin. Graph 3 ()3; fxxx  the graph supports symmetry with respect to the origin.

85. If ()6,fx  then ()6,fx Since ()(),fxfx  the function is symmetric with respect to the yaxis. Graph ()6;fx  the graph supports symmetry with respect to the y-axis.

86. If ()||,fxx  then ()|()|()||. fxxfxx    Since ()(),fxfx  the function is symmetric with respect to the y-axis. Graph ()||;fxx  the graph supports symmetry with respect to the y-axis.

87. If 3 1 (), 4 fx x  then 33 11 ()()4()4 fxfx xx 

and 33 11 ()(). 44fxfx xx

Since ()(), fxfx  the function is symmetric with respect to the origin. Graph 3 1 (); 4 fx x  the graph supports symmetry with respect to the origin.

88. If 2 ()(), fxxfxx    then 22 ()()()(). fxxfxxfxx

Since ()(),fxfx  the function is symmetric with respect to the y-axis. Graph 2 ();fxx  the graph

2.2: Vertical and Horizontal Shifts of Graphs

1. The equation 2 yx  shifted 3 units upward is 2 3. yx

2. The equation 3 yx  shifted 2 units downward is 3 2. yx

3. The equation yx  shifted 4 units downward is 4. yx

4. The equation 3 yx  shifted 6 units upward is 3 6. yx

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5. The equation ||yx  shifted 4 units to the right is |4|.yx

6. The equation ||yx  shifted 3 units to the left is |3|.yx

7. The equation 3 yx  shifted 7 units to the left is 3 (7).yx

8. The equation yx  shifted 9 units to the right is 9. yx

9. The equation 2 yx  shifted 2 units downward and 3 units right is  2 32. yx

10. The equation 2 yx  shifted 4 units upward and 1 unit left is  2 14. yx

11. The equation yx  shifted 3 units upward and 6 units to the left is 63. yx

12. The equation ||yx  shifted 1 unit downward and 5 units to the right is |5|1.yx

13. The equation 2 yx  shifted 500 units upward and 2000 units right is  2 2000500. yx

14. The equation 2 yx  shifted 255 units downward and 1000 units left is  2 1000255. yx

15. Shift the graph of f 4 units upward to obtain the graph of g.

16. Shift the graph of f 4 units to the left to obtain the graph of g.

17. The equation 2 3 yx is 2 yx  shifted 3 units downward; therefore, graph B.

18. The equation 2 (3)yx is 2 yx  shifted 3 units to the right; therefore, graph C.

19. The equation (3)2yx is 2 yx  shifted 3 units to the left; therefore, graph A.

20. The equation || 4 yx is ||yx  shifted 4 units upward; therefore; graph A.

21. The equation |4| 3 yx is ||yx  shifted 4 units to the left and 3 units downward; therefore, graph B.

22. The equation |4| 3 yx is ()yfx  shifted 4 units to the right and 3 units downward; therefore, graph C.

23. The equation (3)3yx is 3 yx  shifted 3 units to the right; therefore, graph C.

24. The equation 3 (2)4yx is 3 yx  shifted 2 units to the right and 4 units downward; therefore, graph A.

25. The equation 3 (2)4yx is   ,.ab shifted 2 units to the left and 4 units downward; therefore, graph B.

26. Using 21 YYk  and 0. x  we get 19154. kk   

27. Using 21 YYk  and 0, x  we get 532. kk   

28. Using 21 YYk  and 0 x  , we get 5.541.51.5. k   

29. From the graphs, (6,2) is a point on 1Y and (6,1) a point on 2Y Using 21 YYk  and 6, x  we get 123. kk   

30. From the graphs, (4,3) is a point on 1Y and (4,8) a point on 2Y Using 21 YYk  and4, x  we get 835. kk   

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31. For the equation 2 , yx  the Domain is (,)  and the Range is [0,).  Shifting this 3 units downward gives us: (a) Domain: (,)  (b) Range: [3,). 

32. For the equation 2 , yx  the Domain is (,)  and the Range is [0,).  Shifting this 3 units to the right gives us: (a) Domain: (,)  (b) Range: [0,)  .

33. For the equation ||,yx  the Domain is (,)  and the Range is [0,).  Shifting this 4 units to the left and 3 units downward gives us: (a) Domain: (,)  (b) Range: [3,) 

34. For the equation ||,yx  the Domain is (,)  and the Range is [0,).  Shifting this 4 units to the right and 3 units downward gives us: (a) Domain: (,)  (b) Range: [3,). 

35. For the equation 3 , yx  the Domain is (,)  and the Range is (,).  Shifting this 3 units to the right gives us: (a) Domain: (,)  (b) Range: (,) 

36. For the equation 3 , yx  the Domain is (,)  and the Range is (,)  . Shifting this 2 units to the right and 4 units downward gives us: (a) Domain: (,)  (b) Range: (,) 

37. For the equation 2 , yx  the Domain is (,)  and the Range is [0,).  Shifting this 1 unit to the right and 5 units downward gives us: (a) Domain: (,)  (b) Range: [5,) 

38. For the equation 2 , yx  the Domain is (,)  and the Range is [0,).  Shifting this 8 units to the left and 3 units upward gives us: (a) Domain: (,)  (b) Range: [3,) 

39. For the equation , yx  the Domain is [0,).  and the Range is [0,).  Shifting this 4 units to the right gives us: (a) Domain: [4,).  (b) Range: [0,) 

40. For the equation , yx  the Domain is [0,).  and the Range is [0,).  Shifting this 1 units to the left and 10 units downward gives us: (a) Domain: [1,).  (b) Range: [10,) 

41. For the equation 3 , yx  the Domain is (,)  and the Range is (,)  . Shifting this 1 unit to the right and 4 units upward gives us: (a) Domain: (,)  (b) Range: (,) 

42. For the equation 3 , yx  the Domain is (,)  and the Range is (,)  . Shifting this 7 units to the left and 10 units downward gives us: (a) Domain: (,)  (b) Range: (,) 

43. The graph of ()yfx  is the graph of the equation 2 yx  shifted 1 unit to the right. See Figure 43.

44. The graph of 2 yx is the graph of the equation yx  shifted 2 units to the left. See Figure 44.

45. The graph of 3 1 yx is the graph of the equation 3 yx  shifted 1 unit upward. See Figure 45.

Figure 43

Figure 44

Figure 45

46. The graph of |2|yx is the graph of the equation ||yx  shifted 2 units to the left. See Figure 46.

47. The graph of (1)3yx is the graph of the equation 3 yx  shifted 1 unit to the right. See Figure 47.

48. The graph of || 3 yx is the graph of the equation ||yx  shifted 3 units downward. See Figure 48.

Figure 46

Figure 47

Figure 48

49. The graph of 2 1 yx is the graph of the equation yx  shifted 2 units to the right and 1 unit downward. See Figure 49.

50. The graph of 3 4 yx is the graph of the equation yx  shifted 3 units to the left and 4 units downward. See Figure 50.

51. The graph of ()fx is the graph of the equation 2 yx  shifted 2 units to the left and 3 units upward. See Figure 51.

Figure 49

Figure 50

Figure 51

52. The graph of (4)2 4 yx is the graph of the equation 2 yx  shifted 4 units to the right and 4 units downward. See Figure 52.

53. The graph of |4| 2 yx is the graph of the equation ||yx  shifted 4 units to the left and 2 units downward. See Figure 53.

54. The graph of (3)3 1 yx is the graph of the equation 3 yx  shifted 3 units to the left and 1 unit downward. See Figure 54.

Figure 52

Figure 53

Figure 54

55. Since h and k are positive, the equation is 2 yx  shifted to the right and down; therefore, B.

56. Since h and k are positive, the equation is 2 yx  shifted to the left and down; therefore, D.

57. Since h and k are positive, the equation is 2 yx  shifted to the left and up; therefore, A.

58. Since h and k are positive, the equation is 2 yx  shifted to the right and up; therefore, C.

59. The equation ()2yfx is ()yfx  shifted up 2 units or add 2 to the y-coordinate of each point as follows: (3,2)(3,0);(1,4)(1,6);(5,0)(5,2).    See Figure 59.

60. The equation ()2yfx is ()yfx  shifted down 2 units or subtract 2 from the y-coordinate of each point as follows: (3,2)(3,4);(1,4)(1,2);(5,0)(5,2).    See Figure 60.

Figure 59

Figure 60

61. The equation (2)yfx is ()yfx  shifted left 2 units or subtract 2 from the x-coordinate of each point as follows: (3,2)(5,2);(1,4)(3,4);(5,0)(3,0).    See Figure 61.

62. The equation (2)yfx is ()yfx  shifted right 2 units or add 2 to the x-coordinate of each point as follows: (3,2)(1,2);(1,4)(1,4);(5,0)(7,0).   See Figure 62.

Figure 61

Figure 62

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63. The graph is the basic function 2 yx  translated 4 units to the left and 3 units up; therefore, the new equation is 2 (4)3.yx The equation is now increasing for the interval: (a)  4,  and decreasing for the interval: (b)  ,4 .

64. The graph is the basic function yx  translated 5 units to the left; therefore, the new equation is 5. yx

The equation is now increasing for the interval: (a)  5,  and does not decrease; therefore: (b) none.

65. The graph is the basic function 3 yx  translated 5 units down; therefore, the new equation is 3 5. yx

The equation is now increasing for the interval: (a) (,)  and does not decrease; therefore: (b) none.

66. The graph is the basic function ||yx  translated 10 units to the left; therefore, the new equation is | 10|. yx The equation is now increasing for the interval: (a)  10,  and decreasing for the interval: (b)  ,10

67. The graph is the basic function yx  translated 2 units to the right and 1 unit up; therefore, the new equation is 2 1. yx The equation is now increasing for the interval: (a)  2,  and does not decrease; therefore: (b) none.

68. The graph is the basic function 2 yx  translated 2 units to the right and 3 units down; therefore, the new equation is 2 (2)3.yx The equation is now increasing for the interval: (a)  2,  and decreasing for the interval: (b)  ,2 .

69. The translation is 3 units to the left and 1 unit up; therefore, the new equation is |3| 1. yx The form || yxhk  will equal |3| 1 yx when: 3 h  and 1. k 

70. The equation 2 yx  has a Domain: (,)  and a Range: [0,).  After the translation the Domain is still: (,)  but now the Range is (38,)  , a positive or upward shift of 38 units. Therefore, the horizontal shift can be any number of units, but the vertical shift is up 38. This makes h any real number and 38. k 

71. (a) O(9)2649.4(9)13,26037,104.6  ; In 2014, organic food sales totaled $37,104.6 million.

(b) We will use the point (2005, 13260) and the slope of 2649.4 in the point slope form for the equation of a line. 11()136202649.4(2005)2649.4(2005)13620 yymxxyxyx  (c) 2649.4(20142005)1362037,104.6 y  (d) 12,887 26,5072649.4(2005)1362012,8872649.4(2005)20052009.86 2649.4 xxxx 

The organic food sales reached $26,507 around 2010.

72. (a) (3)23.8(3)248176.6 S  ; In 2013, 176.6 million units were sold.

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(b) We will use the point (2010, 248) and the slope of 23.8 in the point slope form for the equation of a line. 11()24823.8(2010)23.8(2010)248 yymxxyxyx 

(c) 23.8(20132010)24823.8(3)248176.6 y  ; In 2013, 176.6 million units were sold.

(d) 12923.8(2010)24811923.8(2010)520102015 xxxx 

73. 2 (2011)13(20112006)11513(25)115440 U  ; The average U.S. household spent $440 on Apple products in 2011.

74. The formula for ()Wx can be found by shifting 2 ()13(2006)115Uxx to the right 4 units. 2 ()13(2010)115Wxx ; 2 ()13(20152010)11513(25)115440Wx 

In 2015, the average worldwide household spending on Apple products was $440, which equaled U.S. spending 4 years earlier.

75. (a) Enter the year in 1L and enter tuition and fees in 2L . The year 2000 corresponds to 0 x  and so on. The regression equation is 397.23526.yx

(b) Since 0 x  corresponds to 2000, the equation when the exact year is entered is 397.2(2000)3526yx

(c) 397.2(20142000)35269087 y 

76. (a) Enter the year in 1L and enter the percent of women in the workforce in 2L The year 2015 corresponds to 0 x  and so on. The regression equation is 404.129136.25.yx

(b) Since 0 x  corresponds to 2015, the equation when the exact year is entered is  404.1220159136.25.yx

(c)  404.12202520159136.2513,177. y 

77. See Figure 77.

Figure 77

78. 2(2)4 2 312 mm   

79. Using slope-intercept form yields: 111 22(3)22624 yxyxyx

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80. (1,26)  and (3,26)(1,4)   and (3,8)

81. 844 2 312 mm   

82. Using slope-intercept form yields: 222 42(1)42222 yxyxyx      .

83. Graph 1 24yx and 2 22yx See Figure 87. The graph 2y can be obtained by shifting the graph of 1y upward 6 units. The constant 6, comes from the 6 we added to each y-value in Exercise 84. [-10,10] by [-10,10] Xscl = 1 Yscl = 1

Figure 83

84. c; c; the same as; c; upward (or positive vertical)

2.3: Stretching, Shrinking, and Reflecting Graphs

1. The function 2 yx  vertically stretched by a factor of 2 is 2 2.yx 

2. The function 3 yx  vertically shrunk by a factor of one half is 3 1 2 yx 

3. The function yx  reflected across the y-axis is yx 

4. The function 3 yx  reflected across the x-axis is 3 yx 

5. The function yx  vertically stretched by a factor of 3 and reflected across the x-axis is 3.yx 

6. The function yx  vertically shrunk by a factor of one-half and reflected across the y-axis is 1 . 3 yx 

7. The function 3 yx  vertically shrunk by a factor of 0.25 and reflected across the y-axis is 3 0.25()yx  or 3 0.25.yx 

8. The function yx  vertically shrunk by a factor of 0.2 and reflected across the x-axis is 0.2 yx 

9. Graph 1 yx  , 2 3 yx ( 1y shifted up 3 units), and 3 3 yx ( 1y shifted down 3 units). See Figure 9.

10. Graph 3 1 yx  , 3 2 4 yx ( 1y shifted up 4 units), and 3 3 4 yx ( 1y shifted down units).

See Figure 10.

11. Graph 1 yx  , 2 3 yx ( 1y shifted right 3 units), and 3 3 yx ( 1y shifted left 3 units). See Figure 11.

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Figure 9

Figure 10

Figure 11

12. Graph 1 , yx  2 3 yx ( 1y shifted down 3 units), and 3 3 yx ( 1y shifted up 3 units).

See Figure 12.

13. Graph 1 , yx  2 6 yx ( 1y shifted left 6 units), and 3 6 yx ( 1y shifted right 6 units). See Figure 13.

14. Graph 1 yx  , 2 2 yx  ( 1y stretched vertically by a factor of 2), and 3 2.5 yx  ( 1y stretched vertically by a factor of 2.5). See Figure 14

Figure 12

Figure 13

Figure 14

15. Graph 3 1 , yx  3 2 yx  ( 1y reflected across the x-axis), and 3 3 2 yx  ( 1y reflected across the x-axis and stretched vertically by a factor of 2). See Figure 15

16. Graph 2 1 , yx  2 2 (2) 1 yx ( 1y shifted right 2 units and up 1 unit), and 2 3 (2)yx ( 1y shifted left 2 units and reflected across the x-axis). See Figure 16

17. Graph 1 , yx  2 211yx ( 1y reflected across the x-axis, stretched vertically by a factor of 2, shifted right 1 unit, and shifted up 1 unit), and 3 1 4 2 yx ( 1y reflected across the x-axis, shrunk by factor of 1 , 2 and shifted down 4 units). See Figure 17

Figure 15

Figure 16

17

18. Graph 1 , yx  2 yx  ( 1y reflected across the x-axis), and 3 yx  ( 1y reflected across the y-axis). See Figure 18.

19. Graph 2 1 1 yx (which is 2 yx  shifted down 1 unit),

( 1y shrunk vertically by A factor of 1 2 ), and  2 3 21yx ( 1y stretched vertically by a factor of 22 or 4). See Figure 19.

20. Graph 1 3 yx  (which is yx  reflected across the x-axis and shifted up 3 units), 2 33 yx  stretched vertically by a factor of 3), and 3 1 3 3 yx  ( 1y shrunk vertically by a factor of 1 3 ). See Figure 20

Figure 18

Figure 19

Figure 20

21. Graph 3 1 , yx  3 2 yx  ( 1y reflected across the y-axis), and 3 3 (1)yx ( 1y reflected across the y-axis and shifted right 1 unit). See Figure 21.

22. Graph 3 1 , yx  3 2 2 yx  ( 1y reflected across the x-axis and shifted up 2 units), and 3 3 1 yx  ( 1y shifted up 1 unit). See Figure 22.

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Figure

21

22

23. The graph 2 () yfxx  has been reflected across the x-axis, shifted 5 units to the right, and shifted 2 units downward; therefore, the equation of ()gx is 2 ()(5)2.gxx

24. The graph 3 () yfxx  has been shifted 4 units to the right and shifted 3 units upward; therefore, the equation of ()gx is 3 ()(4)3gxx .

25. The graph () yfxx  has been reflected across the y-axis and shifted 1 unit upward; therefore, the equation of ()gx is ()1gxx

26. The graph 3 () yfxx  has been reflected across the x-axis and shifted 2 units to the right; therefore, the equation of ()gx is 3 ()2gxx .

27. 4; x

28. 6; x

29. 2; left; 1 ; 4 ; 3; x downward (or negative)

30. 2 ; ; ; 6 5 yx upward (or positive)

31. 3; right; 6

32. 2; left; 0.5

33. The function 2 yx  is vertically shrunk by a factor of 1 2 and shifted 7 units down; therefore, 2 1 7 2 yx .

34. The function 3 yx  is vertically stretched by a factor of 3 , reflected across the x-axis, and shifted 8 units upward; therefore, 38yx .

35. The function yx  is shifted 3 units right, vertically stretched by a factor of 4.5, and shifted 6 units down; therefore, 4.536.yx

36. The function 3 yx  is shifted 2 units left, vertically stretched by a factor of 1.5 , and shifted 8 units upward; therefore, 3 1.528.yx

37. The function ()32fxx is () fxx  shifted 3 units right and 2 units upward. See Figure 37.

Figure
Figure

38. The function ()23fxx is () fxx  shifted 2 units left and 3 units downward. See Figure 38.

39. The function ()22 fxxx  is () fxx  stretched vertically by a factor of 2 . See Figure 39.

Figure 37

Figure 38

Figure 39

40. The function 2 1 ()(2) 2 fxx is 2 () fxx  shifted 2 units left and shrunk vertically by a factor of 1 2

See Figure 40.

41. The function ()22 fxxx  is () fxx  stretched vertically by a factor of 2. See Figure 41.

42. The function 1 () 2 fxx  is () fxx  shrunk vertically by a factor of 1 2 . See Figure 42

Figure 40

Figure 41

Figure 42

43. The function ()1 fxx  is () fxx  reflected across the x-axis and shifted 1 unit upward. See Figure 43.

44. The function ()221fxx is () fxx  shifted 2 units right, stretched vertically by a factor of 2, and shifted 1 unit downward. See Figure 44.

45. The function ()1(1) fxxx  is () fxx  reflected across both the x-axis, the y-axis and shifted 1 unit right. See Figure 45.

Figure 43

Figure 44

Figure 45

46. The function ()1fxx is () fxx  reflected across the y-axis and shifted 1 unit downward.

See Figure 46.

47. The function ()(1)fxx is () fxx  reflected across the y-axis and shifted 1 unit left. See Figure 47.

48. The function ()2(3) fxx is () fxx  reflected across the y-axis, shifted 3 units right, and shifted 2 units upward. See Figure 48.

Figure 46

Figure 47

Figure 48

49. The function ()(1)3fxx is 3 () fxx  shifted 1 unit right. See Figure 49.

50. The function ()(2)3fxx is 3 () fxx  shifted 2 unit left. See Figure 50.

51. The function 3 () fxx  is 3 () fxx  reflected across the x-axis. See Figure 51.

Figure 49

Figure 50

Figure 51

52. The function 3 ()()1fxx is 3 () fxx  reflected across the y-axis and shifted 1 unit upward. See Figure 52.

Figure 52

Figure 53

Figure 54

53. The function ()2fxx is () fxx  shifted two units to the right and reflected across the y-axis. See Figure 53.

54. The function ()(1)2fxx is 2 () fxx  shifted one unit to the left and reflected across the y-axis. See Figure 54.

55. The function ()3fxx is () fxx  shifted three units to the left and reflected across the y-axis. See Figure 55.

56. The function ()21fxx is () fxx  stretched vertically by a factor of 2, shifted one units to the left and reflected across the y-axis. See Figure 56.

56

57. (a) The equation ()yfx  is ()yfx  reflected across the x-axis. See Figure 57a.

(b) The equation ()yfx  is ()yfx  reflected across the y-axis. See Figure 57b.

(c) The equation 2()yfx  is ()yfx  stretched vertically by a factor of 2. See Figure 57c.

(d) From the graph (0)1. f 

58. (a) The equation ()yfx  is ()yfx  reflected across the x-axis. See Figure 58a.

(b) The equation ()yfx  is ()yfx  reflected across the y-axis. See Figure 58b.

(c) The equation 3()yfx  is ()yfx  stretched vertically by a factor of 3. See Figure 58c.

(d) From the graph (4)1 f 

Figure 58a

Figure 58b

Figure 58c

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Figure 55
Figure
Figure 57a
Figure 57b
Figure 57c

59. (a) The equation ()fx is ()yfx  reflected across the x-axis. See Figure 59a.

(b) The equation ()yfx  is ()yfx  reflected across the y-axis. See Figure 59b.

(c) The equation (1)yfx is ()yfx  shifted 1 unit to the left. See Figure 59c.

(d) From the graph, there are two x-intercepts,  1,0 and  4,0

Figure 59a

Figure 59b

Figure 59c

60. (a) The equation ()yfx  is ()yfx  reflected across the x-axis. See Figure 60a.

(b) The equation ()yfx  is ()yfx  reflected across the y-axis. See Figure 60b.

(c) The equation 1 () 2 yfx  is 20,5. shrunk vertically by a factor of 1 . 2 See Figure 60c.

(d) From the graph ()0fx  for the interval: ,0.

Figure 60a

Figure 60b

Figure 60c

61. (a) The equation ()yfx  is ()yfx  reflected across the x-axis. See Figure 61a.

(b) The equation 1 3 yfx

is ()yfx  stretched horizontally by a factor of 3. See Figure 61b.

(c) The equation 0.5()yfx  is ()yfx  shrunk vertically by a factor of 0.5. See Figure 61c.

(d) From the graph, symmetry with respect to the origin.

Figure 61a

Figure 61b

Figure 61c

62. (a) The equation (2)yfx  is ()yfx  stretched horizontally by a factor of 1 2 See Figure 62a

(b) The equation ()yfx  is ()yfx  reflected across the y-axis. See Figure 62b.

(c) The equation 3()yfx  is ()yfx  stretched vertically by a factor of 3. See Figure 62c.

(d) From the graph, symmetry with respect to the y-axis.

Figure 62a

Figure 62b

Figure 62c

63. (a) The equation ()1yfx is ()yfx  shifted 1 unit upward. See Figure 63a.

(b) The equation 30 is ()yfx  reflected across the x-axis and shifted 1 unit down. See Figure 63b.

(c) The equation 1 2

is

yfx  stretched vertically by a factor of 2 and horizontally by a factor of 2. See Figure 63c.

Figure 63a

Figure 63b

Figure 63c

64. (a) The equation ()2yfx is ()yfx  shifted 2 units downward. See Figure 64a

(b) The equation (1)2yfx is ()yfx  shifted 1 unit right and 2 units upward. See Figure 64b.

(c) The equation 2()yfx  is ()yfx  stretched vertically by a factor of 2. See Figure 64c.

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64a

Figure 64b

64c

65. (a) The equation (2)1yfx is x shrunk horizontally by a factor of  2 and shifted 1 unit upward. See Figure 65a.

(b) The equation 1 21 2 yfx



is ()yfx  stretched vertically by a factor of 2, stretched horizontally by a factor of 2, and shifted 1 unit upward. See Figure 65b.

(c) The equation 1 (2) 2 yfx is ()yfx  shrunk vertically by a factor of 1 2 and shifted 2 units to the right. See Figure 65c.

Figure 65a

Figure 65b

Figure 65c

66. (a) The equation (2)yfx  is ()yfx  shrunk horizontally by a factor of 1 2 See Figure 66a

(b) The equation

stretched horizontally by a factor of 2, and shifted 1 unit downward. See Figure 66b.

(c) The equation 2()1yfx is ()yfx  stretched vertically by a factor of 2 and shifted 1 unit downward. See Figure 66c.

Figure
Figure

67. (a) If (r, 0) is the x-intercept of ()yfx  and ()yfx  is ()yfx  reflected across the x-axis, then (r, 0) is also the x-intercept of ()yfx 

(b) If (r, 0) is the x-intercept of ()yfx  and ()yfx  is ()fx reflected across the y-axis, then  ,0 r is the x-intercept of ()yfx 

(c) If (r, 0) is the x-intercept of ()yfx  and ()yfx  is ()yfx  reflected across both the x-axis and y-axis, then ,0, r is the x-intercept of ()yfx  .

68. (a) If (0,) b is the y-intercept of ()yfx  and ()yfx  is ()yfx  reflected across the x-axis, then  0, b is the y-intercept of ()yfx 

(b) If (0,) b is the y-intercept of ()yfx  and ()yfx  is ()yfx  reflected across the y-axis, then (0,) b is also the y-intercept of ()yfx 

(c) If (0,) b is the y-intercept of ()yfx  and 5()yfx  is ()yfx  stretched vertically by a factor of 5, then (0,5) b is the y-intercept of 5()yfx 

(d) If (0,) b is the y-intercept of ()yfx  and 3()yfx  is ()yfx  reflected across the x-axis and stretched vertically by a factor of 3, then (0,3) b is the y-intercept of 3()yfx 

69. Since (2)yfx is ()yfx  shifted 2 units to the right, the domain of (2)fx is   12,22 or   1,4 , and the range is the same:  0,3

70. Since 5(1) fx  is ()fx shifted 1 unit to the left, the domain of 5(1) fx  is  11,21 or  2,1 and, stretched vertically by a factor of 5, the range is   5(0),5(3) or  0,15 .

71. Since ()fx is ()fx reflected across the x-axis, the domain of ()fx is the same:   1,2, and the range is   3,0.

72. Since (3)1fx  is ()fx shifted 3 unit to the right, (3)1fx  is  13,23 or   2,5, an shifted 1 unit upward, the range is  01,31  or   1,4

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Figure 66a
Figure 66b
Figure 66c

73. Since (2)fx is ()fx shrunk horizontally by a factor of 1 2 , the domain of (2)fx is 11 22(1),(2) 

or 1 ,1 2

, and the range is the same:  0,3

74. Since 2(1) fx is ()fx shifted 1 unit to the right, the domain of 2(1) fx is  11,21 or   0,3, and stretched vertically by a factor of 3, the range is   2(2),2(3) or   0,6.

75. Since 1 3 4 fx 

is ()fx stretched horizontally by a factor of 4 , the domain of 1 3 4 fx

is 

4(1),4(2) or  4,8 , and stretched vertically by a factor of 3, the range is   3(0),3(3) or   0,9

76. Since 2(4) fx is ()fx shrunk horizontally by a factor of 1 4 , the domain of 2(4) fx is 11 44(1),(2) 

or 11 42,;

and reflected across the x-axis while being stretched vertically by a factor of 2, the range is

2(0),2(3)0,6  or   6,0

77. Since ()fx is ()fx reflected across the y-axis, the domain of ()fx is 

(1),(2)1,2  or

2,1; and the range is the same:  0,3

78. Since 2() fx is ()fx reflected across the y-axis, the domain of 2() fx is 

(1),(2)1,2  or   2,1, and reflected across the x-axis while being stretched vertically by a factor of 2, the range is

2(0),2(3)0,6  or 

6,0

79. Since (3)fx is ()fx reflected across the y-axis and shrunk horizontally by a factor of 1 3 , the domain of (3)fx is 1112 (1),(2), 3333

or 21 , 33

, and the range is the same:  0,3 .

80. Since 1 (3) 3 fx is ()fx shifted 3 units to the right, the domain of 1 (3) 3 fx is  13,23 or   2,5, and shrunk vertically by a factor of 1 , 3 the range is

11

or

0,1.

81. Since yx  has an endpoint of (0, 0), and the graph of 10205yx is the graph of yx  shifted 20 units right, stretched vertically by a factor of 10, and shifted 5 units upward, the endpoint of 10205yx is   020,1005  or 20,5. Therefore, the domain is [20,)  , and the range is [5,). 

82. Since yx  has an endpoint of (0, 0), and the graph of 21518yx is the graph of yx  shifted 15 units left, reflected across the x-axis, stretched vertically by a factor of 2, and shifted 18 units downward, the endpoint of 21518yx is   015,2018 or  15,18. Therefore, the domain is [15,)  , and the range, because of the reflection across the x-axis, is   ,18.

83. Since yx  has an endpoint of (0, 0), and the graph of .5105yx is the graph of yx  shifted 10 units left, reflected across the x-axis, shrunk vertically by a factor of .5, and shifted 5 units upward, the endpoint of .5105yx is   010,.505  or 10,5. Therefore, the domain is [10,)  , and the range, because of the reflection across the x-axis, is   ,5.

84. Using ex. 81, the domain is [,), h  and the range is   ,. k 

85. The graph of ()yfx  is ()yfx  reflected across the x-axis; therefore, ()yfx  is decreasing for the interval ,.ab

86. The graph of ()yfx  is ()yfx  reflected across the y-axis; therefore, ()yfx  is decreasing for the interval  , ba .

87. The graph of ()yfx  is ()yfx  reflected across both the x-axis and y-axis; therefore, ()yfx  is increasing for the interval  , ba

88. The graph of ()ycfx  is ()yfx  reflected across the x-axis; therefore, ()ycfx  is decreasing for the interval ,.ab

89. (a) the function is increasing for the interval:  1,2 .

(b) the function is decreasing for the interval: ,1.

(c) the function is constant for the interval: 2,. 

90. (a) the function is increasing for the interval: ,1.

(b) the function is decreasing for the interval: 1,2.

(c) the function is constant for the interval: 2,. 

91. (a) the function is increasing for the interval: 1,. 

(b) the function is decreasing for the interval: 2,1.

(c) the function is constant for the interval: ,2.

92. (a) the function is increasing for the interval: ,3.

(b) the function is decreasing for the interval: 3,.

(c) the function is not constant for any interval.

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93. From the graph, the point on 2y is approximately 8,10.

94. From the graph, the point on 2y is approximately  27,15.

95. Use two points on the graph to find the slope. Two points are  2,1 and 1,1; therefore, the slope is

The stretch factor is 2 and the graph has been shifted 2 units to the left and 1 unit down; therefore, the equation is 221.yx

96. Use two points on the graph to find the slope. Two points are  1,2 and 5,0; therefore, the slope is 0221 5142

The shrinking factor is 1 , 2 the graph has been reflected across the x-axis, shifted 1 unit to the right, and shifted 2 units upward; therefore, the equation is 1 12. 2 yx

97. Use two points on the graph to find the slope. Two points are  0,2 and 1,1; therefore, the slope is 123 3. 101 mm    The stretch factor is 3, the graph has been reflected across the x-axis, and shifted 2 units upward; therefore, the equation is 32.yx

98. Use two points on the graph to find the slope. Two points are  1,2 and 0,1; therefore, the slope is

The stretch factor is 3 and the graph has been shifted 1 unit to the left and 2 units down; therefore, the equation is 312.yx

99. Use two points on the graph to find the slope. Two points are (0,-4) and (3,0); therefore, the slope is 4044 0333

The stretch factor is 4 3 and the graph has been shifted 4 units down.; therefore, the equation is 4 4 3

100. Use two points on the graph to find the slope. Two points are (0,4) and (-3,5); therefore, the slope is 5411

The stretch factor is 1 3 and the graph has been shifted 3 units to the left and 5 units up.; therefore, the equation is 1 35 3 yx .

101. Since ()yfx  is symmetric with respect to the y-axis, for every (,) xy on the graph, (,) xy is also on the graph. Reflection across the y-axis reflect onto itself and will not change the graph. It will be the same.

102. The graph is no longer symmetric to the origin since it has been moved away from the origin.

Reviewing Basic Concepts (Sections 2.1—2.3)

1. (a) The function () fxx  shifted up one unit yields the function ()1fxx . Therefore, this function has a domain of   ,  and a range of of   1,  . The function is increasing from   0,  and decreasing from   ,0

(b) The function 2 () fxx  shifted to the right 2 units yields the function 2 ()2fxx . Therefore, this function has a domain of   ,  and a range of of   0,  . The function is increasing from   2,  and decreasing from   ,2

(c) The function () fxx  reflected over the x-axis yields the function () fxx  . Therefore, this function has a domain of   0,  and a range of of   ,0 . The function is never increasing and decreasing from   0, 

2. (a) If ()yfx  is symmetric with respect to the origin, then another function value is (3)6. f 

(b) If ()yfx  is symmetric with respect to the y-axis, then another function value is (3)6. f 

(c) If ()(),() fxfxyfx  is symmetric with respect to both the x-axis and y-axis, then another function value is (3)6. f 

(d) If (),yfx  ()yfx  is symmetric with respect to the y-axis, then another function value is (3)6. f 

3. (a) The equation 

(b) The equation 2 7 yx is 2 yx  shifted 7 units downward: D.

(c) The equation 2 7 yx  is 2 yx  stretches vertically by a factor of 7: E.

(d) The equation (7)2yx is 2 yx  shifted 7 units to the left: A.

(e) The equation

stretches horizontally by a factor of 3: C.

4. (a) The equation 2 2 yx is 2 yx  shifted 2 units upward: B.

(b) The equation 2 2 yx is 2 yx  shifted 2 units downward: A.

(c) The equation  2 2 yx is

(d) The equation (2)2yx

yx  shifted 2 units to the left: G.

shifted 2 units to the right: C.

(e) The equation 2 2 yx  is 2 yx  stretched vertically by a factor of 2: F.

(f) The equation 2 yx  is 2 yx  reflected across the x-axis D.

(g) The equation  2 21 yx is 2 yx  shifted 2 units to the right and 1 unit upward: H.

(h) The equation  2 21 yx is 2 yx  shifted 2 units to the left and 1 unit upward: E.

5. (a) The equation 4 yx is yx  shifted 4 units upward. See Figure 5a.

(b) The equation 4 yx is yx  shifted 4 units to the left. See Figure 5b.

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(c) The equation 4 yx is yx  shifted 4 units to the right. See Figure 5c.

(d) The equation 24 yx is yx  shifted 2 units to the left and 4 units down. See Figure 5d.

(e) The equation 24 yx is yx  reflected across the x-axis, shifted 2 units to the right, and 4 units upward. See Figure 5e.

5d

5e

6. (a) The graph is the function () fxx  reflected across the x-axis, shifted 1 unit left and 3 units upward Therefore, the equation is 13. yx

(b) The graph is the function () gxx  reflected across the x-axis, shifted 4 units left and 2 units upward. Therefore, the equation is 42. yx

(c) The graph is the function () gxx  stretches vertically by a factor of 2, shifted 4 units left and 4 units downward. Therefore, the equation is 244.yx

(d) The graph is the function () fxx  shrunk vertically by a factor of 1 , 2 shifted 2 units right and 1 unit downward. Therefore, the equation is 1 21. 2 yx

7. (a) The graph of ()gx is the graph ()fx shifted 2 units upward. Therefore, 2. c 

(b) The graph of ()gx is the graph ()fx shifted 4 units to the left. Therefore, 4. c 

8. The graph of yFxh  is a horizontal translation of the graph of yFx  The graph of  yFxh  is not the same as the graph of  , yFxh  because the graph of  yFxh  is a vertical translation of the graph of  . yFx 

9. (a) If f is even, then ()().fxfx  See Figure 9a.

(b) If f is odd, then ()(). fxfx  See Figure 9b.

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Figure 5a
Figure 5b
Figure 5c
Figure
Figure

Figure 9a

Figure 9b

10. (a) (7)8.25(7)27.585.25 R  , In 2017, Google’s ad revenues were $85.25 billion.

(b) Using the point (2010, 27.5) and the slope of 8.255 with the point slope formula we will have 27.58.25(2010)8.25(2010)27.5 yxyx 

(c) 8.25(20172010)27.58.25(7)27.585.25 y  , In 2017, Google’s ad revenues were $85.25 billion.

(d) 41.5 698.25(2010)27.541.58.25(2010)20102015 8.25 xxxx 

11. (a) ()0: {3,4} fx 

(b) ()0:fx  for the intervals (,3)(4,). 

(c) ()0:fx  for the interval (3,4).

12. (a) ()0: {2} fx 

(b) ()0:fx  for the interval 2,. 

(c) ()0:fx  for the interval ,2.

13. (a) ()0: {4,5} fx 

(b) ()0:fx  for the intervals ( ] [ ) ,45, -¥-È¥

(c) ()0:fx  for the interval [4,5].

14. (a) ()0:fx  never; therefore: 

(b) ()0:fx  for the interval [1,). 

(c) ()0:fx  never; therefore:  .

2.4: Absolute Value Functions

1. We reflect the graph of ()yfx  across the x-axis for all points for which 0. y  Where 0, y  the graph remains unchanged. See Figure 1. The domain and range of ()fx is   ,  and   1,  respectively. The domain and range of ()fx is   ,  and   0,  respectively.

2. We reflect the graph of ()yfx  across the x-axis for all points for which 0. y  Where 0, y  the graph remains unchanged. See Figure 2. The domain and range of ()fx is   ,  and  ,1 respectively. The domain and range of ()fx is   ,  and   0,  respectively.

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3. We reflect the graph of ()yfx  across the x-axis for all points for which 0. y  Where 0, y  the graph remains unchanged. See Figure 3. The domain and range of ()fx is   ,  and 

,  respectively. The domain and range of ()fx is 

,  and   0,  respectively.

Figure 1

Figure 2

Figure 3

4. We reflect the graph of ()yfx  across the x-axis for all points for which 0. y  Where 0, y  the graph remains unchanged. See Figure 4. The domain and range of ()fx is 

,  and

,  respectively. The domain and range of ()fx is 

and

0,  respectively.

,

5. Since for all y,y  0, the graph remains unchanged. That is, ()yfx  has the same graph as ().yfx  The domain and range of ()fx is   0,  and   0,  respectively. The domain and range of ()fx is

0,  and

 0,  respectively.

6. We reflect the graph of ()yfx  across the x-axis for all points for which 0. y  Where 0, y  the graph remains unchanged. See Figure 6. The domain and range of ()fx is   0,  and   0,  respectively. The domain and range of ()fx is   0,  and   0,  respectively.

7. We reflect the graph of ()yfx  across the x-axis for all points for which 0. y  Where 0, y  the graph remains unchanged. See Figure 7. The domain and range of ()fx is     ,00,   and     ,00,   respectively. The domain and range of ()fx is     ,00,   and   0,  respectively.

8. We reflect the graph of ()yfx  across the x-axis for all points for which 0. y  Where 0, y  the graph remains unchanged. See Figure 8. The domain and range of ()fx is   0,  and   1,  respectively.

The domain and range of ()fx is   0,  and   0,  respectively.

Figure 4
Figure 6

9. We reflect the graph of ()yfx  across the x-axis for all points for which 0. y  Where 0, y  the graph remains unchanged. See Figure 9. The domain and range of ()fx is   2,  and   1,  respectively. The domain and range of ()fx is   2,  and   0,  respectively.

7

8

9

10. We reflect the graph of ()yfx  across the x-axis for all points for which 0. y  Where 0, y  the graph remains unchanged.

11. If  5, fa  then  55. fa 

12. Since  2 fxx  is an even function,  2 fxx  and  2 fxx  are the same graph.

13. If  2 , fxx  then  22 yfxyxyx      Therefore, the range of yfx  is [0,).

14. If the range of ()yfx  is [–2, ), the range of yfx  is [0,) since all negative values of y are reflected across the x-axis.

15. If the range of ()yfx  is   ,2, the range of ()yfx  is [2,) since all negative values of y are reflected across the x-axis.

16. ()fx is greater than or equal to 0 for any value of x. Since 1 is less than 0, –1 cannot be in the range of f

17 From the graph of 2 (1)2yx the domain of ()fx is (,),  and the range is [2,). 

From the graph of  2 12 yx the domain of ()fx is (,),  and the range is [0,). 

18. From the graph of 1 2 2 yx  the domain of ()fx is (,),  and the range is (,).  From the graph of 1 2 2 yx  the domain of ()fx is (,),  and the range is [0,). 

19. From the graph of  2 12yx the domain of ()fx is (,),  and the range is (,1].  From the graph of  2 12yx the domain of ()fx is (,),  and the range is [1,). 

20. From the graph of 22 yx the domain of ()fx is (,),  and the range is (,2].  From the graph of  22 yx the domain of ()fx is (,),  and the range is [2,). 

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Figure
Figure
Figure

21. From the graph, the domain of ()fx is   2,3, and the range is   2,3. For the function (),yfx  we reflect the graph of ()yfx  across the x-axis for all points for which 0 y  , and, where 0, y  the graph remains unchanged. Therefore, the domain of ()yfx  is [–2, 3], and the range is [0, 3].

22. From the graph, the domain of ()fx is [–3, 2], and the range is [–2, 2]. For the function (),yfx  we reflect the graph of ()yfx  across the x-axis for all points for which 0 y  and, where 0, y  the graph remains unchanged. Therefore, the domain of ()yfx  is [–3, 2], and the range is [0, 2].

23. From the graph, the domain of ()fx is [–2, 3], and the range is [–3, 1]. For the function (),yfx  we reflect the graph of ()yfx  across the x-axis for all points for which 0 y  , and, where 0, y  the graph remains unchanged. Therefore, the domain of ()yfx  is [–2, 3], and the range is [0, 3].

24. From the graph, the domain of ()yfx  is [–3, 3], and the range is [–3, –1]. For the function (),yfx  we reflect the graph of ()yfx  across the x-axis for all points for which 0 y  , and, where 0, y  the graph remains unchanged. Therefore, the domain of ()yfx  is [–3, 3], and the range is [1, 3].

25. (a) The function ()yfx  is the function ()yfx  reflected across the y-axis. See Figure 25a.

(b) The function ()yfx  is the function ()yfx  reflected across both the x-axis and y-axis. See Figure 25b.

(c) For the function ()yfx  we reflect the graph of ()yfx  (ex. b) across the x-axis for all points for which 0 y  , and where 0, y  the graph remains unchanged. See Figure 25c.

25a

25b

25c

26. (a) The function ()yfx  is the function ()yfx  reflected across the y-axis. See Figure 26a.

(b) The function ()yfx  is the function ()yfx  reflected across both the x-axis and y-axis. See Figure 26b

(c) For the function ()yfx  we reflect the graph of ()yfx  (ex. b) across the x-axis for all points for which 0 y  , and, where 0, y  the graph remains unchanged. See Figure 26c

Figure
Figure
Figure

26a

26b

27. The graph of ()yfx  can not be below the x-axis; therefore, Figure A shows the graph of (),yfx  while Figure B shows the graph of ().yfx 

28. The graph of ()yfx  can not be below the x-axis; therefore, Figure B shows the graph of (),yfx  while Figure A shows the graph of ().yfx 

29. See Figure 29a and 29b

30. See Figure 30a and 30b.

31. See Figure 31a and 31b.

29a

30b

32. See Figure 32a and 32b.

33. See Figure 33a and 33b.

34. See Figure 34a and 34b.

31a

31b

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Figure
Figure
Figure 26c
Figure
Figure 29b
Figure 30a
Figure
Figure
Figure

35. See Figure 35a and 35b.

36. See Figure 36a and 36b.

37. See Figure 37a and 37b.

38. See Figure 38a and 38b.

39. See Figure 39a and 39b.

40. See Figure 40a and 40b.

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Figure 32a
Figure 32b
Figure 33a
Figure 33b
Figure 34a
Figure 34b
Figure 35a
Figure 35b
Figure 36a
Figure 36b
Figure 37a
Figure 37b

Figure 38a

39b

38b

40a

39a

40b

41. (a) From the graph, 12yy  at the coordinates  1,5 and 6,5; therefore, the solution set is   1,6.

(b) From the graph, 12yy  for the interval 1,6.

(c) From the graph, 12yy  for the intervals  ,16,.

42. (a) From the graph, 12yy  at the coordinates  0,2 and 8,2; therefore, the solution set is   0,8.

(b) From the graph, 12yy  for the intervals  ,08,.

(c) From the graph, 12yy  for the interval 0,8.

43 (a) From the graph, 12yy  at the coordinate 4,1; therefore, the solution set is   4.

(b) From the graph, 12yy  never occurs; therefore, the solution set is the null set. .

(c) From the graph, 12yy  for all values for x except 4; therefore, the solution set is the intervals  ,44,.

44. (a) From the graph, 12yy  never occurs; therefore, the solution set is 

(b) From the graph, 12yy  for all values for x; therefore, the solution set is ,. 

(c). From the graph, 12yy  never occurs; therefore, the solution set is  .

45. The V-shaped graph is that of ().56,fxx since this is typical of the graphs of absolute value functions of the form ()|| fxaxb 

46. The straight line graph is that of ()314gxx which is a linear function.

47. The graphs intersect at 8,10, so the solution set is   8.

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Figure
Figure
Figure
Figure
Figure

48. From the graph, ()() fxgx  for the interval ,8.

49. From the graph, ()() fxgx  for the interval 8,. 

50. If  .563140 xx then .56314. xx Therefore, the solution is the intersection of the graphs, or   8.

51. (a) 4949xx    or 495xx    or 13. x  The solution set is   13,5, which is supported by the graphs of 1 4 yx and 2 9. y 

(b) 4949xx    or 495xx    or 13. x  The solution is  ,13   5,  , which is supported by the graphs of 1 4 yx and 2 9. y 

(c) 49949135. xxx      The solution is 13,5, which is supported by the graphs of |4| x  and 2 9. y 

52. (a) 3535xx    or 358xx    or 2. x  The solution set is   2,8, which supported by the graphs of 1 3 yx and 2 5. y 

(b) 3535xx    or 358xx    or 2. x  The solution is  ,28,  which is supported by the graphs of 1 3 yx and 2 5. y 

(c) 3553528. xxx      The solution is 2,8, which is supported by the graphs of 1 3 yx and 2 5. y 

53. (a) 723723 xx   or 72324 xx   or 2102 xx   or 5. x  The solution set is   2,5, which is supported by the graphs of 1 72 yx  and 2 3. y  (b) 723723 xx   or 72324 xx   or 2102 xx   or 5. x  The solution set is     ,25,, which is supported by the graphs of 1 72 yx  and 2 3. y  (c) 7233723102452 xxxx   

 or 25. x  The solution is   2,5, which is supported by the graphs of 1 72 yx  and 2 3. y 

54. (a) 936936 xx

or 936315 xx

or x = -1. The solution set is   5,1, which is supported by the graphs of 1 93 yx  and 2 6. y

(b) 936936 xx

or 936315

or 335 xx

or 1. x

The solution is     ,51,, which is supported by the graphs of 1 93 yx  and 2 6. y 

(c) 9366936331515 xxxx

or 51. x  The solution is   5,1, which is supported by the graphs of 1 93 yx  and 2 6. y 

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55. (a) 2135212 xx    or 21221 xx    or 1 23 2 xx 

 or 3 2 x  The solution set is 31 22,,

which is supported by the graphs of 1 213yx and 2 5. y  (b) 31 21352212321. 22 xxxx   

The solution is 31 22,,

which is supported by the graphs of 1 213yx and 2 5. y

(c) 2135212 xx    or 212 x 

which is supported by the graphs of 1 213yx and 2

the solution set is ,

which is supported by the graphs of 1 474yx and 2 4.

which is supported by the graphs of 1 57 yx  and 2 0. y 

(b) 570570 xx   or 57075 xx

which is supported by the graphs of 1 57 yx  and 2 0. y 

(c) 55 5700570575. 77 xxxx       The solution set is 5 , 7      which is supported by the graphs of 1 57 yx  and 2 0. y 

58. (a) Absolute value is always positive; therefore, the solution set is , which is supported by the graphs of 1 8 yx   and 2 4. y 

(b) Absolute value is always positive, and so cannot be less than 4; therefore, the solution set is , which is supported by the graphs of 1 8 yx   and 2 4. y 

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(c) Absolute value is always positive, and so is always greater than 4; therefore, the solution set is ,,  which is supported by the graphs of 1 8 yx   and 2 4. y 

59. (a) Absolute value is always positive; therefore, the solution set is , which is supported by the graphs of 1 23.6yx and 2 1. y 

(b) Absolute value is always positive, and so cannot be less than or equal to 1; therefore, the solution set is , which is supported by the graphs of 1 23.6yx and 2 1. y 

(c) Absolute value is always positive, and so is always greater than 1; therefore, the solution is ,,  which is supported by the graphs of 1 23.6yx and 2 1. y 

60. 24210248248 xxx   

or 24824 xx

or 2122 xx 

 and 6. x  Therefore, the solution set is

61. 3434834312434434 xxxx

or 43430

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69. 231231 xx    or 23124 xx    or 2 3(3)45 (3). 9 (3) f g



or 1. x  Therefore, the solution is  ,12,.

70. 431431 xx   or 43133 xx   or 351 xx   or 5 3 x  . Therefore, the solution is

5 ,1, 3

71. 383383 xx   or 38335 xx   or 5 311 3 xx   or 11 3 x  Therefore, the solution is 511 ,,. 33

72. Absolute value is always positive, and so is always greater than 1; therefore, the solution is ,. 

73. 11 6060 33 xx   or 11 606 33 xx   or 1 618 3 xx   or 18. x  Therefore, the solution is every real number except 18:  ,1818,.

74. Absolute value is always positive, and cannot be less than 0; therefore, the solution set is 

75. Absolute value is always positive, and so cannot be less than or equal to 6; therefore, the solution set is 

76. Absolute value is always positive, and so cannot be less than 4; therefore, the solution set is 

77. Absolute value is always positive, and so is always greater than 5; therefore, the solution is ,. 

78. To solve such an equation, we must solve the compound equation axbcxd  or axbcxd  . The solution set consists of the union of the two individual solution sets.

79. (a) 3127178

(b) Graph 1 31yx

and 2 27.yx

which is for the interval

(c) Graph 1 31

See Figure 79. From the graph, ()() fxgx

when 12yy

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(b) Graph 1 4 yx and 2 712.yx See Figure 80. From the graph, ()() fxgx  when 12yy  which is for the interval 8 ,1. 3

(c) Graph 1 4 yx and 2 712.yx See Figure 80. From the graph, ()() fxgx  when 12yy  which is for the interval  8 ,1,. 3

[-20,20] by [-10,50] [-10,10] by [-4,16] Xscl = 2 Yscl = 5 Xscl = 1 Yscl = 1

79

Figure 80 81. (a) 2 25332 3 xxxx

the solution

(b) Graph 1225andy3 yxx  . See Figure 81. From the graph, ()() fxgx  when 12yy  , which it is for the

(c) Graph 1225andy3 yxx  See Figure 81. From the graph, ()() fxgx  when 12yy  , which it is for the interval 2 ,8 3

. Therefore, the solution set is 35 , 28

(b) Graph 1 51yx and 2 34.yx See Figure 82. From the graph, ()() fxgx  when 12yy  , which it is for the interval 35 ,,. 28

(c) Graph 1 51yx and 2 34yx . See Figure 82. From the graph, ()() fxgx  when 12yy  , which it is for the interval 35 28,.

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Figure

Figure 81

(b) Graph 1 3 yx and 2 1 8 3 yx . See Figure 84. From the

(c) Graph 1 3 yx and 2 1 8 3 yx . See Figure 84. From the graph, ()() fxgx  when 12yy  , which it is for the interval 3315

(b). Graph 1 41yx or 2 y46.

See Figure 85. From the graph ()() fxgx  when 12yy  , which it is for the interval 7 , 8

(c) Graph 1 41yx or 2 y46. x  See Figure 85. From the graph, ()() fxgx  when 12yy  , which is for the interval 7 , 8

. [-10,10] by [-10,10] [-10,10] by [-10,10] [-10,10] by [-10,10] Xscl = 1 Yscl = 1 Xscl = 1 Yscl = 1 Xscl= 1 Yscl = 1

Figure 84

Figure 85

86. (a) 696393 xx   

Figure 86

Therefore, the solution set is 1 2

(b) Graph 12 69 and63yxyx  . See Figure 86. From the graph, ()() fxgx  when 12yy  , which it is for the interval 1 ,. 2

(c) Graph 12 69 and63yxyx  . See Figure 86. From the graph, ()() fxgx  when 12yy  , which it is for the interval 1 ,. 2

87. (a) 0.2510.7530.5048 xxxx      or 0.251(0.753) xx  0.2510.753 xx 

2. x  Therefore, the solution set is   2,8.

(b) Graph 12 .251 and.753.yxyx  See Figure 87. From the graph, ()() fxgx  when 12yy  , which it is for the interval (2,8).

(c) Graph 12 .251 and.753.yxyx  See Figure 87. From the graph, ()() fxgx  when 12yy  , which it is for the interval  ,28,.

88. (a) .402.60520735 xxxx      or .402(.605) xx  .402.6053 xxx    . Therefore, the solution set is   3,35.

(b) Graph 12 .402 andy.605yxx  . See Figure 88. From the graph, ()() fxgx  when 12yy  , which it is for the interval (3,35).

(c ) Graph 12 .402 andy.605yxx  . See Figure 88. From the graph, ()() fxgx  when 12yy  , which it is for the interval  ,335, 

Copyright © 2019 Pearson Education, Inc.

[-20,20] by [-4,16] [-30,50] by [-5,30] [-10,10] by [-10,10] [-10,10] by [-10,10]

Xscl = 2 Yscl = 1 Xscl = 5 Yscl = 5 Xscl= 1 Yscl = 1 Xscl = 1 Yscl = 1

Figure 87

Figure 88

Figure 89

Figure 90

89. (a) 310(310)310310 xxxx     there are an infinite number of solutions.

Therefore, the solution set is  ,  .

(b) Graph 12 310 andy310yxx  . See Figure 89. From the graph, ()() fxgx  when 12yy  , for which there is no solution.

(c ) Graph 12 310 andy310yxx  . See Figure 89. From the graph, ()() fxgx  when 12yy  , for which there is no solution.

90. (a) 56(56)5656 xxxx     there are an infinite number of solutions.

Therefore, the solution set is  ,  .

(b) Graph 12 56 andy56yxx  . See Figure 90. From the graph, ()() fxgx  when 12yy  , for which there is no solution.

(c ) Graph 12 56 andy56yxx  . See Figure 90. From the graph, ()() fxgx  when 12yy  , for which there is no solution.

91. Graph 1 16yxx and 2 11. y  See Figure 91 From the graph, the lines intersect at (3,11) and (2,9). Therefore, the solution set is   3,8.

92. Graph 1 221yxx and 2 9. y  See Figure 92. From the graph, the lines intersect at (4,9) and (2,9). Therefore, the solution set is   4,2.

93. Graph 1 4 yxx and 2 8. y  See Figure 93. From the graph, the lines intersect at (2,8) and (6,8). Therefore, the solution set is   2,6.

94. Graph 1 .52.254yxx  and 2 9. y  See Figure 94. From the graph, the lines intersect at (20,9) and (4,9). Therefore, the solution set is   20,4.

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[-10,10] by [-4,16] [-10,10] by [-4,16] [-10,10] by [-4,16] [-30,10] by [-4,16] Xscl = 1 Yscl = 1 Xscl = 1 Yscl = 1 Xscl= 1 Yscl = 1 Xscl = 5 Yscl= 1

Figure 91

Figure 92

Figure 93

Figure 94

95. (a) 50222250222872. TTT     

(b) The average monthly temperatures in Boston vary between a low of 28 F and a high of 72 F. The monthly averages are always within 22 of 50 F.

96. (a) 10363610362646. TTT      .

(b) The average monthly temperatures in Chesterfield vary between a low of 26 F and a high of 46 F.

The monthly averages are always within 36 of 10 F.

97. (a) 61.512.512.561.512.54974. TTT   

(b) The average monthly temperatures in Buenos Aires vary between a low of 49 F (possibly in July) and a high of 74 F (possibly in January). The monthly averages are always within 12.5 of 61.5 F.

98. (a) 43.58.58.543.58.53552. TTT     

(b) The average monthly temperatures in Punta Arenas vary between a low of 35 F and a high of 52 F.

The monthly averages are always within 8.5 of 43.5 F.

99. (a) The max circumference is 10.1 and the minimum circumference is 9.9. This can be represented as 100.1 C  (b)

102. (a)

103. (a)   4840.04 x 

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Chapter 2: Analysis of Graphs of Functions

(b)   4840.044120.040.044120.0411.96412.04 xxxx  2.993.01 x 

104. (a)

 512100.1 x  (b)   512100.15220.10.15220.121.9522.1 xxxx  4.384.42 x 

105. (a)

40.50.001 x  (b)

 40.50.0014.50.0010.0014.50.0014.4994.501 xxxx 

106. (a)   71.20.002 x  (b) 

71.20.0025.80.0020.0025.80.0025.8025.798 xxxx  5.7985.802 x 

107. 8.01.51.58.01.56.59.5; xxx      therefore, the range is the interval   6.5,9.5.

108. If 680780 730 2   is the midpoint, then 680780680730730780730 FF  

507305073050 FF  

109. (a) 11612599dd

(b) 1712012017or12017137orP=103 PPPP

.

110. If 98148 123 2   is the midpoint, then 9814898123123123148123 xxx

  2512325|123|25 xx   (or 12325 x  ); and if 1626 2  21  is the midpoint, then 16261621212621 xx    5215 x   215 x  (or 215 x  ).

11. If 2761 xx then 27(61)0. xx Graph 1 27(61),yxx See Figure 111. The x -intercept is 2; therefore, the solution set is   2.

112. If 3121 xx  then 312(1)0. xx  Graph 1 312(1),yxx  See Figure 112. The equation is 0,  or the graph intersects or is above the x-axis, for the interval:   2.75,6.5.

113. If 4.56xx then  4.560.xx Graph  1 4.56,yxx See Figure 113. The equation is 0,  or the graph is above the x-axis, for the interval: (,). 

114. If 2834 xx then  28340. xx Graph  1 2834,yxx See Figure 114. The equation is 0,  or the graph is above the x-axis, for the interval: (,). 

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[-10,10] by [-10,10] [-10,10] by [-10,10] [-10,10] by [-10,10] [-10,10] by [-4,16]

Xscl = 1 Yscl = 1 Xscl = 1 Yscl = 1 Xscl= 1 Yscl = 1 Xscl = 1 Yscl= 1

Figure 111

Figure 112

113

114

115. If 34314 xx then  343140. xx Graph  1 34314,yxx See Figure 115. The equation is 0,  or the graph is below the x-axis, never or for the solution set: .

116. If 13610xx  then  136100.xx  Graph

1 13610yxx  See Figure 100. The equation is 0,  or the graph intersects or is below the x-axis, never or for the solution set: 

[-10,10] by [-10,10] [-10,10] by [-10,10] Xscl = 1 Yscl = 1 Xscl = 1 Yscl = 1

2.5: Piecewise-Defined Functions

1. (a) From the graph, the speed limit is 40 mph.

(b) From the graph, the speed limit is 30 mph for 6 miles.

(c) From the graph, (5)40mph; f  (13)30mph; f  and (19)55mph. f 

(d) From the graph, the graph is discontinuous at 4,6,8,12,and16. x  The speed limit changes at each discontinuity.

2. (a) From the graph, the Initial amount was: $1,000; and the final amount was: $600.

(b) From the graph, (10)$900;(50)$600. ff The function f is not continuous.

(c) From the graph, the discontinuity shows 3 drops or withdrawals.

(d) From the graph, the largest drop or largest withdrawal of $300 occurred after 15 minutes.

(e) From the graph, the one increase or deposit was $200.

3. (a) From the graph, the Initial amount was: 50,000 gal.; and the final amount was: 30,000 gal.

(b) From the graph, during the first and fourth days.

(c) From the graph, (2)45,000gal;(4)40,000gal ff .

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Figure
Figure
Figure 115
Figure 116

(d) From the graph, between days 1 and 3 the water dropped: 50,00040,00010,000 5,000 22  gal./day.

4. (a) From the graph, when x = 1 the tank had 20 gallons of gas.

(b) Since x = 0 represents 3:15 pm and 3 hours later the tank was filled to 20 gallons, the time was 6:15 pm

(c) When the graph is horizontal, the engine is not running; when the graph is decreasing, the engine is burning gasoline; and when the graph is increasing, gasoline is being put into the tank.

(d) When the graph is decreasing, gasoline was burned the fastest between 1 and 2.9 hours

5. (a)

7. (a) (5)2(5)3 f

(0)(0)0 f

8. (a) (5)2(5)10 f

(c) (0)3(0)11 f

(b) (1)3(1)14 f

9. Yes, continuous. See Figure 9.

10. Yes, continuous. See Figure 10.

11. Not continuous. See Figure 11

Figure 9

12. Not continuous. See Figure 12.

13. Not continuous. See Figure 13.

14. Not continuous. See Figure 14.

Figure 12

Figure 10

Figure 11

Figure 13

Figure 14

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15. Not continuous. See Figure 15.

16. Not continuous. See Figure 16.

17. Yes, continuous. See Figure 17.

15

18. Not continuous. See Figure 18.

19. Yes, continuous. See Figure 19.

20. Yes, continuous. See Figure 20.

18

16

17

19

20

21. Look for a 2 yx  graph if 0; x  and a linear graph if 0 x  . Therefore: B.

22. Look for a yx  graph if 1; x  and a reflected 2 yx  graph if 1 x  . Therefore: A.

23. Look for a horizontal graph above the x-axis if 0; x  and a horizontal graph below the x-axis if 0 x  .Therefore: D.

24. Look for a yx  graph if 0; x  and a reflected 2 yx  graph if 0 x  .Therefore: C.

25. Graph 1 (1)(3)(2)(3),yxxx  See Figure 25.

26. Graph 1 (6)(3)(36)(3),yxxxx  See Figure 26.

27. Graph 1 (4)(2)(12)(2),yxxxx  See Figure 27.

28. Graph 1 (21)(0)()(0),yxxxx  See Figure 28.

Figure
Figure
Figure
Figure
Figure
Figure

Figure 25

Figure 26

Figure 27

29. Graph 1 (2)(4)()(4and5)(3)(5), yxxxxxxx

See Figure 29.

Figure 28

30. Graph 1 (2)(3)(31)(3and2)(4)(2), yxxxxxxx  See Figure 30.

31. Graph 1 11 222(2)(2),yxxxx

See Figure 31.

32. Graph 32 1 (5)(0)()(0),yxxxx  See Figure 32.

Figure 29

33. From the graph, the function is

34. From the graph, the function

35. From the graph, the

36. From the graph, the function

37. From the graph, the function is

38. From the graph, the function is

39. There is an overlap of intervals since the number 4 satisfies both conditions. To be a function, every x-value is used only once.

Figure 30
Figure 31
Figure 32

40. The value (4) f cannot be found since using the first formula, (4)11, f  and using the second formula, (4)16. f  To have two different values violates the definition of function.

41. The graph of yx = is shifted 1.5 units downward.

42. The graph of yx = is reflected across the y-axis.

43. The graph of yx = is reflected across the x-axis.

44. The graph of yx = is shifted 2 units to the left.

45. Graph  1.5, yx=- See Figure 45.

46. Graph  , yx =- See Figure 46.

47. Graph  , yx =- See Figure 47.

48. Graph 2, yx=+ See Figure 48.

Figure 45

Figure 46

Figure 47

Figure 48

49. (a) In 2012 there were 200 million users (2012, 200) and in 2015 there were 700 million users (2015, 700). The slope is given as 700200500 201520123  . The equation is given as

500 2002012 3 Wx

 500 2012200 3 Wx between 2012 and 2015 inclusively. In 2018 there were 1000 million users (2018, 1000). The slope is given as 1000700300 100 201820153 

The equation is  7001002015Wx 100(2015)700Wx between 2015 and 2018.

(b) See Figure 49. The function is continuous on this time period.

(c) From 2012 to 2015, the number of active WhatsApp users increased , on average, by about 166.7 million users per year. From 2015 to 2018, this increase slowed to 100 million per year, on average.

50. (a) In January 2014 there were 46 million users (0, 46) and in July 2015 there were 94 million users (17, 94). The slope is given as 9446488 180183  . The equation is given as 8 46 3 Sx between 0 and 18 months inclusively.

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In January 2017 there were 160 million users (36, 160). The slope is given as 160946611 3618183  . The equation is given as 1111 160(36)28 33 SxSx  between 18 and 36 months.

(b) See Figure 50. The function is continuous on the time interval.

(c) From January 2014 to July 2015, the number of daily active Snapchat users increased, on average, by about 2.7 million per month. From July 2015 to January 2017, this increased to 3.7 million per month, on average.

Figure 49

Figure 50

51. When 03 x  the slope is 5, which means the inlet pipe is open and the outlet pipe is closed; when 35 x  the slope is 2, which means both pipes are open; when 58 x  the slope is 0, which means both pipes are closed; when 810 x  the slope is 3 , which means the inlet pipe is closed and the outlet pipe is open.

52. (a) Since for each year given the shoe size is 1 size smaller than his age, the formula is 1 yx

(b) From the table and question, graph the function. See figure 50.

53. (a) (1.5)0.70,(3)0.91, ff It costs $0.70 to mail 1.5 oz and $0.91 to mail 3 oz.

(b) Domain:   0,5; Range:  0.49,0.70,0.91,1.12,1.33 . See figure 53.

54. (a) (1967)0.0475(1967)93.30.1325 f  , In 1967, the average Super Bowl ad cost $0.1325 million.

(b) (1998)0.475(1998)93.31.605 f  , In 1998, the average Super Bowl ad cost $1.605 million.

(c) (2013)0.179(2013)356.0375.01 f  , In 2013, the average Super Bowl ad cost $5.01 million

(d) The function ()fx is continuous on its domain since there are no breaks in the graph of ()fx .

(e) See Figure 54.

Figure 52

Figure 53

Figure 54

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55. (a) From the table, graph the piecewise function. See Figure 55.

(b) The likelihood of being a victim peaks from age 16 up to age 20, then decreases.

56. (a) From the table, graph the piecewise function. See Figure 56.

(b) Housing starts increased, decreased, and increased, with the maximum occurring during the 1960’s.

57. (a) From the graph, the highest speed is 60 mph and the lowest speed is 20 mph.

(b) There are approximately 4 miles of highway with a speed of 55 mph.

(c) Fromt the graph, (4)40;(12)20;(18)7. fff

56. From the table, graph the piecewise function. See Figure 56.

Figure 55

Figure 56

Figure 58

59. (a) A 3.5 minute call would round up to 4 minutes. A 4 minute call would cost: .503(.25)$1.25. 

(b) We use a piecewise defined function where the cost increases after each whole number as follows: .50if 01 .75if 12 () 1.00if 23 1.25if 34 1.50if 45 x x fxx x x 

. Another possibility is

.50if 01 () .50.251if 15 x fx xx

60. (a) Since cost is rounded down to the nearest 2 foot interval, we can use the greatest integer function. The function is ().80 2 fxx =

(b) Graph ().80 2 fxx =

(c)

from 618. x ££

from 618. x ££ See Figure 60.

8.5 (8.5).80.804.25.80(4)(8.5)$3.20 2 ff====

15.2 (15.2).80.807.6.80(7)(15.2)$5.60 2 ff   

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[6,18] by [0,8]

Xscl = 1 Yscl = 1

Figure 60

61. For x in the interval   0,2,25. y  For x in the interval   2,3,25328. y  For x in the interval   3,4,28331 y  and so on. The graph is a step function. In this case, the first step has a different width. See Figure 61.

62. Sketch a piecewise function that measures miles over minutes. The first piece increases at a rate of 40 mph or 402 603  miles per minute, for 30 minutes (the time it takes to travel 20 miles at that rate). The second piece stays at a constant distance of 20 miles for the 2 hour period at the park. The third piece decreases (returns home) at a rate of 20 mph or 201 603  miles per minute, for 60 minutes (the time it takes to travel 20 miles at that rate). See Figure 62.

63. Sketch a piecewise function that fills a tank at a rate of 5 gallons a minute for the first 20 minutes (the time it takes to fill the 100 gallon tank) and then drains the tank at a rate of 2 gallons per minute for 50 minutes (the time it takes to drain the 100 gallon tank). See Figure 63.

64. (a) Use the points (1990, 1.3) and (2007, 2.7) to find the slope. 2.71.31.4 0.082 2007199017

11()1.30.082(1990)0.082161.88 yymxxyxyx

Use the points (2007, 2.7) and (2016, 1.5) to find the slope. 1.52.71.2 0.13 201620079 m

1.50.13(2016)0.13263.58 yxyx

(b) Use the two functions above to create 0.082161.8819902007 () 0.13263.5820072016

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Figure 61
Figure 62
Figure 63

2.6: Operations and Composition

1. 22(25)25. xxxxE  

2. 22(25)25. xxxxB  

3. 232 (25)25. xxxxF  

4. 2 . 25 x D x 

5. 22 (25)42025 xxxA  

6. 22 (2()5)25. xxC  

7.  2 (3)((3))(2(3)1)(5)53(5)40fgfgff 

8.  2 (2)((2))((2)3(2))(2)2(2)15gfgfgg 

9.  222 ()(())(21)(21)3(21)44163422 fgxfgxfxxxxxxxx  

10.  222 ()(())(3)2(3)1261 gfxgfxgxxxxxx  

11.  2 3(3)(3)((3)3(3))(2(3)1)23fgfg 

12.  2 (5)(5)(5)53(5)(2(5)1)1fgfg 

13. 

14.

15.

16.

2 (4)(4)(4)43(4)2(4)1196fgfg

2 13(1) (1)2 (1) (1)2(1)13

2 43(4) (4)28 (4)4 (4)2(4)17

22. Because the two compositions are opposites of each other.

23. (a)

()()(41)(63)102 fgxxxx  , ()()(41)(63)24 fgxxxx  22 ()()(41)(63)2412632463 fgxxxxxxxx 

(b) All values can replace x in all three equations; therefore, the domain is (,)  in all cases.

(c) 41 (); 63 fx x gx

all values can replace x, except 1 ; 2 therefore, the domain is 11 ,,. 22

(d)  

()()()4(63)1241212411; fgxfgxxxx   all values can be input for x; therefore, the domain is (,) 

(e)

24. (a)

()()()6(41)32463243; gfxgfxxxx   all values can replace x; therefore, the domain is (,) 

()()(92)(52)711 fgxxxx  , ()()(92)(52)37 fgxxxx 

22 ()()(92)(52)4518104104918 fgxxxxxxxx 

(b) All values can replace x in all three equations; therefore, the domain is (,)  in all cases.

(c) 92 (); 52 fx x gx

all values can replace x, except 2 5 ; therefore, the domain is 22 ,,. 55

(d)   ()()()92(52)9104105; fgxfgxxxx   all values can replace x; therefore, the domain is (,) 

(e)   ()()()5(92)2451021043; gfxgfxxxx   all values can replace x; therefore, the domain is (,) 

25. (a) ()()32 fgxxx  , ()()32 fgxxx  , ()()3(2) fgxxx 

(b) All values can replace x in all three equations; therefore, the domain is (,)  in all cases.

(c) 3 (); 2 fx x gx   

all values can replace x, except 0; therefore, the domain is (,0)(0,)  

(d)   ()()()(2)323; fgxfgxxx   all values can replace x; therefore, the domain is (,).  (e)    ()()()2323; gfxgfxxx   all values can replace x; therefore, the domain is (,) 

26. (a) ()()24(1)241 fgxxxxx  , ()()24(1)241 fgxxxxx  ()()24(1)(1)24 fgxxxxx 

(b) All values can replace x in all three equations; therefore, the domain is (,)  in all cases.

(c) 24 () 1 fx x gx

all values can replace x, except 1; therefore, the domain is (,1)(1,).  

(d)   ()()()2(1)422; fgxfgxxx   all values can replace x, so the domain is (,). 

(e)   ()()()241; gfxgfxx  all values can replace x, so the domain is (,). 

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27. (a) 3333 ()()4(5)45 fgxxxxx  , 3333 ()()4(5)45 fgxxxxx   3 3 ()()4(5) fgxxx

(b) All values can replace x in all three equations; therefore, the domain is (,)  in all cases.

(c) 3 3 4 () (5) fx x gx     

all values can replace x, except 3 5 , so the domain is 33 (,5)(5,).  

(d)  3 33 3 ()()()(5)49; fgxfgxxx   all values can replace x, so the domain is (,). 

(e)  3 3 ()()()45459; gfxgfxxxx   all values can replace x, so the domain is (,).  28. (a)

3333 ()()63(21)6321 fgxxxxx 

3333 ()()63(21)6321 fgxxxxx 

3 3 ()()63(21) fgxxx

(b) All values can replace x in all three equations; therefore, the domain is (,)  in all cases.

(c) 3 3 63 (); (21) fx x gx

all values can replace x, except 3 1 2 , so the domain is 33 11 ,, 22

(d)

(e)

3 33 3 ()()()63(21)63; fgxfgxxx   all values can replace x, so the domain is (,). 

3 3 ()()()26312(63)1613; gfxgfxxxx   all values can replace x; therefore, the domain is (,). 

29. (a) 22 ()()3(1)31 fgxxxxx  , 22 ()()3(1)31 fgxxxxx    2 ()()3(1) fgxxx

(b) All values can replace x in all three equations; therefore, the domain is (,)  in all cases.

(c) 2 3 (); (1) fx x gx

all values can replace x, except 1; therefore, the domain is  ,11,  

(d)  222 ()()()(1)321324; fgxfgxxxxxx   all values can replace x; therefore, the domain is (,). 

(e)    22 ()()()3131; gfxgfxxx  all values can replace x; therefore, the domain is, (,). 

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30. (a) ()()24()2422 fgxxxxx  , ()()24()2422 fgxxxxx    ()()24()2422 fgxxxxx 

(b) All values can replace x in all three equations; therefore, the domain is (,)  in all cases.

(c) 2 24 ();fx x gx    

all values can replace x, except 0; therefore, the domain is  ,00,. 

(d)  22 ()()()24()24; fgxfgxxx   all values can replace x; therefore, the domain is (,). 

(e)    22 ()()()2424; gfxgfxxx   all values can replace x; therefore, the domain is (,). 

31. (a) From the graph, 4(2)2. 

(b) From the graph, 1(3)4. 

(c) From the graph, (0)(4)0. 

(d) From the graph, 11 . 33 

32. (a) From the graph, 022. 

(b) From the graph, 213.

(c) From the graph, (2)(1)2. 

(d) From the graph, 4 2. 2 

33. (a) From the graph, 033. 

(b) From the graph, 145.

(c) From the graph, (2)(1)2. 

(d) From the graph, 3 0  undefined.

34. (a) From the graph, 312.

(b) From the graph, 202.

(c) From the graph, (3)(1)3. 

(d) From the graph, 3 3. 1 

35. (a) From the table, 7(2)5. 

(b) From the table, 1055. 

(c) From the table, (0)(6)0. 

(d) From the table, 5 0  undefined.

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36. (a) From the table, 549. 

(b) From the table, 000. 

(c) From the table, (4)(2)8. 

(d) From the table, 8 8. 1 

37. See Table 37.

38. See Table 38.

39. (a)   ()(4)(4), fgfg   so from the graph find (4)0. g  Now find (0)4; f  therefore, ()(4)4. fg  

(b)

 ()(3)(3), gfgf   so from the graph find (3)2. f  Now find 22; g  therefore, ()(3)2. gf  

(c)   ()(2)(2), ffff   so from the graph find (2)0. f  Now find (0)4; f  therefore, ()(2)4. ff  

40. (a)   ()(2)(2), fgfg   so from the graph find (2)2. g  Now find (2)4; f  therefore, ()(2)4. fg  

(b)   ()(0)(0), gggg   so from the graph find (0)2. g  Now find (2)2; g  therefore, ()(0)2. gg  

(c)   ()(4)(4), gfgf   so from the graph find (4)2. f  Now find (2)2; g  therefore, ()(4)2. gf  

41. (a)   ()(1)(1), fgfg   so from the graph find (1)2. g  Now find (2)3; f  therefore, ()(1)3. fg  

(b)

 ()(2)(2), gfgf  so from the graph find (2)3. f  Now find (3)2; g  therefore, ()(2)2. gf  

(c)   ()(2)(2), gggg  so from the graph find (2)1. g  Now find (1)0; g  therefore, ()(1)0. gg  

42. (a)   ()(2)(2), fgfg  so from the graph find (2)4. g  Now find (4)2; f  therefore, ()(2)2. fg  

(b)

(c)

43. (a)

(b)

  ()(1)(1), gfgf   so from the graph find (1)1. f  Now find (1)1; g  therefore, ()(1)1. gf  

  ()(0)(0), ffff   so from the graph find (0)0. f  Now find (0)0; f  therefore, ()(0)0. ff  

  ()(1)(1), gfgf   so from the table find (1)4. f  Now find (4)5; g  therefore, ()(1)5. gf  

  ()(4)(4), fgfg   so from the table find (4)5. g  (5) f is undefined; therefore, ()(4) fg  is undefined.

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Figure 37
Figure 38

2: Analysis of Graphs of Functions

(c)   ()(3)(3), ffff   so from the table find (3)1. f  Now find (1)4; f  therefore, ()(3)4. ff  

44. (a)   ()(1)(1), gfgf   so from the table find (1)2. f  Now find (2)4; g  therefore, ()(1)4 gf   . (b)   ()(4)(4), fgfg   so from the table we find  4 g is undefined; therefore, ()(4) fg  is undefined.

(c)   ()(3)(3), ffff   so from the table find (3)6. f  Now find (6)7; f  therefore, ()(3)7. ff  

45. From the table, (3)4 g  and (4)2. f 

46. From the table, (6)7 f  and (7)0. g 

47. Since 312 YYY   and 1, X  we solve   12 (1). YY First solve 2 2 (1)1, Y  now solve  1 2153; Y  therefore, 3 3. Y 

48. Since 312 YYY   and 2, X  we solve   12 (2). YY First solve 2 2 (2)4, Y  now solve 1 2(4)53; Y  therefore, 3 3. Y 

49. Since 312 YYY   and 7, X  we solve   12 (7). YY First solve  2 2 749, Y  now solve 1 2(49)593; Y  therefore, 3 93. Y 

50. Since 312 YYY   and 8, X  we solve   12 (8). YY  First solve  2 2 864, Y  now solve 1 2(64)5123; Y  therefore, 3 123. Y 

51. (a)

  23 ()()()(31); fgxfgxxx   all values can be input for x; therefore, the domain is (,).  (b)   32363 ()()()()3()131; gfxgfxxxxx   all values can be input for x; therefore, the domain is (,). 

(c)   339 ()()()(); ffxffxxx   all values can be input for x; therefore, the domain is (,). 

52. (a)  22 11 ()()()22; fgxfgx xx

 all values can be input for x, except 0; therefore, the domain is (,0)(0,).  

(b)  2 1 ()()(); (2) gfxgfx x   all values can be input for x, except 2; therefore, the domain is (,2)(2,).  

(c)   ()()()2(2); ffxffxxx   all values can be input for x; therefore, the domain is (,). 

53. (a)   2 ()()()11; fgxfgxxx   all values can be input for x; therefore, the domain is (,). 

(b)    22 ()()()1()1; gfxgfxxx   only values where 2 1 x  can be input for x; therefore, the domain is   1,1.

(c)   224 ()()()(); ffxffxxx   all values can be input for x; therefore, the domain is (,). 

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54. (a)   4242 ()()()(34)232; fgxfgxxxxxxx   all values can be input for x; therefore, the domain is (,). 

(b)   42 ()()()(2)(2)3(2)4; gfxgfxxxx  all values can be input for x; therefore, the domain is (,). 

(c)   ()()()(2)24; ffxffxxx   all values can be input for x; therefore, the domain is (,). 

55. (a) 11 ()()[()]; (5)151 fgxfgx

all values can be input for x, except 1 ; 5 therefore, the domain is 11 ,,. 55

(b) 15 ()()[()]5; 11 gfxgfx xx

all values can be input for x, except, −1; therefore, the domain is (.1)(1,).  (c) 1111 ()()[()]; 112 1 2 1 111 1 ffxffxx

 all values can be input for x, except those that make 1 0 2 x x    or undefined. That would be −1 and −2; therefore, the domain is (,2)(2,1)(1,). 

56. (a) 22 ()()[()](4)444; fgxfgxxx   only values where 2 4 x  can be input for x; therefore, the domain is [−2, 2].

(b) 2 ()()[()]4(4); gfxgfxx   only values where 2 (4)4 x  can be input for x; therefore, the domain is [−6, −2].

(c) ()()[()](4)48; ffxffxxx   all values can be input for x; therefore, the domain is (,). 

57. (a) 3232 ()()[()]2(45)18101; fgxfgxxxxx   all values can be input for x; therefore, the domain is (,). 

(b) 32 ()()[()]4(21)5(21) gfxgfxxx  322 4(81261)5(441) xxxxx  32232 3248244(20205)322841; xxxxxxxx  all values can be input for x; therefore, the domain is (,). 

(c) ()()[()]2(21)143; ffxffxxx   all values can be input for x; therefore, the domain is (,). 

58. (a) (23)32 ()()[()]; 23 fxx gxfgxx    all values can be input for x; therefore, the domain is (,). 

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(b) 3 ()()[()]23(3)3; 2 x gfxgfxxx

all values can be input for x; therefore, the domain is (,).

all values can be input for x; therefore, the domain is (,). 

59. (a)  ()(())(5)5,fgxfgxf

all values can be input for x, thefore the domain is  ,

.

(b)  ()(())(5)5,gfxgfxg  all values can be input for x, therefore the domain is

(c)  ()(())(5)5,ffxffxf  all values can be input for x, therefore the domain is  , 

60. (a) 1 ()()[()] 1 fgxfgx x   all values can but 1 can be input for x; therefore, the domain is

,11,. 

(b) 11 ()()[()]1 gfxgfxx xx 

all values but 0 can be input for x; therefore, the domain is  ,00,. 

(c) 1 ()()[()] 1 ffxffxx x   all values can be input for x; therefore, the domain is (,). 

61. (a) 1 ()()[()] 2 fgxfgx x   all values can but 4 can be input for x; therefore, the domain is

,44,. 

(b) 112 ()()[()] 22 2 gfxgfxx xx x

all values bigger than 2 can be input for x; therefore, the domain is

all values but 5 2 can be input for x; therefore, the domain is 55 ,,. 22

62. (a) 11 ()()[()] 36146 fgxfgx xx

all values can but 2 3 can be input for x; therefore, the domain is 22 ,,. 33

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(b) ()()[()]3631611 gfxgfx xx

all values but –1 can be input for x; therefore, the domain is  ,11,.

()()[()]

all values but 2 can be input for x; therefore, the domain is

63. Let () fxc

then

Therefore, there is only one value in the range of  ()ffx 

64. Let () fxk  and g()xmxb

Therefore, there is only one value k in the range of  ()fgx 

65. 1 ()()[()]4(2)222 4

66.

69. Graph 3 3 12 6, 6, yxyx and 3 yx  in the same viewing window. See Figures 69. The graph of 2 y can be obtained by reflecting the graph of 1y across the line 3 .yx 

70. Graph  12 1 53, 3, 5 yxyx  and 3 yx  in the same viewing window. See Figures 70. The graph of 2 y can be obtained by reflecting the graph of 1y across the line 3 .yx  [-10,10] by [-10,10] [-10,10] by [-10,10] Xscl = 1 Yscl = 1 Xscl = 1 Yscl = 1

Figure 69

Figure 70

71. (a) 2222 ()4()()424; fxxfxhxhxxhh 

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(b)

Chapter 2: Analysis of Graphs of Functions

 2222 ()()448 fxfhxhxh 

72. (a) 222222 ()5()5()()5(2)()5105; fxxxfxhxhxhxxhhxhxxhhxh  (b)

73. (a)

 2222 ()()5555 fxfhxxhhxxhh 

 222222 ()3()3()()332233 fxxxfxhxhxhxhxxhhxxhxhh  ; (b)

74. (a)

  2222 ()()3333 fxfhxxhhxxhh 

 ()()()23333223223 fxxfxhxhxxhhxhxhxhxh  (b) 33 ()() fxfhxh 

75. Using  fxhfx h  gives:

76. Using  fxhfx h  gives:

77. Using

gives:

443434

Using

 = 2 1 4 1 2 4. 2 xhhh xh h  

79. Using ()() fxhfx h  gives: 3332233223 22 ()3333 33. xhxxxhxhhxxhxhh xxhh hhh  

80.

81. Using ()() fxhfx h  gives:

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82. Using ()() fxhfx h  gives:

83. Using ()() fxhfx h  gives:

84. Using ()() fxhfx h  gives:

85. Using ()() fxhfx h  gives:

2(22)2 11 222222 1 2()2 2() xxhh xhxxxhxxh hhhxxh

86. Using ()() fxhfx h  gives:

222222 22 ()22 11 () xxhxxxhhxhh xhxxxhxxhxxh

87. One possible solution is 2 () fxx  and ()62.gxx Then 2 ()()[()](62). fgxfgxx 

88. One possible solution is 2 () fxx  and 2 ()1112. gxxx  Then 22 ()()[()](1112). fgxfgxxx  

89. One possible solution is () fxx  and 2 ()1.gxx Then 2 ()()[()]1. fgxfgxx 

90. One possible solution is 3 () fxx  and ()23.gxx Then 3 ()()[()](23). fgxfgxx 

91. One possible solution is ()12fxx and ()6. gxx  Then ()()[()]612. fgxfgxx 

92. One possible solution is 3 ()4fxx and ()23.gxx Then 3 ()()[()]234. fgxfgxx 

93. (a) There are 5280 feet in a mile. ()5280 Fxx 

(b) There are 12 inches in a foot. ()12 Ixx 

(c) ()()(())I(5280)12(5280)63,360 IFxIFxxxx  

(d) The number of inches in a mile.

94. (a) There are 2000 pounds in a ton. ()2000 Txx 

(b) There are 16 ounces in a pound. ()16 Oxx 

(c)   ()(())(2000)16(2000)32,000 OTxOTxOxxx  

(d) The number of ounces in a ton.

95. (a) With a cost of $10 to produce each item and a fixed cost of $500, the cost function is ()10500.Cxx

(b) With a selling price of $35 for each item, the revenue function is ()35. Rxx 

(c) The profit function is ()()()()35(10500)()25500. PxRxCxPxxxPxx 

(d) A profit is shown when ()02550002550020. Pxxxx        Therefore, 21 items must be produced and sold to realize a profit.

(e) Graph 1 25500.yx The smallest whole number for which ()0Px  is 21. Use a window of [0,30] by [1000,500], for example.

96. (a) With a cost of $11 to produce each item and a fixed cost of $180, the cost function is ()11180.Cxx

(b) With a selling price of $20 for each item, the revenue function is ()20. Rxx 

(c) The profit function is ()()()()20(11180)()9180. PxRxCxPxxxPxx  

(d) A profit is shown when ()091800918020. Pxxxx        Therefore, 21 items must be produced and sold to realize a profit.

(e) Graph 1 9180.yx The smallest whole number for which ()0Px  is 21. Use a window of [5,30] by [200,200], for example.

97. (a) With a cost of $100 to produce each item and a fixed cost of $2700, the cost function is ()1002700.Cxx

(b) With a selling price of $280 for each item, the revenue function is ()280. Rxx 

(c) The profit function is ()()()()280(1002700)()1802700. PxRxCxPxxxPxx

(d) A profit is shown when ()018027000180270015. Pxxxx

Therefore, 16 items must be produced and sold to realize a profit.

(e) Graph 1 1802700,yx the smallest whole number for which ()0Px  is 16. Use a window of [0,30] by [3000,500], for example.

98. (a) With a cost of $200 to produce each item and a fixed cost of $1000, the cost function is ()2001000.Cxx

(b) With a selling price of $240 for each item, the revenue function is ()240. Rxx 

(c) The profit function is ()()()()240(2001000)()401000. PxRxCxPxxxPxx     

(d) A profit is shown when ()0401000040100025. Pxxxx        Therefore, 26 items must be produced and sold to realize a profit.

(e) Graph 1 401000,yx the smallest whole number for which ()0Px  is 26. Use a window of [5,40] by [1200,600], for example.

99. (a) If 3 4 (), 3 Vrr   then a 3 inch increase would be 3 4 ()(3), 3 Vrr   and the volume gained would be 4433 ()(3). 33 Vrrr 

(b) Graph 33 1 44 (3) 33 yxx  in the window [0,10] by [0,1500]. See Figure 99. Although this appears to be a portion of a parabola, it is actually a cubic function.

(c) From the graph in exercise 91b, an input value of 4 x  results in a gain of 1168.67. y 

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(d) 33 444413722561116 (4)(43)(4)(343)(64)3721168.67. 3333333 V  

100. If 2 ()4,Srr   then doubling the radius would give us a surface area gained function of 22222 ()4(2)416412. Srrrrrr  

101. (a) If x  width, then 2 x  length. Since the perimeter formula is 22 PWL  our perimeter function is ()2()2(2)24()6. PxxxxxPxx    This is a linear function.

(b) Graph ()6 Pxx  in the window [0,10] by [1,100]. See Figure 101b. From the graph when 4,24.xy The 4 represents the width of a rectangle and 24 represents the perimeter.

(c) If 4 x  is the width of a rectangle then 28 x  is the length. See Figure 101c. Using the standard perimeter formula yields 2(4)2(8)24. P  This compares favorably with the graph result in part b.

(d) (Answers may vary.) If the perimeter y of a rectangle satisfying the given conditions is 36, then the width x is 6. See Figure 101d.

99

101b

102. (a) If 4 xs  then . 4 x s 

(b) If 2 ys  and , 4 x s  then

()(). 416 xx yxyx

(c) Sine x is the perimeter and 2 (6)369 6,(6)2.25. 16164 xy

(d) Show that the point (6,2.25) is on the graph

A square with perimeter 6 will have area 2.25 square u nits.

103. (a) 33222 (2)(2)(4)(2)3 44 AxxxAxx   

(b) 2 33 ()(16)(256)()643 44 AxAx

square units.

(c) On the graph of 2 3 , 4 yx  locate the point where 16 x  to find 110.85, y  an approximation for 643.

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Figure
Figure
Figure 101c
Figure 101d

104. (a) If 2 () Arr   and ()2 rtt  then 22 ()()[()][2](2)4. ArtArtAttt  

(b) ()() Art  is a composite function that expresses the area of the circular region covered by the pollutants as a function of time t (in hours).

(c) Since 0 t  is 8 A.M., noon would be 22 4.()(4)4(4)64mi.tAr 

(d) Graph 2 1 4 yx   and show that for 4,201xy (an approximation for 64). 

105. (a) (2100)42,A  ; The average age of a person in 2100 is projected to be 42 years. (2100)430 T  , In 2100, the living world’s population will have a combined life experience of 430 billion years.

(b) (2100)430 10.2 (2100)42 T A  , The world population will be about 10.2 billion in 2100.

(c) P(x) gives the world’s population during year x.

106. (a) The domain is the years. See table below. Therefore, D= { 2013, 2014, 2015, 2016 }.

(b) The function h computes the number of acres of soybeans harvested in the U.S. in millions of acres.

107. (a) ()(2010)13.074.387.3 fg

(b) The function ()() fgx  computes the total SO2 and Carbon Monoxide during year x

(c) Add functions f and g

108. (a) The function h is the addition of functions f and g

(b) The function h is the addition of functions f and g. Therefore, ()()(). hxfxgx 

109. (a) The function h is the subtraction of function f from g. Therefore, ()()(). hxgxfx  (b) (2000)(2000)(2000)403010 hgf

, (2014)(2014)(2014)1008020 hgf

(c) Using the points (2014, 100) and (2000, 40), the slope is 100406030 20142000147 m  The equation is  30 1002014 7 yx Using the points (2014, 80) and (2000, 30), the slope is

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. The difference between the two equations is given as

110. (a) Graph

in the window [1982, 1994] by [0, 1] See Figure 11-a.

Approximately 59% of the people who contracted AIDS during this time period died.

(b) Divide number of deaths by number of cases for each year. The results compare favorably with the graph. See Figure 110b. Xscl = 1 Yscl = 1

110a

Reviewing Basic Concepts (Sections 2.4—2.6)

1.

Therefore, the solution set is 12,4.

(b) 111 242424 222 xxxx

Therefore, the solution interval is (,12)(4,).  (c) 111 2442462124. 222 xxxx

Therefore, the solution interval is [-12,4].

2. For the graph of |()|,yfx  we reflect the graph of ()yfx  across the x-axis for all points for which 0. y  Where 0, y  the graph remains unchanged. See Figure 2.

Figure
Figure 110b
Figure 2
Figure 5a

3. 3 2413241353or24(13)24315 5 xxxxxxxxxxx 

The solution set is 3 ,5 5

4. (a) (3)2(3)33 f  (b) 2 (0)(0)44 f  (c) 2 (2)(2)48 f 

5. (a) See Figure 5a.

(b) Graph 2 1 ()*(0)(4)*(0)yxxxx  in the window [-10,10] by [-10,10]. See Figure 5b. [-10,10] by [-10,10] Xscl = 1 Yscl = 1

5b

6. (a) 22 ()()(34)()34. fgxxxxx  Therefore, 2 ()(1)(1)3(1)46. fg

(b) 22 ()()(34)()34. fgxxxxx  Therefore, 2 ()(3)(3)3(3)422. fg

(c) 232 ()()(34)()34. fgxxxxx  Therefore, 32 ()(1)3(2)4(2)24168. fg 

(d) 2 34 ().fx x gx

Therefore, 2 3(3)45 (3). 9 (3)

(e) 22 ()()[()]3()4()()34 fgxfgxxfgxx

   (f) 22 ()()[()](34)()()92416 gfxgfxxgfxxx

7. (a) ()()[()]4 fgxfgxx

(b) (g)()[f()]4fxgxx  

8. One of many possible solutions for ()()() fgxhx   is 4 () fxx  and ()2.gxx Then 4 ()()[()](2). fgxfgxx 

9. 22222 2()3()5(235)2(2)335235 xhxhxxxxhhxhxx hh   2222 242335235423 423. xxhhxhxxxhhh xh hh

10. (a) At 4% simple interest the equation for interest earned is 1 .04.yx 

(b) If he invested x dollars in the first account, then he invested x+500 in the second account. The equation for the amount of interest earned on this account is 22.025(500).02512.5. yxyx   

(c) It represents the total interest earned in both accounts for 1 year.

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Figure

Section 2.R 139

(d) Graph 1212.04(.02512.5).04.02512.5 yyxxyyxx    in the window [0,1000] by [0,100]. See Figure 10. An input value of 250, x  results in $28.75 earned interest.

(e) At 12 250,.04(250).025(250)12.5106.2512.5$28.75.xyy [0,1000] by [-20,100] Xscl = 100 Yscl = 10

Figure 10

11. If the radius is r, then the height is 2r and the equation is 222222 (2)455.SrrrrrrrrSr    

Chapter 2 Review Exercises

The graphs for exercises 1–10 can be found in the “Function Capsule” boxes located in section 2.1 in the text.

1. True Both 2 () fxx  and ()||fxx  have the interval: [0,)  as the range.

2. True Both 2 () fxx  and ()||fxx  increase on the interval: [0,)  .

3. False The function () fxx  has the domain: [0,)  and 3 () fxx  the domain: (,) 

4. False The function 3 () fxx  increases on its entire domain.

5. True The function () fxx  has a domain and range of: (,) 

6. False The function () fxx  is not defined on (,0)  , so certainly cannot be continuous.

7. True All of the functions show increases on the interval: [0,) 

8. True Both () fxx  and 3 () fxx  have graphs that are symmetric with respect to the origin.

9. True Both 2 () fxx  and ()||fxx  have graphs that are symmetric with respect to the y-axis.

10. True No graphs are symmetric with respect to the x-axis.

11. Only values where 0 x  can be input for x, therefore the domain of () fxx  is: [0,) 

12. Only positive solution are possible in absolute value functions, therefore the range of () fxx  is: [0,) 

13. All solution are possible in cube root functions, therefore the range of 3 () fxx  is: (,) 

14. All values can be input for x, therefore the domain of 2 () fxx  is: (,) 

15. The function 3 () fxx  increases for all inputs for x, therefore the interval is: (,) 

16. The function ()||fxx  increases for all inputs where 0 x  , therefore the interval is: [0,) 

17. The equation is the equation yx  . Only values where 0 x  can be input for x, therefore the domain of yx  is: [0,) 

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18. The equation 2 yx  is the equation yx  Square root functions have both positive and negative solutions and all solution are possible, therefore the range of yx  is: (,) 

19. The graph of ()(3)1fxx is the graph yx  shifted 3 units to the left and 1 unit downward.

See Figure 19.

20. The graph of 1 ()1 2 fxx is the graph reflected yx  across the x-axis, vertically shrunk by a factor of 1 2 , and shifted 1 unit upward. See Figure 20.

21. The graph of 2 ()(1)2fxx is the graph 2 yx  shifted 1 unit to the left and 2 units downward.

See Figure 21.

Figure 19

Figure 20

Figure 21

22. The graph of 2 ()23fxx is the graph 2 yx  reflected across the x-axis, vertically stretched by a factor of , and shifted 3 units upward. See Figure 22.

23. The graph of  3 2 fxx is the graph 3 yx  reflected across the x-axis and shifted 2 units upward.

See Figure 23.

24. The graph of 3 3 fxx is the graph 3 yx  shifted 3 units to the right. See Figure 24.

Figure 22

Figure 23

Figure 24

25. The graph of  1 2 fxx  is the graph yx  horizontally stretched by a factor of 2. See Figure 25.

26. The graph of  21 fxx is the graph yx  shifted 2 units to the right and 1 unit upward.

See Figure 26.

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27. The graph of  3 2 fxx  is the graph 3 yx  vertically stretched by a factor of . See Figure 27.

Figure 25

Figure 26

Figure 27

28. The graph of  3 2 fxx is the graph 3 yx  shifted 2 units downward. See Figure 28.

29. The graph of  |2|1fxx is the graph ||yx  shifted 2 units right and 1 unit upward. See Figure 29.

30. The graph of  |23|fxx is the graph ||yx  horizontally shrunk by a factor of 1 2 , shifted

or 3 2 units to the left, and reflected across the y-axis. See Figure 30.

Figure 28

Figure 29

Figure 30

31. (a) From the graph, the function is continuous for the intervals:   ,2,2,1 and  1, 

(b) From the graph, the function is increasing for the interval:  2,1

(c) From the graph, the function is decreasing for the interval:  ,2

(d) From the graph, the function is constant for the interval:  1, 

(e) From the graph, all values can be input for x, therefore the domain is:  , 

(f) From the graph, the possible values of y or the range is:

32. 22444xyyxyx      and 4 yx .

21,12, 

33. From the graph, the relation is symmetric with respect to the x-axis, y-axis, and origin. The relation is not a function since some inputs x have two outputs y.

34. If  3 6 Fxx ,then  3 3 66FxxFxx

and

3366FxxFxx 

. Since  FxFxFx  the function has no symmetry and is neither an ever nor an odd function.

35. If  ||4fxx ,then ||4||4fxxfxx

and  ||4fxx  Since  fxfx  the function is symmetric with respect to the y-axis and is an even function.

36. If  5 fxx then  5 fxx and  5 fxx  .Since  fxfxfx  , the function has no symmetry and is neither an even nor an odd function.

37. If 2 5 yx then 5 yx . Since  5 fxx is the reflection of  5 fxx across the x-axis, the relation has symmetry with respect to the x-axis. Also, one x inputs can produce two y outputs the relation is not a function.

38. If  42 321fxxx then  42 42 321321 fxxxfxxx 

 and  42 321fxxx  Since  fxfx  the function is symmetric with respect to the y-axis and is an even function.

39. True, a graph that is symmetrical with respect to the x-axis means that for every  , xy there is also  , xy

40. True, since an even function and one that is symmetric with respect to the y-axis both contain the points  , xy and  , xy .

41. True, since an odd function and one that is symmetric with respect to the origin both contain the points  , xy and  , xy

42. False, for an even function, if  , ab is on the graph, then  , ab is on the graph and not  , ab . For example,  2 fxx  is even, and (2, 4) is on the graph, but (2, 4) is not.

43. False, for an odd function, if  , ab is on the graph, then  , ab is on the graph and not  , ab For example,  3 fxx  is odd, and  2,8 is on the graph, but  2,8 is not.

44. True, if  ,0x is on the graph of  0 fx  then  ,0x is on the graph.

45. The graph of  2 348yx is the graph of 2 yx  shifted 4 units to the left, vertically stretched by a factor of 3, reflected across the x-axis, and shifted 8 units downward.

46. The equation yx  reflected across the y-axis is: () fxx  then reflected across the x-axis is: yx  now vertically shrunk by a factor of 2 3 is: 2 3 yx  , and finally shifted 4 units upward is: 2 4 3 yx .

47. Shift the function f upward 3 units. See Figure 47.

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48. Shift the function f to the right 2 units. See Figure 48.

49. Shift the function f to the left 3 units and downward 2 units. See Figure 49.

Figure 47

Figure 48

49

50. For values where  0 fx  the graph remains the same. For values where  0 fx  reflect the graph across the x-axis. See Figure 50.

51. Horizontally shrink the function f by a factor of 1 4 . See Figure 51.

52. Horizontally stretch the function f by a factor of 2. See Figure 52.

Figure 50

Figure 51

Figure 52

53. The function is shifted upward 4 units, therefore the domain remains the same:   3,4 and the range is increased by 4 and is:   2,9

54. The function is shifted left 10 units, therefore the domain is decreased by 10 and is   13,6; and the function is stretched vertically by a factor of 5, therefore the range is multiplied by 5 and is:   10,25.

55. The function is horizontally shrunk by a factor of 1 , 2 therefore the domain is divided by 2 and is: 3 ,2; 2

and the function is reflected across the x-axis, therefore the range is opposites of the original and is: 

5,2.

56. The function is shifted right 1 unit, therefore the domain is increased by 1 and is:   2,5; and the function is also shifted upward 3 units, therefore the range is increased by 3 and is:   1,8.

57. We reflect the graph of yfx  across the x-axis for all points for which 0. y  Where 0, y  the graph remains unchanged. See Figure 57.

Figure

58. We reflect the graph of yfx  across the x-axis for all points for which 0.  Where 0, y  the graph remains unchanged. See Figure 58.

59. Since the range is   2,0, y  so the graph remains unchanged.

60. Since the range is   2,0, y  so we reflect the graph across the x-axis. See Figure 60.

4312431249 4

or 15 4312415, 4 xxx

therefore the solution set

62. 2641263. xx

Since an absolute value equation can not have a solution less than zero, the solution

63. 53115311482

therefore the solution set is: 7 ,2. 3

64. 257257221 xxxx

or 2572126, xxx

63

therefore the solution set is:

65. 2577257122261, xxxx  

therefore the interval is:

66. 257257221 xxxx +³+³³³ or 2572126, xxx +£-£-£- therefore the solution is the interval: ( ] [ ) ,61,.-¥-¥ 

67. 12 51205120512 5 xxxx   

  or 12 5120512, 5 xxx

    therefore the solution is the interval: 1212 ,, 55

68. Since an absolute value equation can not have a solution less than zero, the solution set is: 

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Figure 57
Figure 58
Figure 60

69. 11 2311212312031103110311 3 xxxxxx            or 3110393, xxx 

 therefore the solution set is: 11 3,. 3

70. 21312131500 xxxxxx        or  213122, xxxx      therefore the solution set is:   0,2.

71. The x-coordinates of the points of intersection of the graphs are 6 and 1.Thus,  6,1 is the solution set of 12yy  The graph of 1y lies on or below the graph of 2y between 6 and 1, so the solution set of 12yy  is  6,1 The graph of 1y lies above the graph of 2y everywhere else, so the solution set of 12yy  is

72. Graph 1 13yxx and {3,1} See Figure 72. The intersections are 3 x  and 5, x  therefore the solution set is:   3,5. Check:

3133826826888

and

5153862862888

[-10,10] by [-4,16] Xscl = 1 Yscl = 1

73. Initially, the car is at home. After traveling 30 mph for 1 hr, the car is 30 mi away from home. During the second hour the car travels 20 mph until it is 50 mi away. During the third hour the car travels toward home at 30 mph until it is 20 mi away. During the fourth hour the car travels away from home at 40 mph until it is 60 mi away from home. During the last hour, the car travels 60 mi at 60 mph until it arrives home.

74. See Figure 74

75. See Figure 75

76. See Figure 76

Figure 72

Figure 74

Figure 75

77. Graph  1 31*24*2yxxxx  in the window

Figure 76

78. See Figure 78.

[-10,10] by [-10,10] [-5,5] by [-5,5] Xscl = 1 Yscl = 1 Xscl = 1 Yscl = 1

Figure 77 Figure 78

79. From the graphs ()(1)235 fg

80. From the graphs ()(0)143 fg

81. From the graphs ()(1)(0)(3)0 fg 

82. From the graphs 3 (2) 2 f g   

83. From the graphs ()(2)[(2)](2)3 fgfgf 

84. From the graphs ()(2)[(2)](3)2 gfgfg 

85. From the graphs ()(4)[(4)](2)2 gfgfg 

86. From the graphs ()(2)[(2)](2)3 fgfgf 

87. From the table ()(1)718 fg

88. From the table ()(3)910 fg

89. From the table ()(1)(3)(2)6 fg 

90. From the table 5 (0) 0 f g     , which is undefined.

91. From the tables ()(2)[(2)](1)2 gfgfg 

92. From the graphs ()(3)[(3)](2)1 fgfgf 

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93. 2()9(29)229292 2 xhxxhxh hhh  

94. 22222 ()5()3(53)255353 xhxhxxxxhhxhxx hh   2 25 25 xhhh xh h  

95. One of many possible solutions for ()()() fgxhx  is: 2 () fxx  and 3 ()3 gxxx  Then   2 3 ()3 fgxxx  

96. One of many possible solutions for ()()() fgxhx  is 1 ()fx x  and ()5gxx Then 1 ()()[()] 5 fgxfgx x 

97. If 3 4 () 3 Vrr   , then a 4 inch increase would be: 3 4 ()(4) 3 Vrr   , and the volume gained would be: ()(4)4433 33 Vrrr 

98. (a) Since , 2 d hdr and the formula for the volume of a can is: 2 Vrh   , the function is: 2 3 ()()24 dd VddVd  

(b) Since ,,2 2 d hdrcr   and the formula for the surface area of a can is: 2 22 Arhr  , the function is: 2 22 2 3 ()22()()2222 dddd SddSddSd 

99. The function for changing yards to inches is: ()36 fxx  and the function for changing miles to yards is: ()1760 gxx  The composition of this which would change miles into inches is: [()]36[1760()]()()63,360 fgxxfgxx    .

100. If x  width, then length 2 x  A formula for Perimeter can now be written as: 22 Pxxxx  and the function is: ()6 Pxx  This is a linear function.

Chapter 2 Test

1. (a) D, only values where 0 x  can be input into a square root function.

(b) D, only values where 0 y  can be the solution to a square root function.

(c) C, all values can be input for x in a squaring function.

(d) B, only values where 3 y  can be the solution to 2 ()3fxx

(e) C, all values can be input for x in a cube root function.

(f) C, all values can be a solution in a cube root function.

(g) C, all values can be input for x in an absolute value function.

(h) D, only values where 0 y  can be the solution to an absolute value function.

(i) D, if 2 xy  then yx  and only values where 0 x  can be input into a square root function.

(j) C, all values can be a solution in this function.

2. (a) This is ()fx shifted 2 units upward. See Figure 2a.

(b) This is ()fx shifted 2 units to the left. See Figure 2b.

(c) This is ()fx reflected across the x-axis. See Figure 2c.

(d) This is ()fx reflected across the y-axis. See Figure 2d.

(e) This is ()fx vertically stretched by a factor of 2 . See Figure 2e.

(f) We reflect the graph of ()yfx  across the x-axis for all points for which $2.75 Where 0 y  the graph remains unchanged. See Figure 2f.

Figure 2a

Figure 2d

Figure 2b

Figure 2e

Figure 2c

Figure 2f

3. (a) Since (2)yfx  is ()yfx  horizontally shrunk by a factor of 1 2 , the point (2,4) on ()yfx  becomes the point (1,4) on the graph of (2)yfx  .

(b) Since 1 2 yfx 

is ()yfx  horizontally stretched by a factor of 2, the point (2,4) on ()yfx  becomes the point (4,4) on the graph of 1 2 yfx

4. (a) The graph of 2 ()(2)4fxxx is the basic graph 2 () fxx  reflected across the x-axis, shifted 2 units to the right, and shifted 4 units upward. See Figure 4a.

(b) The graph of ()2 fxx  is the basic graph () fxx  reflected across the y-axis and vertically stretched by a factor of 2 . See Figure 4b.

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5. (a) If the graph is symmetric with respect to the y-axis, then (,)(,) xyxy  therefore (3,6)(3,6) 

(b) If the graph is symmetric with respect to the x-axis, then (,)(,) xyxy  therefore (3,6)(3,6)  .

(c) See Figure 5. We give an actual screen here. The drawing should resemble it.

= 1 Yscl = 1

5

6. (a) Shift the graph of 3 yx  to the left 2 units, vertically stretch by a factor of 4 , and shift 5 units downward.

(b) Graph yx  reflected across the x-axis, vertically shrunk by a factor of 1 2 shifted 3 units to the right, and shifted up 2 units. See Figure 6. From the graph the domain is: (,)  ; and the range is: (,2]  .

6

7. (a) From the graph, the function is increasing for the interval: (,3) 

(b) From the graph, the function is decreasing for the interval: (4,) 

(c) From the graph, the function is constant for the interval:  3,4

(d) From the graph, the function is continuous for the intervals:  (,3),3,4,(4,)  .

(e) From the graph, the domain is: (,) 

(f) From the graph, the range is: (,2) 

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Figure 4a
Figure 4b
Xscl
Figure
Figure

8. (a) 484484441 xxxx    

or 4844123 xxx

, therefore the solution set is: {3,1}

(b) 4844484124431 xxxx

, therefore the solution is: (3,1) .

9.

(c) 484484441 xxxx

or 4844123 xxx

therefore the solution is: (,3)(1,)   .

(b) 2 232 () 21 fxx x gx

(c) The domain can be all values for x, except any that make ()gx =0. Therefore 1

(d)  22 ()()2(21)3(21)22(441)632 fgxfgxxxxxx

-++-+=-+

(e)  222 ()()2(232)14641463 gfxgfxxxxxxx 

(f) 22222 2()3()2(232)2(2)332232 xhxhxxxxhhxhxx hh   2222 242332232423 423 xxhhxhxxxhhh xh

10. (a) ()()[()]211 fgxfgxx  D:

(b) (g)()[f()]2112 fxgxxx 

11. (a) See Figure 11a.

(b) Graph

See Figure 11b.

(c) The graph is not connected at x = 1 and thus f is not continuous when x =1.

Figure 11a

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[-4.7,4.7] by [-5.1,5.1]

[0,10] by [0,6]

Section 2.T 151

[0,1,000] by [-4,000,4000]

Xscl = 1 Yscl = 1 Xscl = 1 Yscl = 1 Xscl = 50 Yscl = 500

12. (a) See Figure 12.

(b) Set 5.5 x  then $2.75 is the cost of a 5.5 minute international call. See the display at the bottom of the screen.

13. (a) With an initial set-up cost of $3300 and a production cost of $4.50 the function is: ()33004.50 Cxx 

(b) With a selling price of $10.50 the revenue function is: ()10.50 Rxx 

(c) ()()()()10.50(33004.50)()63300 PxRxCxPxxxPxx   

(d) To make a profit ()0,Px  therefore 63300063300550 xxx      Tyler needs to sell 551 before he earns a profit.

(e) Graph 1 63300,yx See Figure 13. The first integer x-value for which ()0Px  is 551

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Figure 11b
Figure 12
Figure 13

A Graphical Approach to College Algebra

Seventh Edition

Chapter R Review: Basic Algebraic Concepts

Section R.2 Review of Exponents and Polynomials

Review of Exponents and Polynomials (1 of 2)

Product Rule

For all positive integers m and n and every real number a,

Review of Exponents

Zero Exponent

For any nonzero real number

Example 2: Using the Definition of a to the Zeroth Power

Evaluate each power.

Reviewing Exponents and Polynomials

Power Rules

For all positive integers m and n and all real numbers a and b,

Example 3: Using the Power Rules

Simplify each expression.

Terminology for Polynomials (1 of 4)

An algebraic expression results from adding, subtracting, multiplying, dividing (except by 0), or finding roots or powers of any combination of variables and constants.

Terminology for Polynomials (2 of 4)

A term is the product of a real number and one or more variables raised to powers.

The real number in a term is called the numerical coefficient, or just the coefficient.

In the term 4 3, m  the coefficient is −3.

Terminology for Polynomials (3 of 4)

Like terms are terms with the same variables raised to the same powers.

33313,4,xxx  and are like terms, 2 66yy and are not.

Terminology for Polynomials (4 of 4)

A polynomial is a term or a finite sum of terms with only nonnegative integer exponents permitted on the variables.

Adding and Subtracting Polynomials

Polynomials are added by adding coefficients of like terms.

Polynomials are subtracted by subtracting coefficients of like terms.

Example 4: Adding and Subtracting Polynomials (1 of 2)

Add or subtract, as indicated.

Example 4: Adding and

Polynomials (2 of 2) Add or subtract, as indicated.

Multiplying Polynomials (1 of 2)

The associative and distributive properties, together with the properties of exponents, can be used to multiply polynomials.

Polynomials can be multiplied my multiplying each term of the first polynomial by each term of the second, then combining like terms.

Multiplying Polynomials (2 of 2)

A binomial is a polynomial with two terms.

The FOIL method can be used to multiply two binomials.

The FOIL method says to multiply the First terms, the Outside terms, the Inside terms and the Last terms, then combine like terms.

Example 6: Using FOIL Method (1 of 2)

Find each product.

Review of Exponents and Polynomials (2 of 2)

Special Products

Product of the Sum and Difference of Two Terms

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