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 | 1. The restricted domain of example 7.2 may be puzzling. Consider the following analogy. Suppose you have an airplane flight from New York to Los Angeles with a stop for refueling in Minneapolis. If bad weather has closed the airport in Minneapolis, explain why your flight will be canceled (or at least rerouted) even if the weather is great in New York and Los Angeles. |
| | 2. Explain why the graphs of y = 4(x2 -1) and y = in Figures 0.66c and 0.67c appear “thinner” than the graph of y = x2 - 1. |
| | 3. As illustrated in example 7.9, completing the square can be used to rewrite any quadratic function in the form a(x - d)2 + e. Using the transformation rules in this section, explain why this means that all parabolas (with a > 0 ) will look essentially the same. |
| | 4. Explain why the graph of y = f (x + 4) is obtained by moving the graph of y = f (x) four units to the left, instead of to the right. |
| In exercises 5-10, find the compositions f  g and g f and identify their respective domains.
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| 5. 
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| 6. 
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| 7. f (x) = ex, g(x) = ln x
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| 8. 
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| 9. f (x) = x2 + 1, g(x) = sin x
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| 10. 
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| In exercises 11-26, identify functions f (x) and g(x) such that the given
function equals ( f g)(x).
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| 11.  | 12.  |
| 13.  | 14.  |
| 15. (4x + 1)2 + 3 | 16. 4(x + 1)2 + 3 |
| 17. sin 3x | 18. sin x3 |
| 19. cos 4x | 20. 4cos x |
| 21.  | 22. e4x - 2 |
| 23.  | 24.  |
| 25. ln (3x -5) | 26. ln 3x - 5
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| In exercises 27-34, use the graph of y = f (x) given in the figure to graph the indicated
function.
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| 27. f (x) - 3 | 28. f (x + 2) |
| 29. f (x -3) | 30. f (x) + 2 |
| 31. f (2x) | 32. 3f (x) |
| 33. 4f (x) - 1 | 34. 3f (x + 2)
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| In exercises 35-42, use the graph of y = f (x) given in the figure to graph the indicated function.
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| 35. f (x -4) | 36. f (x + 3) |
| 37. f (2x) | 38. f (2x -4) |
| 39. f (3x + 3) | 40. 3f (x) |
| 41. 2f (x) - 4 | 42. 3f (x) + 3
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| In exercises 43-48, complete the square and explain how to transform the graph of y = x2 into the graph of the given function.
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| 43. f (x) = x2 + 2x + 1 | 44. f (x) = x2 - 4x + 4 |
| 45. f (x) = x2 + 2x + 4 | 46. f (x) = x2 - 4x + 2 |
| 47. f (x) = 2x2 + 4x + 4 | 48. f (x) = 3x2 - 6x + 2
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| In exercises 49-52, graph the given function and compare to the graph of y = x2 - 1.
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| 49. f (x) = - 2(x2 -1) | 50. f (x) = - 3(x2 -1) |
| 51. f (x) = - 3(x2 -1) + 2 | 52. f (x) = - 2(x2 -1) - 1
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| In exercises 53-56, graph the given function and compare to the graph of y = (x -1)2 -1 = x2 - 2x.
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| 53. f (x) = ( - x)2 - 2( - x)
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| 54. f (x) = ( - 2x)2 - 2( - 2x)
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| 55. f (x) = ( - x)2 - 2( - x) + 1
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| 56. f (x) = ( - 3x)2 - 2( - 3x) - 3
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| 57. Based on exercises 49-52, state a rule for transforming the graph of y = f (x) into the graph of y = cf(x) for c < 0.
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| 58. Based on exercises 53-56, state a rule for transforming the graph of y = f (x) into the graph of y = f (cx) for c < 0.
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| 59. After a car goes over a bump in the road, the car's vertical position is given by f (t ) = e - t /4cos t . Describe what the passenger feels. |
| 60. After a car goes over a bump in the road, the car's vertical position is given by f (t ) = e - 4t cos t . Describe what the passenger feels. Is this passenger happier than the one in exercise 59? |
| 61. Iterations of functions are important in a variety of applications. To iterate f (x) , start with an initial value x0 and compute x1 = f (x0) , x2 = f (x1) , x3 = f (x2) and so on. For example, with f (x) = cos x and x0 = 1 , the iterates are x1 = , x2 = cos x1 cos 0.54 0.86, x3 cos 0.86 0.65 and so on. Keep computing iterates and show that they get closer and closer to 0.739085. Then pick your own x0 (any number you like) and show that the iterates with this new x0 also converge to 0.739085. |
| 62. Referring to exercise 61, show that the iterates of a function can be written as x1 = f (x0) , x2 = f (f (x0)) , x3 = f (f (f (x0))), and so on. Graph y = cos (cos x) , y = cos (cos (cos x)) and y = cos (cos (cos (cos x))). The graphs should look more and more like a horizontal line. Use the result of exercise 61 to identify the limiting line. |
| 63. Compute several iterates of f (x) = sin x (see exercise 61) with a variety of starting values. What happens to the iterates in the long run? |
| 64. Repeat exercise 63 for f (x) = x2. |
| 65. In cases where the iterates of a function (see exercise 61) repeat a single number, that number is called a fixed point. Explain why any fixed point must be a solution of the equation f (x) = x. Find all fixed points of f (x) = cos x by solving the equation cos x = x. Compare your results to that of exercise 61. |
| 66. Find all fixed points of f (x) = sin x (see exercise 65). Compare your results to those of exercise 63. |
 | 67. You have explored how completing the square can transform any quadratic function into the form y = . We concluded that all parabolas with a > 0 look alike. To see that the same statement is not true of cubic polynomials, graph y = x3 and y = x3 - 3x. In this exercise, you will use completing the cube to determine how many different cubic graphs there are. To see what “completing the cube” would look like, first show that (x + a)3 = x3 + . Use this result to transform the graph of y = x3 into the graphs of (a) y = x3 - 3x2 + 3x - 1 and (b) y = x3 - 3x2 + 3x + 2. Show that you can't get a simple transformation to y = x3 - 3x2 + 4x - 2. However, show that y = x3 - 3x2 + 4x - 2 can be obtained from y = x3 + x by basic transformations. Show that the following statement is true: any cubic ( y = ax3 + bx2 + cx + d ) can be obtained with basic transformations from y = ax3 + kx for some constant k. |
| | 68. In many applications, it is important to take a section of a graph (e.g., some data) and extend it for predictions or other analysis. For example, suppose you have an electronic signal equal to f (x) = 2x for 0 x 2. To predict the value of the signal at x = - 1 , you would want to know whether or not the signal was periodic. If the signal is periodic, explain why f ( -1) = 2 would be a good prediction. In some applications, you would assume that the function is even. That is, f (x) = f ( - x) for all x. In this case, you want f (x) = for - 2 x 0. Graph the even extension  (a) f (x) = x2 + 2x + 1 , 0 x 2 and (b) f (x) = e - x , 0 x 2. |
| | 69. Similar to the even extension discussed in exercise 68, applications sometimes require a function to be odd; that is, f ( - x) = - f (x). For f (x) = x2 , 0 x 2 , the odd extension requires that for - 2 x 0, so that  Graph y = f (x) and discuss how to graphically rotate the right half of the graph to get the left half of the graph. Find the odd extension for (a) f (x) = x2 + 2x , 0 x 2 and (b) f (x) =
0 x 2. |
