 |
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| 1. Explain in terms of graphs (see
Figures 0.41a,0.41b, and
0.41c) why a
polynomial of odd degree must have at least one real zero. |
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| 2. Explain in terms of graphs why a
cubic polynomial is more likely to have one zero than two zeros, but a fourth-degree
polynomial is more likely to have two zeros than one zero. |
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| 3. In Example 4.1 we factored the
quadratic polynomial, but in Example 4.2 we used the
quadratic formula. Explain how you know when to factor and when to use the quadratic formula. |
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| 4. If you substitute x = 1 into x3 - 4x2 + 2x + 1 , you get 0. Explain why this means that x - 1 is a factor of
x3 - 4x2 + 2x + 1
< ><>. Explain how to use synthetic division or long division to get the remaining (quadratic) factor. |
| In exercises 5-20, factor and/or use the
quadratic formula to find all zeros of the given
function.
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| 5. f (x) = x2 - 4x + 3 | 6. f (x) = x2 + x - 12 |
| 7. f (x) = x2 - 2x - 15 | 8. f (x) = x2 + 6x + 8 |
| 9. f (x) = x2 - 4x + 2 | 10. f (x) = x2 + 2x - 2 |
| 11. f (x) = 2x2 + 4x - 1 | 12. f (x) = 2x2 - 6x + 3 |
| 13. f (x) = x3 - 3x2 + 2x | 14. f (x) = x3 - 2x2 - x + 2 |
| 15. f (x) = x3 + x2 - 4x - 4 | 16. f (x) = x3 + 4x2 + 3x
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| 17. f (x) = x4 - 1 | 18. f (x) = x4 - 3x2 + 2 |
| 19. f (x) = x6 + x3 - 2 | 20. f (x) = x5 - 1
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| In exercises 21-30, use a graphing calculator or computer graphing utility to estimate all zeros.
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| 21. f (x) = x3 - 3x + 1 | 22. f (x) = x3 - 4x2 + 2 |
| 23. f (x) = x3 - x + 1 | 24. f (x) = x3 + 2x2 - 1 |
| 25. f (x) = x4 + 3x3 - x + 1 | 26. f (x) = x4 - 2x + 1 |
| 27. f (x) = 4x5 + 8x - 1 | 28. f (x) = x5 - 10x + 2
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| 29. f (x) = x4 - 7x3 - 15x2 - 10x - 1410
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| 30. f (x) = x6 - 4x4 + 2x3 - 8x - 2
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| In exercises 31-40, determine the number of (real) solutions. Solve for the intersection points exactly if possible and estimate the points if necessary.
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| 31.  | 32.  |
| 33. x3 - 3x2 = 1-3x | 34. x3 + 1 = - 3x2 - 3x |
| 35. (x2 -1)2/3 = 2x + 1 | 36. (x + 1)2/3 = 2 - x |
| 37.  | 38.  |
| 39. cos x = x2 - 1 | 40. sin x = x2 + 1
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| 41. Show that the
quadratic formula is correct by showing that
 and  are solutions of the equation ax2 + bx + c = 0. |
| 42. Find a formula for and sketch the graph of a fourth-degree
polynomial with (a) no zeros; (b) one zero; (c) two zeros; (d) three zeros; (e) four zeros. |
| 43. As discussed in
Example 4.5, you should always compare your equation-solving results to a graph to see if you have made any mistakes. Consider the following steps in solving the equation  . Square both sides of the equation to get x2 - 2x + 1 = 7 - x and combine terms to get x2 - Finally, factor to get (x -3)(x + 2) = 0 and conclude that the solutions are x = 3 and x = - 2. Use a graph to show that x = - 2 is not actually a solution. We say that x = - 2 is an extraneous solution, produced by squaring the original equation. |
| 44. For the equation in exercise 43, show that when you substitute x = - 2 into both sides of the original equation, the two sides are not equal. Explain why squaring both sides of that equation makes it appear that x = - 2 is a solution. |
| In exercises 45 and 46, find the true solution and an
extraneous solution produced by squaring (see exercises 43 and 44).
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| 45.  | 46. 
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| In exercises 47-52, use factoring and the
quadratic formula to find all real and complex zeros. Sketch a graph and show that the number of real zeros corresponds to the number of
x - intercepts.
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| 47. f (x) = x2 + 2x + 5 | 48. f (x) = x2 + x + 1 |
| 49. f (x) = x2 + 2x - 5 | 50. f (x) = x2 + x - 1 |
| 51. f (x) = x3 - 1 | 52. f (x) = x4 - 1
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| In exercises 53-56, find all fixed points of the given
function. (A fixed point of a function
f (x) is a solution of the equation f (x)=x. For example, to find the fixed points of
f (x)=x2, solve the equation x2=x. The fixed points are
x=0 and x=1.)
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| 53. f (x) = x2 - 1 | 54. f (x) = x2 - 2 |
| 55. f (x) = x3 | 56. f (x) = 4x(1 - x)
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| 57. Show that if c >  , the
function f (x) = x2 + c has no real fixed points (see exercises 53 and 54). However, if c < , the
function f (x) = x2 + c has two fixed points. Because of the change in the number of fixed points, the value c = is called a bifurcation point of the
function f (x) = x2 + c. |
| 58. Graph f (x) = x2 + and g(x) = x and explain why c = , which we identified as a bifurcation point in exercise 57, is called a tangent bifurcation. |
| 59. A baseball thrown from third base to first base follows the path of a parabola. Suppose the ball is released at the point (0,5). Based on gravity and a release angle of 8° , an equation describing the path is y = 5 + 0.14x -16.3x2/v2, where v is the speed (in ft/s) of the throw. Find the value of v so that the throw hits the first baseman's glove at the point (130,5). (Substitute x = 130 and y = 5 and solve for v. Convert to mph by multiplying by 3600/5280.) |
| 60. Using the equation of exercise 59, find the speed (value of v ) needed for an outfielder to hit a catcher's glove at the point (260,5). This speed is faster than any human can throw. Given this, how must the outfielder adjust the angle of the throw to get the throw to the catcher? |
| 61. A ball tossed h feet in the air will stay in the air for  seconds. For example, T(1) = 0.5 means that a ball tossed 1 foot in the air will come down half a second later. Solve T(h) = 4.5 to find how high a ball would have gone if it remained in the air for 4.5 seconds. |
| 62. Professional jugglers sometimes claim that 10 balls is the limit for a human being to juggle. Suppose that it takes half a second to catch a ball and throw it back in the air. If you were juggling 10 balls, explain why you would want that ball to remain in the air for 4.5 seconds. Based on exercise 61, how high would you have to toss each ball? Determine how much higher you would have to toss each ball to add an 11th ball. |
| 63. If you have used the built-in equation solver on a graphing calculator or computer, you may have wondered how it works. In this exercise, we introduce a simple method for finding zeros. Consider the
function f (x) = Show that f (0) = - 1 and f (1) = 3. Since f (0) is negative and f (1) is positive, explain why you would suspect that f (x) has a zero between x = 0 and x = 1. In the method of bisections, you next try the midpoint x =  . Show that f () is negative and explain why you would now look between x = and x = 1 for the zero. Show that f ( ) is positive and explain why the zero is between x = and x = . Continue in this fashion for three more steps and give your best estimate of the location of the zero. Note that while this method is tedious for you to use by hand, a computer would have no trouble in carrying this out. |
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| 64. In exercise 63, you explored the method of bisections. The main assumption that bisections depends on is that if f (a) is negative and f (b) is positive, then f (x) = 0 for some x between a and b. To see that this assumption is not always
valid, take  and show that f ( -1) is negative and
f (1) is positive. Show graphically that f (x) has no zeros. Briefly describe how the graph gets from the point ( -1, -1) to the point (1,1) without passing through the x - axis. Sketch other graphs containing the points ( -1, -1) and (1,1) and no x -
intercepts. The property that these graphs share (we call them discontinuous) is precisely defined in
Chapter 1. |
