# Business Calc Homework w answers_Part_31 - Section 4.3 30....

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30. y 95 ( x 2 1) 2 ( x 2 2)( x 2 4) y 05 } d d x } [( x 2 1) 2 ( x 2 2 6 x 1 8)] 5 ( x 2 1) 2 (2 x 2 6) 1 ( x 2 2 6 x 1 8)(2)( x 2 1) 5 ( x 2 1)[( x 2 1)(2 x 2 6) 1 2( x 2 2 6 x 1 8)] 5 ( x 2 1)(4 x 2 2 20 x 1 22) 5 2( x 2 1)(2 x 2 2 10 x 1 11) Note that the zeros of y 0 are x 5 1 and x 55 } 10 6 4 ˇ 1 w 2 w }5 } 5 6 2 ˇ 3 w } < 1.63 or 3.37. The zeros of y 0 can also be found graphically, as shown. [ 2 3, 7] by [ 2 8, 4] (a) Local maximum at x 5 2 (b) Local minimum at x 5 4 (c) Points of inflection at x 5 1, at x < 1.63, and at x < 3.37. 31. 32. y x y = f ( x ) y = f ( x ) y = f ′′ ( x ) P 0 y x y = f ( x ) y = f ( x ) y = f ′′ ( x ) P 10 6 ˇ 1 w 0 w 2 w 2 w 4 w (2 w )( w 1 w 1 w ) w }}} 4 Section 4.3 151 Intervals x , 11 , x , 22 , x , 4 Sign of y 91 1 2 Behavior of y Increasing Increasing Decreasing Increasing 1 4 , x Intervals x , , x , 1.63 1.63 , x , 3.37 Sign of y 02 1 2 Behavior of y Concave down Concave up Concave down Concave up 1 3.37 , x

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33. (a) Absolute maximum at (1, 2); absolute minimum at (3, 2 2) (b) None (c) One possible answer: 34. (a) Absolute maximum at (0, 2); absolute minimum at (2, 2 1) and ( 2 2, 2 1) (b) At (1, 0) and ( 2 1, 0) (c) One possible answer: (d) Since f is even, we know f (3) 5 f ( 2 3). By the continuity of f , since f ( x ) , 0 when 2 , x , 3, we know that f (3) # 0, and since f (2) 52 1 and f 9 ( x ) . 0 when 2 , x , 3, we know that f (3) .2 1. In summary, we know that f (3) 5 f ( 2 3), 2 1 , f (3) # 0, and 2 1 , f ( 2 3) # 0. 35. 36. 37. (a) v ( t ) 5 s 9 ( t ) 5 2 t 2 4 (b) a ( t ) 5 v 9 ( t ) 5 2 (c) It begins at position 3 moving in a negative direction. It moves to position 2 1 when t 5 2, and then changes direction, moving in a positive direction thereafter. 38. (a) v ( t ) 5 s 9 ( t ) 2 2 2 t (b) a ( t ) 5 v 9 ( t ) 2 (c) It begins at position 6 and moves in the negative direction thereafter. 39. (a) v ( t ) 5 s 9 ( t ) 5 3 t 2 2 3 (b) a ( t ) 5 v 9 ( t ) 5 6 t (c) It begins at position 3 moving in a negative direction. It moves to position 1 when t 5 1, and then changes direction, moving in a positive direction thereafter. 40. (a) v ( t ) 5 s 9 ( t ) 5 6 t 2 6 t 2 (b) a ( t ) 5 v 9 ( t ) 5 6 2 12 t (c) It begins at position 0. It starts moving in the positive direction until it reaches position 1 when t 5 1, and then it changes direction. It moves in the negative direction thereafter. 41. (a) The velocity is zero when the tangent line is horizontal, at approximately t 5 2.2, t 5 6, and t 5 9.8. (b) The acceleration is zero at the inflection points, approximately t 5 4, t 5 8, and t 5 11. 42. (a) The velocity is zero when the tangent line is horizontal, at approximately t 0.2, t 5 4, and t 5 12. (b) The acceleration is zero at the inflection points, approximately t 5 1.5, t 5 5.2, t 5 8, t 5 11, and t 5 13. 43. No. f must have a horizontal tangent at that point, but f could be increasing (or decreasing), and there would be no local extremum. For example, if f ( x ) 5 x 3 , f 9 (0) 5 0 but there is no local extremum at x 5 0.
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## This document was uploaded on 10/31/2011 for the course MAC 2311 at University of Florida.

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Business Calc Homework w answers_Part_31 - Section 4.3 30....

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