Chapter 5 Solutions Manual

# Chapter 5 Solutions Manual - February 8, 2005 10:32...

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February 8, 2005 10:32 L38-ch05 Sheet number 1 Page number 81 black 81 CHAPTER 5 The Derivative in Graphing and Applications EXERCISE SET 5.1 1. (a) f 0 > 0 and f 0 > 0 y x (b) f 0 > 0 and f 0 < 0 y x (c) f 0 < 0 and f 0 > 0 y x (d) f 0 < 0 and f 0 < 0 y x 3. A : dy/dx < 0 ,d 2 y/dx 2 > 0 B : dy/dx > 0 2 y/dx 2 < 0 C : dy/dx < 0 2 y/dx 2 < 0 5. An inﬂection point occurs when f 0 changes sign: at x = 1 , 0 , 1 and 2. 7. (a) [4 , 6] (b) [1 , 4] and [6 , 7] (c) (1 , 2) and (3 , 5) (d) (2 , 3) and (5 , 7) (e) x =2 , 3 , 5 9. (a) f is increasing on [1 , 3] (b) f is decreasing on ( −∞ , 1] , [3 , + ] (c) f is concave up on ( −∞ , 2) , (4 , + ) (d) f is concave down on (2 , 4) (e) points of inﬂection at x , 4 11. f 0 ( x )=2( x 3 / 2) f 0 ( x )=2 (a) [3 / 2 , + ) (b) ( −∞ , 3 / 2] (c) ( −∞ , + ) (d) nowhere (e) none 13. f 0 ( x ) = 6(2 x +1) 2 f 0 ( x ) = 24(2 x (a) ( −∞ , + ) (b) nowhere (c) ( 1 / 2 , + ) (d) ( −∞ , 1 / 2) (e) 1 / 2 15. f 0 ( x )=12 x 2 ( x 1) f 0 ( x )=36 x ( x 2 / 3) (a) [1 , + ) (b) ( −∞ , 1] (c) ( −∞ , 0), (2 / 3 , + ) (d) (0 , 2 / 3) (e) 0 , 2 / 3 17. f 0 ( x )= 3( x 2 3 x ( x 2 x 3 f 0 ( x 6 x (2 x 2 8 x +5) ( x 2 x 4 (a) [ 3 5 2 , 3+ 5 2 ] (b) ( −∞ , 3 5 2 ], [ 3+ 5 2 , + ) (c) (0 , 2 6 2 ), (2 + 6 2 , + ) (d) ( −∞ , 0), (2 6 2 , 2+ 6 2 ) (e) 0 , 2 6 / 2 , 6 / 2

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February 8, 2005 10:32 L38-ch05 Sheet number 2 Page number 82 black 82 Chapter 5 19. f 0 ( x )= 2 x +1 3( x 2 + x +1) 2 / 3 f 0 ( x 2( x + 2)( x 1) 9( x 2 + x 5 / 3 (a) [ 1 / 2 , + ) (b) ( −∞ , 1 / 2] (c) ( 2 , 1) (d) ( −∞ , 2) , (1 , + ) (e) 2 , 1 21. f 0 ( x 4( x 2 / 3 1) 3 x 1 / 3 f 0 ( x 4( x 5 / 3 + x ) 9 x 7 / 3 (a) [ 1 , 0] , [1 , + ) (b) ( −∞ , 1] , [0 , 1] (c) ( −∞ , 0), (0 , + ) (d) nowhere (e) none 23. f 0 ( x xe x 2 / 2 f 0 ( x )=( 1+ x 2 ) e x 2 / 2 (a) ( −∞ , 0] (b) [0 , + ) (c) ( −∞ , 1), (1 , + ) (d) ( 1 , 1) (e) 1 , 1 25. f 0 ( x x x 2 +4 f 0 ( x x 2 4 ( x 2 +4) 2 (a) [0 , + ) (b) ( −∞ , 0] (c) ( 2 , +2) (d) ( −∞ , 2) , (2 , + ) (e) 2 , +2 27. f 0 ( x 2 x 1+( x 2 1) 2 f 0 ( x 2 3 x 4 2 x 2 2 [1+( x 2 1) 2 ] 2 (a) [0 + ) (b) ( −∞ , 0] (c) ( 1+ 7 3 , 1 7 3 ) , ( 1 7 3 , 1+ 7 3 ) (d) ( −∞ , 1+ 7 3 ) , ( 1 7 3 , 1 7 3 ) , ( 1+ 7 3 , + ) (e) four: ± p 1 ± 7 3 29. f 0 ( x )=cos x + sin x f 0 ( x sin x +cos x (a) [ π/ 4 , 3 4] (b) ( π, 4] , [3 4 ) (c) ( 3 4 ,π/ 4) (d) ( 3 4) , ( 4 ) (e) 3 4 4 1.5 –1.5 ^6 31. f 0 ( x 1 2 sec 2 ( x/ 2) f 0 ( x 1 2 tan( x/ 2) sec 2 ( x/ 2)) (a) nowhere (b) ( π,π ) (c) ( 0) (d) (0 ) (e) 0 10 –10 Cc
February 8, 2005 10:32 L38-ch05 Sheet number 3 Page number 83 black Exercise Set 5.1 83 33. f ( x ) = 1 + sin 2 x f 0 ( x )=2cos2 x f 0 ( x )= 4 sin 2 x (a) [ π, 3 π/ 4] , [ 4 ,π/ 4] , [3 4 ] (b) [ 3 4 , 4] , [ 4 , 3 4] (c) ( 2 , 0) , ( 2 ) (d) ( 2) , (0 2) (e) 2 , 0 2 2 0 Cc 35. (a) 2 4 x y (b) 2 4 x y (c) 2 4 x y 37. (a) g ( x ) has no zeros: There can be no zero of g ( x ) on the interval −∞ <x< 0 because if there were, say g ( x 0 ) = 0 where x 0 < 0, then g 0 ( x ) would have to be positive between x = x 0 and x =0,say g 0 ( x 1 ) > 0 where x 0 <x 1 < 0. But then g 0 ( x ) cannot be concave up on the interval ( x 1 , 0), a contradiction. There can be no zero of g ( x )on0 4 because g ( x ) is concave up for 0 4 and thus the graph of g ( x ), for 0 4, must lie above the line y = 2 3 x + 2, which is the tangent line to the curve at (0 , 2), and above the line y =3( x 4)+3=3 x 9 also for 0 4 (see ±gure). The ±rst condition says that g ( x ) could only be zero for x> 3 and the second condition says that g ( x ) could only be zero for x< 3, thus g (

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## This note was uploaded on 09/04/2008 for the course CALCULUS calculus taught by Professor Calculus during the Fall '08 term at UNC Charlotte.

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Chapter 5 Solutions Manual - February 8, 2005 10:32...

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