Pre-Calc Homework Solutions 188

# Pre-Calc Homework Solutions 188 - 188 Chapter 4 Review...

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7. Note that the domain is ( 2 1, 1). y 5 (1 2 x 2 ) 2 1/4 y 952} 1 4 } (1 2 x 2 ) 2 5/4 ( 2 2 x ) 5 } 2(1 2 x x 2 ) 5/4 } y 05 5 5 The second derivative is always positive, so the function is concave up on its domain ( 2 1, 1). Graphical support: [ 2 1.3, 1.3] by [ 2 1, 3] (a) [0, 1) (b) ( 2 1, 0] (c) ( 2 1, 1) (d) None (e) Local minimum at (0, 1) (f) None 8. y 95 52 } ( 2 x x 3 3 2 1 1 1 ) 2 } y 052 5 } 6 ( x x 2 3 ( x 2 3 1 1) 3 2) } Graphical support: [ 2 4.7, 4.7] by [ 2 3.1, 3.1] (a) ( 2‘ , 2 2 2 1/3 ] < ( 2‘ , 2 0.794] (b) [ 2 2 2 1/3 , 1) < [ 2 0.794, 1) and (1, ) (c) ( 2‘ , 2 2 1/3 ) < ( 2‘ , 2 1.260) and (1, ) (d) ( 2 2 1/3 , 1) < ( 2 1.260, 1) (e) Local maximum at 1 2 2 2 1/3 , } 2 3 } ? 2 2 1/3 2 < ( 2 0.794, 0.529) (f) 1 2 2 1/3 , } 1 3 } ? 2 1/3 2 < ( 2 1.260, 0.420) 9. Note that the domain is [ 2 1, 1]. y 952 } ˇ 1 w 1 2 w x w 2 w } Since y 9 is negative on ( 2 1, 1) and y is continuous, y is decreasing on its domain [ 2 1, 1]. y 05 } d d x } [ 2 (1 2 x 2 ) 2 1/2 ] 5 } 1 2 } (1 2 x 2 ) 2 3/2 ( 2 2 x ) } (1 2 x x 2 ) 3/2 } Graphical support: [ 2 1.175, 1.175] by 3 2} p 4 } , } 5 4 p } 4 (a) None (b) [ 2 1, 1] (c) ( 2 1, 0) (d) (0, 1) (e) Local (and absolute) maximum at ( 2 1, p ); local (and absolute) minimum at (1, 0) (f) 1 0, } p 2 } 2 ( x 3 2 1)(6 x 2 ) 2 (2 x 3 1 1)(6 x 2 ) }}}} ( x 3 2 1) 3 ( x 3 2 1) 2 (6 x 2 ) 2 (2 x 3 1 1)(2)( x 3 2 1)(3
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## This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.

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