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Pre-Calc Homework Solutions 101

Pre-Calc Homework Solutions 101 - Section 3.7 70 For h 1 dG...

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70. For h 5 1: [ 2 2, 3] by [ 2 5, 5] For h 5 0.7: [ 2 2, 3] by [ 2 5, 5] For h 5 0.3: [ 2 2, 3] by [ 2 5, 5] As h 0, the second curve (the difference quotient) approaches the first ( y 5 2 2 x sin ( x 2 )). This is because 2 2 x sin ( x 2 ) is the derivative of cos ( x 2 ), and the second curve is the difference quotient used to define the derivative of cos ( x 2 ). As h 0, the difference quotient expression should be approaching the derivative. 71. (a) Let f ( x ) 5 ) x ) . Then } d d x } ) u ) 5 } d d x } f ( u ) 5 f 9 ( u ) } d d u x } 5 1 } d d u } ) u ) 21 } d d u x } 2 5 } ) u u ) } u 9 . The derivative of the absolute value function is 1 1 for positive values, 2 1 for negative values, and undefined at 0. So f 9 ( u ) 5 h But this is exactly how the expression } ) u u ) } evaluates. (b) f 9 ( x ) 5 3 } d d x } ( x 2 2 9) 4 ? } ) x x 2 2 2 2 9 9 ) } 5 } (2 x ) x )( 2 x 2 2 2 9 ) 9) } g 9 ( x ) 5 } d d x } ( ) x ) sin x ) 5 ) x ) } d d x } (sin x ) 1 (sin x ) } d d x } ) x ) 5 ) x ) cos x 1 } x s ) x in ) x } Note: The expression for g 9 ( x ) above is undefined at x 5 0, but actually g 9 (0) 5 lim h 0 } g (0 1 h h ) 2 g (0) } 5 lim h 0 } ) h ) s h in h } 5 0.
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