Explanation Given data: The given functions are f ( x ) = 3 x − 2 x − 1 , f ( x ) = 1 1 + x , f ( x ) = sin π x 2 , and f ( x ) = e x ( x − 2 ) Method/Formula used: The critical points of a function y = f ( x ) are values c such that f ′ ( c ) = 0 or the points where f ′ ( c ) does not exist. Calculation: Consider function f ( x ) = 3 x − 2 x − 1 ⇒ f ′ ( x ) = ( x − 1 ) ⋅ 3 − ( 3 x − 2 ) ⋅ 1 ( x − 1 ) 2 = 3 x − 3 − 3 x + 2 ( x − 1 ) 2 = − 1 ( x − 1 ) 2 Clearly, the function f ′ ( x ) does not exist when ( x − 1 ) 2 = 0 ⇒ ( x − 1 ) = 0 ⇒ x = 1 i.e., at point x = 1 . Therefore, x = 1 is the critical point of the function f ( x ) = 3 x − 2 x − 1 . Consider function f ( x ) = 1 1 + x ⇒ f ′ ( x ) = − 1 ( 1 + x ) 2 ⋅ 1 2 x . The derivative f ′ ( x ) does not exist when ( 1 + x ) 2 ⋅ 2 x = 0 ⇒ ( 1 + x ) 2 = 0 ⇒ x = − 1 ⇒ x = 1 , and x = 0 ⇒ x = 0 =0 Therefore, x = 1 , and x = 0 the critical point of the function f ( x ) = 1 1 + x

14th Edition

ISBN: 9781464140051

Author: TAALMAN, Laura, Kohn, Peter

Publisher: W. H. Freeman

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The critical points of given functions.

Solution Summary: The author explains the critical points of a function y=f(x) which are values c such that f′(c)=0 or the points where

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