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Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 4.1 - Problem 57e
Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 4.1 - Problem 57e

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# Solved: Critical points of functions with unknown

ISBN: 9780321570567 2

## Solution for problem 57E Chapter 4.1

Calculus: Early Transcendentals | 1st Edition

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Problem 57E

Critical points of functions with unknown parameters ?Find the critical points off. Assume a and b are constants.

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Solution 57E Step-1 Critical value definition ; Let f be a continuous function defined on an open interval containing a 1 number ‘c’.The number ‘c’ is critical value ( or critical number ). If f (c) = 0 or f (c) is undefined. A critical point on that graph of f has the form (c,f(c)). Step-2 a). The given function is f(x) = xx a . Clearly the function contains the root value So , x - a 0 That is , x a Therefore , x belongs to [a, ) Therefore, the given function is continuous on [a, ). Now , we have to find out the critical points of f on the interval. Now , f(x) = xx a for the critical points , we have to differentiate the function both sides with respect to x. d d dx f(x) = dx( x x a) f (x) = x a d(x+ x d x a , since d (uv) = v d (u)+u d (v). dx dx dx dx dx = x a(1) + x 1 d (x-a) , since d (x) = 1 2xa dx dx 2x = x a + x 1 (1). 2xa ince , from the definition f (c) = 0 = c a + c 1 2ca 1 c a + c 2ca = 0. 2c2a+c That is , 2ca = 0. Therefore , 3c2a = 0. 2ca Hence , (3c - 2a) =0 2a C = 3 2a 2a 2a 2a At x = 3 , the value of the function is f( 3 ) = 3 23 a = ( 3 )( 3a) Case (1) ; I f a 0 , then a , is not defined. 2a Case (2) ; If a < 0 , then a is defined .so , in this cases the critical value is x = 3 . That is the only critical value is x = 23 . 2a Therefore ,the given function attains the critical value at x = 3 , and the critical point is (a , (a)( a )). 3 3 3

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