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# Absolute maxima and minima a. Find the | Ch 4.1 - 54E ISBN: 9780321570567 2

## Solution for problem 54E Chapter 4.1

Calculus: Early Transcendentals | 1st Edition

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Problem 54E

Absolute maxima and minima a. Find the critical points off on the given interval. b. Determine the absolute extreme values off on the given interval. c. Use a graphing utility to confirm your conclusions. ? ? ? ? f(? ) = x? ln (? /5); [0.1, 5]

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Solution 54E Step-1 Critical value definition; Let f be a continuous function defined on an open interval containing a number 1 1 ‘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 Absolute extreme value definition; When an output value of a function is a maximum or a minimum over the entire domain of the function, the value is called the absolute maximum or the absolute minimum. Let f be a function with domain D and let c be a fixed constant in D. Then the output value f(c) is the 1. Absolute maximum value of f on D if and only if f(x) f(c) , for all x in D. 2. Absolute minimum value of f on D if and only if f(c) f(x) , for all x in D. Step_3 a). The given function is f(x) = x ln( ) ,on [0.1,5] .Clearly the function is a 5 exponential function and it is continuous for all of x . Now , we have to find out the critical points of f on the given interval. x Now , f(x) = x ln5 ) , for the critical points we have to differentiate the function both sides with respect to x. d d x dx f(x) = dx( x ln5 )) 1 d x x d d d(v) d(u) f (x) = x dx( ln( )5)+ ln( ) 5 dx ( ), sincedx (uv) = u dx +v dx = x( 1 ) d (x/5)+ ln( ) (1) (x/5) dx 5 = x( 5 )(1 ) + ln( ) , since d ln(x) = 1/x. x 5 5 dx x = 1 + ln( ) 5 Since , from the definition...

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##### ISBN: 9780321570567

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