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# Absolute maxima and minima a. Find the | Ch 4.1 - 37E ISBN: 9780321570567 2

## Solution for problem 37E Chapter 4.1

Calculus: Early Transcendentals | 1st Edition

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Problem 37E

Absolute maxima and minima a. Find the critical points of f on the given interval. b. Determine the absolute extreme values off on the given interval. c. Use a graphing utility to confirm your conclusions. x f(x) = (2x) , on [0.1,1]

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STEP_BY_STEP SOLUTION Step-1 Let f be a continuous function defined on an open interval containing a number ‘c’.The number ‘c’ is critical value ( or critical number ). If f (c) = 0 1 or f (c) is undefined. A critical point on that graph of f has the form (c,f(c)). Step-2 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 be a funct ion withdomain D and let c be afixed constant in D . Then he 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) = (2x) , on [0.1,1]. Clearly the function is a 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) = (2x) Now, take log on both sides .Because the base and power contains the algebraic functions .So in this cases we can take log. x m ln(f(x)) = ln((2x) ) = x ln(2x), since ln(a ) = m ln(a) then differentiate the function both sides with respect to x. d d dx ( ln(f(x)) ) = dx(x ln(2x)) 1 d (f(x)) = x d ln(2x) + ln(2x) d (x)...............(1) f(x) dx dx dx d d(v) d(u) d 1 since dx (uv)= u dx +v dx and dx (ln(x)) = x f (x) 2x from(1), f(x) = 2x+ln(2x)(1) =...

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##### ISBN: 9780321570567

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