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# Absolute maxima and minima a. Find the | Ch 4.1 - 51E ISBN: 9780321570567 2

## Solution for problem 51E Chapter 4.1

Calculus: Early Transcendentals | 1st Edition

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Problem 51E

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) = sec ?x; [??/4, ?/4]

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Solution 51E Step-1 Critical point 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 Absolute extreme values 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) = sec(x) on [ - , ].Clearly the function is a trigonometric 4 4 function . Now , we have to find out the critical points of f on the given interval. Now , f(x) = sec(x) for the critical values we have to differentiate the function both sides with respect to x. d (f(x)) = d (sec(x)) dx dx 1 d f (x) = sec(x)tan(x) ,since dx sec(x) = sec(x)tan(x). Since , from the definition f (c)=0 = sec(c)tan(c) 1 sin(c) sec(c)tan(c) = 0 cos(c) cos(c) = 0 sin(c) = 0. cos (c) That is , sin(c) = 0 c = n...

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##### ISBN: 9780321570567

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