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Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 4.1 - Problem 9e
Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 4.1 - Problem 9e

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# How do you determine the absolute maximum and minimum

ISBN: 9780321570567 2

## Solution for problem 9E Chapter 4.1

Calculus: Early Transcendentals | 1st Edition

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Problem 9E

How do you determine the absolute maximum and minimum values of a continuous function on a closed interval?

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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 or f (c) is undefined. A critical point on that graph of f has the form (c,f(c)). Step_2 Now , we have to determine the absolute maximum and minimum values of a continuous function on a closed interval. That is ; 1. Verify that the function is continuous on the interval [a, b]. 2. Find all critical points of f(x) that are in the interval (a,b). 3. Add the endpoints a and b of the interval [a,b] to the list of points found in step-2. 4. Compute the value of f at each of the points in this list. 5. The largest value in step 4 is the absolute maximum value of f on [a, b] and the smallest value is the absolute minimum value. Step-3 Example;Determine the absolute extreme values for the following function and interval. 3 2 g(t) = 2t + 3t -12t +4 0n[-4,2]. g (t) = 0 = 6t + 6t-12 6( t + t-2) = 0. 2 ( t + 2t t-2) = 0. t ( t+2) - (t+2) = 0. (t+2)(t-1) = 0 t= -2 , and t=1. Therefore , critical values are t =-2, 1, and the end points are -4,2. 3 2 At , t= -4 then g(-4) = 2( 4) + 3( 4) -12(-4) +4 = -128+48+48+4 = -28. 3 2 At , t= -2 then g(-2) = 2( 2) + 3( 2) -12(-2) +4 = -16+12+24+4 =24. 3 2 At , t= 1 then g(1) = 2(1) + 3(1) -12(1) +4 = 2+3-12+4 = -3. 3 2 At , t= 2 then g(2) = 2(2) + 3(2) -12(2) +4 = 16+12-24+4 = 8. Hence , the absolute maximum of g(t) is 24 and it occurs at t = -2 ( a critical point), and the absolute minimum of g(t) is -28 and it occurs at t = -4 (an endpoint).

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