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Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 4.1 - Problem 40e
Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 4.1 - Problem 40e

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# Absolute maxima and minima a. Find the | Ch 4.1 - 40E ISBN: 9780321570567 2

## Solution for problem 40E Chapter 4.1

Calculus: Early Transcendentals | 1st Edition

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Problem 40E

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.

Step-by-Step Solution:
Step 1 of 3

STEP_BY_STEP SOLUTION Step-1 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) 1 = 0 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 f be a function with domain D and let c be a fixed constant in D. Then the output value f ) 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 2 x 2 , on [- 2, 2].Clearly the function contains the root , the root value is always positive . So , 2 - x 2 0 2 That is , x 2 -2 x 2 Therefore, the given function is continuous on [-2,2] Now , we have to find out the critical points of f on the given interval. 2 Now , f(x) = x 2 x then differentiate the function both sides with respect to x. 1 d 2 f (x) = dx( x 2 x ) d 2 2 d d d d = x dx ( 2 x )+ 2 x dx (x) , sincedx(uv) = u dx(v) + v dx (u) 1 d 2 2 d 1 = x 2 dx (2 -x ) + 2 x (1) , since dx( x) = x 2 2x = x 1 (-2x) + 2 x2 2 2x2 x + 2 x 22x2 = 2 = 2 2x 2x 2 Since , from the definition f (c) = 0 = 22c 2c2 2(1c ) 2c2 = 0 1 - c = 0 2 c = 1 Hence c = -1 , 1. Clearly -1 , 1 lies between [-2,2] 2 At x = -1 , then f(-1) = (-1) 2 ( 1) = (-1) 2 1 = (-1)(1) = -1 2 At x = 1 , then f(1) = (1) 2 (1) = (1) 2 1 = (1)(1) = 1 Therefore ,f(x) attains the critical values at x = -1 ,1, and the critical points are (-1,-1), (1,1) . Step-4 b). Now , we have to determine the absolute extreme values of f on the given interval ; Here the given interval is [ 2, 2] , and the function attains the critical value at x= -1 , 1.So the endpoints are 2, 2 since from the step -2, the absolute extreme values are ; 2 Therefore , at x = - 2 then f(- 2) = (-2) ( 2) = (-2) 2 2 = (-2) (0) =0 2 At x = -1 , then f(-1) = -1) ( 1) = (-1)2 1 = (-1)(1) = -1 2 At x = 1 , then f(1) = 1) (1) = (1) 2 1 = (1)(1) = 1 2 At = 2 then f( 2) = 2) 2 ( 2) = ( 2) 2 2 = ( 2) (0) =0. Therefore , f(-2) = 0 , f(-1) = -1, f(1) =1 and f(2) = 0. Hence , the largest value of f(x) is 1 ,this attains at x= 1 Therefore , the absolute maximum is f(1) = 1 Hence , the smallest value of f(x) is -1 ,this attains at x = -1 Therefore , the absolute minimum is f(-1) = -1.

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##### ISBN: 9780321570567

The full step-by-step solution to problem: 40E from chapter: 4.1 was answered by , our top Calculus solution expert on 03/03/17, 03:45PM. The answer to “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.” is broken down into a number of easy to follow steps, and 35 words. This textbook survival guide was created for the textbook: Calculus: Early Transcendentals, edition: 1. Since the solution to 40E from 4.1 chapter was answered, more than 399 students have viewed the full step-by-step answer. This full solution covers the following key subjects: absolute, interval, given, graphing, determine. This expansive textbook survival guide covers 85 chapters, and 5218 solutions. Calculus: Early Transcendentals was written by and is associated to the ISBN: 9780321570567.

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