Explanation Given : f ( x ) = x 4 − 3 x 2 − 2 , [ a , b ] = [ − 2 , 2 ] Concept used: Extreme value theorem: If a function f ( x ) is continuous on a closed interval [ a , b ] , then f ( x ) has both a maximum and minimum value on [ a , b ] . Continuity: The function f is continuous at x = c if f ( c ) is defined and if lim → x → c f ( x ) = f ( c ) . Calculation: f ( x ) = x 4 − 3 x 2 − 2 , [ a , b ] = [ − 2 , 2 ] be a given function. And it has both a maximum and a minimum value on [ a , b ] . First find lim → x → − 2 f ( x ) lim → x → − 2 f ( x ) = lim → x → − 2 ( x 4 − 3 x 2 x

5th Edition

ISBN: 9781429241861

Author: Laura Taalman, Peter Kohn

Publisher: W. H. Freeman

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Question

Chapter 1.4, Problem 49S

To determine

To show:A fuction f has both a maximum a mimimum value on [a,b] by using the extreme value theorem and to approximate values M and m by using graphing utility.

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Chapter 1.4, Problem 48S

Chapter 1.4, Problem 50S

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Chapter 1 Solutions

Calculus

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A fuction f has both a maximum a mimimum value on [ a , b ] by using the extreme value theorem and to approximate values M and m by using graphing utility.

Solution Summary: The author illustrates the extreme value theorem and graphing utility for a given function f.

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To show:A fuction f has both a maximum a mimimum value on [a,b] by using the extreme value theorem and to approximate values M and m by using graphing utility.

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Chapter 1 Solutions