   Chapter 4.2, Problem 5E Calculus: Early Transcendental Fun...

7th Edition
Ron Larson + 1 other
ISBN: 9781337552516

Solutions

Chapter
Section Calculus: Early Transcendental Fun...

7th Edition
Ron Larson + 1 other
ISBN: 9781337552516
Textbook Problem

Writing In Exercises 5-6, explain why Rolle's Theorem does not apply to the function even though there exist a and b such that f ( a ) = f ( b ) . f ( x ) = 1 − , | x − 1 | ⌈ 0 , 2 ⌉

To determine
Why Rolle’s Theorem does not apply to the function f(x)=1|x1|, [0,2] even though there exist a and b such that f(a)=f(b).

Explanation

Break the absolute function in a piece-wise function as,

f(x)={x           0<x1(2x)  1x<2

At, x=1 function is defined for both the piece-wise functions and f(x) is continuous but at x=1, there exists two values of f(x) as shown,

f(1)={1           0<x1(1)     1x

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