Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 3 - Applications of Differentiation - 3.2 The Mean Value Theorem - 3.2 Exercises - Page 220: 11

Answer

$c=1$

Work Step by Step

Here, $f(x)=2x^2-3x+1$ in interval $[0,2]$. 1. Since $f(x)$ is a polynomial, it is continuous over the given interval $[0,2]$. 2. Similarly, since $f(x)$ is a polynomial, it is differentiable on the open interval $(0,2)$. Thus, there exists a number $c$ in the open interval $(0,2)$ such that $f'(c)=\frac{f(2)-f(0)}{2-0}$. Here, $f(0)=1$. Similarly,$f(2)=3$. Thus, $$f'(c)=\frac{3-1}{2}=1$$ $$4c-3=1\implies c=1$$
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