Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 4 - Section 4.2 - The Mean Value Theorem - 4.2 Exercises: 5

Answer

C=1

Work Step by Step

1. $f(x)=2x^{2}-4x+5$ since this a polynomial, $f(x)$ is continuous at $[-1,3]$ and $f(x)$ is differentiable at $(-1,3)$ 2. Now since the first condition is satisfied, prove that f(a)=f(b) $f(-1) = 2\times(-1)^{2}-4\times(-1)+5$ $= 2+4+5$ $= 11$ $f(3) = 2\times(3)^{2}-4\times(3)+5$ $= 18-12-5$ $=11$ Therefore, $f(a)=f(b)$ Moving on to the final step: $f'(c)=4c-4$ $f(c)=0 $ therefore, $4c-4=0$ $c=1$ Thus, we were able to find the value of $c$ that satisfies the conclusions of Rolle's theorem.
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