Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 3 - Applications of Differentiation - 3.2 Exercises: 40

Answer

The Mean Value Theorem can be applied; $c=\sqrt[3]{2}.$

Work Step by Step

Since $f(x)$ is a polynomial, it is continuous for all values of $x$ and differentiable at every value of $x.$ $a=0; b=2\to f(a)=0$ and $f(b)=0$ Since $f(x)$ is continuous over $[a, b]$ and differentiable over $(a, b)$, applying the Mean Value Theorem over the interval $[a , b]$ guarantees the existence of at least one value c such that $a\lt c\lt b$ and $f′(c)=\dfrac{f(b)-f(a)}{b-a}.$ $f'(c)=\dfrac{0-0}{2-0}=0.$ $f'(x)=4x^3-8\to x^3=2.\to c=\sqrt[3]{2}.$
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