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: 39

Answer

The Mean Value Theorem can be applied; $c=\dfrac{\sqrt{3}}{3}$ or $c-\dfrac{\sqrt{3}}{3}.$

Work Step by Step

Since $f(x)$ is a polynomial, it is continuous for all values of $x$ and differentiable at every value of $x.$ $a=-1; b=1\to f(a)=-3$ and $f(b)=3.$ Since $f(x)$ is continuous over $[a, b]$ and differentiable over $(a, b)$, applying 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 $f′(c)=\dfrac{f(b)-f(a)}{b-a}.$ $f'(c)=\dfrac{3-(-3)}{1-(-1)}=3.$ $f'(x)=3x^2+2\to f'(x)=3\to 3x^2+2=3\to c=+\dfrac{\sqrt{3}}{3}$ or $c=-\dfrac{\sqrt{3}}{3}.$
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