Calculus 10th Edition

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

Chapter 3 - Differentiation - 3.2 Exercises: 12

Answer

Rolle's Theorem can be applied, $c=2.$

Work Step by Step

Since $f(x)$ is a polynomial, it is continuous for all values of $x$ and differentiable at every value of $x$. Since $f(x)$ is continuous over $[ , ]$ and differentiable over $(, )$, applying Rolle's Theorem over the interval $[, ]$ guarantees the existence of at least one value $c$ such that $\lt c\lt $ and $f'(c)=0.$ Product Rule $(f’(x)=(u(x)(v(x))’=u’(x)v(x)+u(x)v’(x))$ $u(x)=(x-4) ;u’(x)= 1$ $v(x)=(x+2)^2 ;v’(x)=2(x+2) $ $f'(x)=(1)(x+2)^2+(x-4)(2x+4)$ $=3x^2-12=3(x-2)(x+2).$ $f'(x)=0\to 3(x-2)(x+2)=0\to c=2$ or $c=-2$( which is rejected since $c\gt-2.$
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