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 - 3.1 Exercises: 4


$f'(-1)$ is not defined. $f'(-\frac{2}{3})=0.$

Work Step by Step

Product Rule $(f’(x)=(u(x)(v(x))’=u’(x)v(x)+u(x)v’(x))$ $u(x)=-3x ;u’(x)=-3 $ $v(x)=\sqrt{x+1} ;v’(x)=\dfrac{1}{2\sqrt{x+1}}. $ $f'(x)=(-3x)(\dfrac{1}{2\sqrt{x+1}})+(-3)(\sqrt{x+1})$ $=\dfrac{1}{2\sqrt{x+1}}(-3x-6(x+1))=\dfrac{-9x-6}{2\sqrt{x+1}}; x\ne-1$ $f'(-1)$ is undefined. $f'(-\frac{2}{3})=\dfrac{-9(-\frac{2}{3})-6}{2\sqrt{(-\frac{2}{3})+1}}=0.$
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