Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 4 - Section 4.1 - Maximum and Minimum Values - 4.1 Exercises - Page 287: 42

Answer

$x = - 1$

Work Step by Step

$$\eqalign{ & h\left( x \right) = {x^{ - 1/3}}\left( {x - 2} \right) \cr & {\text{The domain of the funcion is}} \cr & D = \left\{ {\left. {x \in R} \right|x \ne 0} \right\} \cr & {\text{Differentiate with respect to }}x \cr & h'\left( x \right) = \frac{d}{{dx}}\left[ {{x^{ - 1/3}}\left( {x - 2} \right)} \right] \cr & {\text{Using the product rule for derivatives}} \cr & h'\left( x \right) = {x^{ - 1/3}}\frac{d}{{dx}}\left[ {\left( {x - 2} \right)} \right] + \left( {x - 2} \right)\frac{d}{{dx}}\left[ {{x^{ - 1/3}}} \right] \cr & {\text{Computing derivatives and simplify}} \cr & h'\left( x \right) = {x^{ - 1/3}}\left( 1 \right) + \left( {x - 2} \right)\left( { - \frac{1}{3}{x^{ - 4/3}}} \right) \cr & h'\left( x \right) = {x^{ - 1/3}} - \frac{1}{3}{x^{ - 1/3}} + \frac{2}{3}{x^{ - 4/3}} \cr & h'\left( x \right) = \frac{2}{3}{x^{ - 1/3}} + \frac{2}{3}{x^{ - 4/3}} \cr & h'\left( x \right) = \frac{2}{3}{x^{ - 4/3}}\left( {x + 1} \right) \cr & h'\left( x \right) = \frac{{2\left( {x + 1} \right)}}{{3{x^{4/3}}}} \cr & {\text{The derivative does not exist when the denominator is 0}} \cr & 3{x^{4/3}} = 0 \to x = 0 \cr & x = 0{\text{ is not in the domain of the given function}}{\text{, hence}} \cr & {\text{it is not a critical point}}{\text{.}} \cr & {\text{Find the values of }}x{\text{ where }}p'\left( x \right) = 0 \cr & \frac{{2\left( {x + 1} \right)}}{{3{x^{4/3}}}} = 0 \cr & 2\left( {x + 1} \right) = 0 \cr & x = - 1 \cr & \cr & {\text{Applying the definition of critical numbers:}} \cr & {\text{A critical number of a function is a number }}c{\text{ in the domain }} \cr & {\text{off such that either }}f'\left( c \right) = 0{\text{ or }}f'\left( c \right){\text{ does not exist}}{\text{, so the}} \cr & {\text{critical number is:}} \cr & x = - 1 \cr} $$
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