Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 4 - Applications of the Derivative - 4.1 Maxima and Minima - 4.1 Exercises - Page 243: 35

Answer

$$\eqalign{ & \left( a \right){\text{Critical points }}x = - \frac{4}{5},\,\,\,x = 0 \cr & \left( b \right)\,\,x = 0{\text{ Is the absolute }}\left( {{\text{and local}}} \right){\text{ minimum}} \cr & \,\,\,\,\,\,\,x = - \frac{4}{5}{\text{ is the local maximum}}\,\, \cr} $$

Work Step by Step

$$\eqalign{ & f\left( x \right) = {x^2}\sqrt {x + 1} ,{\text{ on the interval }}\left[ { - 1,1} \right] \cr & \cr & {\text{Differentiate}} \cr & f'\left( x \right) = {x^2}\left( {\frac{1}{{2\sqrt {x + 1} }}} \right) + 2x\sqrt {x + 1} \cr & f'\left( x \right) = \frac{{{x^2} + 4x\left( {x + 1} \right)}}{{2\sqrt {x + 1} }} \cr & f'\left( x \right) = \frac{{5{x^2} + 4x}}{{2\sqrt {x + 1} }} \cr & {\text{Setting the derivative equal to zero}}{\text{, we have}} \cr & 5{x^2} + 4x = 0 \cr & x\left( {5x + 4} \right) = 0 \cr & x = 0,\,\,\,x = - \frac{4}{5} \cr & {\text{Both of which lie in the given interval }}\left[ { - 1,1} \right]. \cr & {\text{These points and the endpoints are candidates for the location}} \cr & {\text{of absolute extrema:}} \cr & {\text{Points }}x = \left\{ { - 1, - \frac{4}{5},0,1} \right\} \cr & \cr & {\text{Evaluating these points}} \cr & f\left( { - 1} \right) = {\left( { - 1} \right)^2}\sqrt {\left( { - 1} \right) + 1} = 0 \cr & f\left( { - \frac{4}{5}} \right) = {\left( { - \frac{4}{5}} \right)^2}\sqrt {\left( { - \frac{4}{5}} \right) + 1} = \frac{{16}}{{25}}\sqrt {\frac{1}{5}} \cr & f\left( 0 \right) = {\left( 0 \right)^2}\sqrt {\left( 0 \right) + 1} = 0 \cr & f\left( 1 \right) = {\left( 1 \right)^2}\sqrt {\left( 1 \right) + 1} = \sqrt 2 \cr & \cr & {\text{The largest of these function values is }}f\left( 1 \right) = \sqrt 2 ,{\text{ which is the}} \cr & {\text{absolute maximum on the interval }}\left[ { - 7,7} \right]. \cr & \cr & {\text{The smallest of these values is }}f\left( 0 \right) = 0\,{\text{which is the absolute}}\,\left( {{\text{and local}}} \right) \cr & {\text{minimum on the interval }}\left[ { - 7,7} \right]. \cr & \cr & {\text{The next graph shows that the critical points corresponds to neither}} \cr & {\text{a local maximum nor a local minimum}}{\text{.}} \cr} $$
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