Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 3 - The Derivative In Graphing And Applications - 3.4 Absolute Maxima And Minima - Exercises Set 3.4 - Page 222: 11

Answer

$$\eqalign{ & {\text{max}} = \frac{{3\sqrt 5 }}{5}{\text{ at }}x = 1 \cr & {\text{min}} = - \frac{{3\sqrt 5 }}{5}{\text{ at }}x = - 1 \cr} $$

Work Step by Step

$$\eqalign{ & f\left( x \right) = \frac{{3x}}{{\sqrt {4{x^2} + 1} }};\,\,\left[ { - 1,1} \right] \cr & f\left( x \right){\text{ is }}\,{\text{continuous }}\,{\text{and}}\,{\text{differentiable everywhere}}{\text{, the }} \cr & {\text{absolute}}\,{\text{ extrema }}\,{\text{must }}\,{\text{occur}}\,{\text{ }}\,{\text{either }}\,{\text{at }}\,{\text{endpoints }}\,{\text{of }} \cr & {\text{the interval or at solutions to the equation }}f'\left( x \right) = 0{\text{ in}} \cr & {\text{the open interval }}\left( { - 1,1} \right){\text{.}} \cr & \cr & {\text{Differentiating }}f\left( x \right) \cr & f'\left( x \right) = \frac{{3\sqrt {4{x^2} + 1} - 3x\left( {\frac{{8x}}{{2\sqrt {4{x^2} + 1} }}} \right)}}{{{{\left( {\sqrt {4{x^2} + 1} } \right)}^2}}} \cr & f'\left( x \right) = \frac{{3\sqrt {4{x^2} + 1} - \frac{{12{x^2}}}{{\sqrt {4{x^2} + 1} }}}}{{4{x^2} + 1}} \cr & f'\left( x \right) = \frac{{3\left( {4{x^2} + 1} \right) - 12{x^2}}}{{{{\left( {4{x^2} + 1} \right)}^{3/2}}}} \cr & f'\left( x \right) = \frac{{12{x^2} + 3 - 12{x^2}}}{{{{\left( {4{x^2} + 1} \right)}^{3/2}}}} \cr & f'\left( x \right) = \frac{3}{{{{\left( {4{x^2} + 1} \right)}^{3/2}}}} \cr & {\text{setting }}f'\left( x \right) = 0 \cr & \frac{3}{{{{\left( {4{x^2} + 1} \right)}^{3/2}}}} = 0 \cr & {\text{Thus}}{\text{, there are no stationary point}}{\text{.}} \cr & {\text{Evaluating }}f\left( x \right){\text{ at the endpoints}}{\text{, }}x = - 1,{\text{ }}x = 1 \cr & f\left( { - 1} \right) = \frac{{3\left( { - 1} \right)}}{{\sqrt {4{{\left( { - 1} \right)}^2} + 1} }} = - \frac{{3\sqrt 5 }}{5} \cr & f\left( 1 \right) = \frac{{3\left( 1 \right)}}{{\sqrt {4{{\left( 1 \right)}^2} + 1} }} = \frac{{3\sqrt 5 }}{5} \cr & {\text{Then we can conclude that the absolute minimum of }}f\left( x \right){\text{ on}} \cr & {\text{the}}\,\,\,{\text{ interval }}\,\,\left[ { - 1,1} \right]{\text{ is }}\, - \frac{{3\sqrt 5 }}{5}\,{\text{ at }}x = - 1.{\text{ }}\,{\text{And}}\,\,{\text{ the}}\,{\text{ absolute }} \cr & {\text{maximum is }}\frac{{3\sqrt 5 }}{5}{\text{ at }}x = 1. \cr & \cr & {\text{max}} = \frac{{3\sqrt 5 }}{5}{\text{ at }}x = 1 \cr & {\text{min}} = - \frac{{3\sqrt 5 }}{5}{\text{ at }}x = - 1 \cr} $$
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