Calculus: Early Transcendentals (2nd Edition)

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

Chapter 4 - Applications of the Derivative - 4.2 What Derivatives Tell Us - 4.2 Exercises - Page 257: 38

Answer

$$\eqalign{ & {\text{Increasing on }}\left( { - \sqrt 2 ,\sqrt 2 } \right) \cr & {\text{Decreasing on }}\left( { - \infty , - \sqrt 2 } \right),\,\left( {\sqrt 2 ,\infty } \right) \cr} $$

Work Step by Step

$$\eqalign{ & f\left( x \right) = {\tan ^{ - 1}}\left( {\frac{x}{{{x^2} + 2}}} \right) \cr & {\text{Derivative}} \cr & f'\left( x \right) = \frac{1}{{1 + {{\left( {x/\left( {{x^2} + 2} \right)} \right)}^2}}}\frac{d}{{dx}}\left[ {\frac{x}{{{x^2} + 2}}} \right] \cr & f'\left( x \right) = \frac{{{{\left( {{x^2} + 2} \right)}^2}}}{{{{\left( {{x^2} + 2} \right)}^2} + {x^2}}}\left( {\frac{{{x^2} + 2 - x\left( {2x} \right)}}{{{{\left( {{x^2} + 2} \right)}^2}}}} \right) \cr & f'\left( x \right) = \frac{{2 - {x^2}}}{{{{\left( {{x^2} + 2} \right)}^2} + {x^2}}} \cr & {\text{Set the derivative to 0}} \cr & 2 - {x^2} = 0 \cr & {\text{Solving the equation we obtain}} \cr & x = - \sqrt 2 {\text{ and }}x = \sqrt 2 \cr & {\text{From the critical values and the domain }}\left( { - \infty ,\infty } \right){\text{ we have}} \cr & \left( { - \infty , - \sqrt 2 } \right),\,\,\left( { - \sqrt 2 ,\sqrt 2 } \right),\,\,\left( {\sqrt 2 ,\infty } \right) \cr & {\text{Now}}{\text{, we will evaluate the critical value and resume in a table}} \cr} $$ \[\begin{array}{*{20}{c}} {{\rm{Interval}}}&{{\rm{Test\ value }}\left( x \right)}&{{\rm{Sign\ of }}f'\left( x \right)}&{{\rm{Behavior\ of }}f\left( x \right)}\\ {\left( { - \infty , - \sqrt 2 } \right)}&{ - 2}& - &{{\rm{Decreasing}}}\\ {\left( { - \sqrt 2 ,\sqrt 2 } \right)}&0& + &{{\rm{Increasing}}}\\ {\left( {\sqrt 2 ,\infty } \right)}&2& - &{{\rm{Decreasing}}}\\ {}&{}&{}&{} \end{array}\] $$\eqalign{ & {\text{From the table we can conlude that the function is:}} \cr & {\text{Increasing on }}\left( { - \sqrt 2 ,\sqrt 2 } \right) \cr & {\text{Decreasing on }}\left( { - \infty , - \sqrt 2 } \right),\,\left( {\sqrt 2 ,\infty } \right) \cr} $$
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