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 256: 26

Answer

\[\begin{align} & \text{Decreasing on }\left( 0,\infty \right) \\ & \text{Increasing on }\left( -\infty ,0 \right) \\ \end{align}\]

Work Step by Step

\[\begin{align} & f\left( x \right)=\frac{{{e}^{x}}}{{{e}^{2x}}+1} \\ & \text{Diferentiate } \\ & f'\left( x \right)=\frac{{{e}^{x}}\left( {{e}^{2x}}+1 \right)-{{e}^{x}}\left( 2{{e}^{2x}} \right)}{{{\left( {{e}^{2x}}+1 \right)}^{2}}} \\ & f'\left( x \right)=\frac{{{e}^{3x}}+{{e}^{x}}-2{{e}^{3x}}}{{{\left( {{e}^{2x}}+1 \right)}^{2}}} \\ & f'\left( x \right)=\frac{{{e}^{x}}-{{e}^{3x}}}{{{\left( {{e}^{2x}}+1 \right)}^{2}}} \\ & \text{Calculate the critical points, set }f'\left( x \right)=0 \\ & \frac{{{e}^{x}}-{{e}^{3x}}}{{{\left( {{e}^{2x}}+1 \right)}^{2}}}=0 \\ & {{e}^{x}}-{{e}^{3x}}=0 \\ & {{e}^{x}}\left( 1-{{e}^{2x}} \right)=0 \\ & 1-{{e}^{2x}}=0 \\ & {{e}^{2x}}=1 \\ & 2x=0 \\ & x=0 \\ & \text{From the critical value we obtain the following intervals}\, \\ & \left( -\infty ,0 \right),\left( 0,\infty \right) \\ & \text{Now, we will evaluate between the critical values and resume } \\ & \text{in a table} \\ & \begin{matrix} \text{Interval} & \text{Test value}\left( x \right) & \text{Sign of }{f}'\left( x \right) & \text{Behavior of }f\left( x \right) \\ \left( -\infty ,0 \right) & -1 & + & \text{Increasing} \\ \left( 0,\infty \right) & 1 & - & \text{Decreasing} \\ \end{matrix} \\ & \text{From the table we can conlude that the function is:} \\ & \text{Decreasing on }\left( 0,\infty \right) \\ & \text{Increasing on }\left( -\infty ,0 \right) \\ & \text{Graph} \\ \end{align}\]
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