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: 65

Answer

$$\eqalign{ & {\text{Concave up on }}\left( { - \infty , - 1} \right){\text{ and }}\left( {1,\infty } \right) \cr & {\text{Concave down on }}\left( { - 1,1} \right) \cr & {\text{The inflection points are: }}x = - 1{\text{ and }}x = 1 \cr} $$

Work Step by Step

$$\eqalign{ & f\left( x \right) = {e^{ - {x^2}/2}} \cr & \cr & {\text{Calculate the second derivative}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {{e^{ - {x^2}/2}}} \right] \cr & f'\left( x \right) = - x{e^{ - {x^2}/2}} \cr & f''\left( x \right) = - \left( { - {x^2}{e^{ - {x^2}/2}} + {e^{ - {x^2}/2}}} \right) \cr & f''\left( x \right) = - {e^{ - {x^2}/2}}\left( {1 - {x^2}} \right) \cr & \cr & {\text{Set the second derivative to 0}} \cr & f''\left( x \right) = 0 \cr & - {e^{ - {x^2}/2}}\left( {1 - {x^2}} \right) = 0 \cr & {x^2} - 1 = 0 \cr & {\text{We see that }}\,f''\left( x \right) = 0{\text{ at }}x = - 1,\,\,\,\,x = 1 \cr & {\text{These points are candidates for inflection points}} \cr & {\text{We need evaluate the intervals }}\left( { - \infty , - 1} \right),\,\,\left( { - 1,1} \right){\text{ and }}\left( {1,\infty } \right) \cr & \cr & {\text{Now}}{\text{, we will evaluate test values and resume in a table}} \cr & {\text{to determinate whether the concavity changes at these points}} \cr} $$ \[\begin{array}{*{20}{c}} {{\text{Interval}}}&{{\text{Test value }}\left( x \right)}&{{\text{Sign of }}f''\left( x \right)}&{{\text{Behavior of }}f\left( x \right)} \\ {\left( { - \infty , - 1} \right)}&{ - 2}&{f''\left( { - 2} \right) > 0}&{{\text{Concave up}}} \\ {\left( { - 1,1} \right)}&0&{f''\left( 0 \right) < 0}&{{\text{Concave down}}} \\ {\left( {1,\infty } \right)}&2&{f''\left( 2 \right) > 0}&{{\text{Concave up}}} \end{array}\] $$\eqalign{ & {\text{From the table we can conclude that the function is:}} \cr & {\text{Concave up on }}\left( { - \infty , - 1} \right){\text{ and }}\left( {1,\infty } \right) \cr & {\text{Concave down on }}\left( { - 1,1} \right) \cr & {\text{The inflection points are: }}x = - 1{\text{ and }}x = 1 \cr} $$
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