Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 5 - Graphs and the Derivative - 5.3 Higher Derivatives, Concavity, and the Second Derivative Test - 5.3 Exercises - Page 284: 45

Answer

Concave upward on $$ (-1, 1 ) , $$ concave downward on $$ (-\infty, -1 ) , \quad (1, \infty ) $$ inflection points at $$ (-1, \ln 2 ), \quad (1, \ln2 ). $$

Work Step by Step

$$ f(x)=\ln(x^{2}+1) $$ The first derivative is $$ \begin{aligned} f^{\prime}(x) &=\frac{2x}{x^{2}+1} \\ \end{aligned} $$ and the second derivative is $$ \begin{aligned} f^{\prime\prime}(x) &=\frac{2(x^{2}+1)-2x (2x)}{(x^{2}+1)^{2}} \\ &=\frac{2x^{2}+2-4x^{2}}{(x^{2}+1)^{2}}\\ &=\frac{2-2x^{2}}{(x^{2}+1)^{2}} \end{aligned} $$ A function $f$ is concave upward on an interval if $f^{\prime\prime}(x) \gt 0$ for all $x$ in the interval, so $$ \begin{aligned} f^{\prime\prime}(x) &=\frac{2-2x^{2}}{(x^{2}+1)^{2}} \gt 0 \\ \Rightarrow \quad\quad\quad\\ 2-2x^{2} & \gt 0\\ -2x^{2} & \gt -2 \\ x^{2} & \lt 1 \\ -1 \lt x & \lt 1 \end{aligned} $$ so $f(x)$ is concave upward on $(-1, 1 )$. Also function $f$ is concave downward on an interval if $f^{\prime\prime}(x) \lt 0$ for all $x$ in the interval, so $$ \begin{aligned} f^{\prime\prime}(x) &=\frac{2-2x^{2}}{(x^{2}+1)^{2}} \lt 0 \\ \Rightarrow \quad\quad\quad\\ 2-2x^{2} & \lt 0\\ -2x^{2} & \lt -2 \\ x^{2} & \gt 1 \\ \Rightarrow \quad\quad\quad\\ x \lt -1 \quad\quad \text {and} & \quad\quad x \gt 1 \end{aligned} $$ so $f(x)$ is concave downward on $(-\infty , -1) , \quad (1, \infty)$. Since the sign of $f^{\prime\prime}(x)$ changes from negative to positive at $ -1 $, the graph changes from concave downward to concave upward at that point, so $(-1 , f(-1)$ is inflection point. Also, the sign of $f^{\prime\prime}(x)$ changes from positive to negative at $1 $, the graph changes from concave upward to concave downward at that point, so $(1 , f(1)$ is inflection point. Now ,we find $f( -1), f(1)$as follows: $$ f(-1)=\ln((-1)^{2}+1)=\ln2,\\ f(1)=\ln((1)^{2}+1)=\ln2, $$ Finally, we have inflection points at $(-1, \ln 2 ), \quad (1, \ln2 ) $.
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