# Chapter 5 - Graphs and the Derivative - 5.3 Higher Derivatives, Concavity, and the Second Derivative Test - 5.3 Exercises - Page 284: 47

Concave upward on $$(-\infty, -e^{-\frac{3}{2} } ), \quad (e^{-\frac{3}{2} }, \infty)$$ concave downward on $$(-e^{-\frac{3}{2} } ,0) , \quad (0,e^{-\frac{3}{2} } ),$$ inflection point at $$(-e^{-\frac{3}{2} } ,-\frac{3e^{-3}}{2\ln(10)} ), \quad (e^{-\frac{3}{2} } ,-\frac{3e^{-3}}{2\ln(10)} ).$$

#### Work Step by Step

$$f(x)=x^{2} \log |x|$$ The first derivative is \begin{aligned} f^{\prime}(x) &=2x \log |x| +x^{2} \frac{1}{x\ln(10)}\\ &=2x \log |x| + \frac{x}{\ln(10)} \end{aligned} and the second derivative is \begin{aligned} f^{\prime\prime}(x) &=2\log |x| + \frac{2x}{x\ln(10)} +\frac{1}{\ln(10)}\\ &=2\log |x| + \frac{3}{\ln(10)}\\ \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) &=2\log |x| + \frac{3}{\ln(10)} \gt 0 \\ \Rightarrow\quad\quad\quad\quad\quad\quad\quad\quad\quad\\ 2\log |x| &\gt -\frac{3}{\ln(10)} \\ \log |x| & \gt -\frac{3}{2\ln(10)} \\ \frac{\ln |x|}{\ln(10)} & \gt -\frac{3}{2\ln(10)} \\ \ln |x| & \gt -\frac{3}{2} \\ |x| & \gt e^{-\frac{3}{2} }\\ \Rightarrow\quad\quad\quad\quad\quad\quad\quad\quad\quad\\ x \lt -e^{-\frac{3}{2} } \quad \text {and} & \quad x \gt e^{-\frac{3}{2} }\\ \end{aligned} so $f(x)$ is concave upward on $(-\infty, -e^{-\frac{3}{2} } ), \quad (e^{-\frac{3}{2} }, \infty)$. 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) &=2\log |x| + \frac{3}{\ln(10)} \lt 0 \\ \Rightarrow\quad\quad\quad\quad\quad\quad\quad\quad\quad\\ 2\log |x| & \lt -\frac{3}{\ln(10)} \\ \log |x| & \lt -\frac{3}{2\ln(10)} \\ \frac{\ln |x|}{\ln(10)} & \lt -\frac{3}{2\ln(10)} \\ \ln |x| & \lt -\frac{3}{2} \\ |x| & \lt e^{-\frac{3}{2} }\\ \Rightarrow\quad\quad\quad\quad\quad\quad\quad\quad\quad\\ -e^{-\frac{3}{2} } & \lt x \lt e^{-\frac{3}{2} }\\ \end{aligned} But $f(x)$ is not defined at $x=0$ so $f(x)$ is concave downward on $(-e^{-\frac{3}{2} } ,0)$ and $(0,e^{-\frac{3}{2} } )$ . Since the sign of $f^{\prime\prime}(x)$ changes from positive to negative at $-e^{-\frac{3}{2} }$, the graph changes from concave upward to concave downward at that point, so $(-e^{-\frac{3}{2} } , f(-e^{-\frac{3}{2} }))$ is inflection point. Also, the sign of $f^{\prime\prime}(x)$ changes from negative to positive at $e^{-\frac{3}{2} }$, the graph changes from concave downward to concave upward at that point, so $(e^{-\frac{3}{2} } , f(e^{-\frac{3}{2} }))$ is inflection point. Now ,we find $f( -e^{-\frac{3}{2} }), f(e^{-\frac{3}{2} })$as follows: \begin{aligned} f(-e^{-\frac{3}{2} }) &=(-e^{-\frac{3}{2} })^{2} \log |-e^{-\frac{3}{2} }| \\ &=e^{-3} \log e^{-\frac{3}{2} }\\ &=-\frac{3e^{-3}}{2\ln(10)} \end{aligned} and \begin{aligned} f(e^{-\frac{3}{2} }) &=(e^{-\frac{3}{2} })^{2} \log |e^{-\frac{3}{2} }| \\ &=e^{-3} \log e^{-\frac{3}{2} }\\ &=-\frac{3e^{-3}}{2\ln(10)} \end{aligned} Finally, we have inflection points at $$(-e^{-\frac{3}{2} } ,-\frac{3e^{-3}}{2\ln(10)} ), \quad (e^{-\frac{3}{2} } ,-\frac{3e^{-3}}{2\ln(10)} ).$$

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