## Calculus with Applications (10th Edition)

Concave upward on $$(-\infty, -3 ), \quad (1, \infty) ,$$ concave downward on $$(-3 , -1) , \quad (-1 , 1)$$ inflection points at $$(-3,9+8 \ln 2 ), \quad (1, 2+8 \ln 2 ).$$
$$f(x)=x^{2}+8 \ln |x+1|$$ The first derivative is \begin{aligned} f^{\prime}(x) &=(2)x+\frac{8(1)}{(x+1) }\\ &=2x+\frac{8}{(x+1) } \end{aligned} and the second derivative is \begin{aligned} f^{\prime\prime}(x) &=2-\frac{8(1)}{(x+1)^{2}} \\ &=2-\frac{8}{(x+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) &=2-\frac{8}{(x+1)^{2}} \gt 0 \\ \Rightarrow\quad\quad\quad\quad\quad\quad\quad\quad\quad\\ -\frac{8}{(x+1)^{2}} & \gt -2\\ 8 & \lt 2(x+1)^{2} \\ 4 & \lt (x+1)^{2} \\ \Rightarrow\quad\quad\quad\quad\quad\quad\quad\quad\quad\\ x+1 \lt -2 \quad\quad \text {and} & \quad\quad x+1 \gt 2\\ \Rightarrow \quad\quad\quad\quad\quad\quad\quad\quad\quad\\ x \lt -3 \quad\quad \text {and} & \quad\quad x \gt 1 \end{aligned} so $f(x)$ is concave upward on $(-\infty, -3 ), \quad (1, \infty)$. 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-\frac{8}{(x+1)^{2}} \lt 0 \\ \Rightarrow \quad\quad\quad\quad\quad\quad\quad\quad\quad\\ -\frac{8}{(x+1)^{2}} & \lt -2\\ 8 & \gt 2(x+1)^{2} \\ 4 & \gt (x+1)^{2} \\ \Rightarrow \quad\quad\quad\quad\quad\quad\quad\quad\quad\\ -2 \lt x+1 & \lt 2\\ \Rightarrow \quad\quad\quad\quad\quad\quad\quad\quad\quad\\ -3 \lt x & \lt 1\\ \end{aligned} But $f(x)$ is not defined at $x=-1$, so $f(x)$ is concave downward on $(-3 , -1)$ and $(-1 , 1)$ . Since the sign of $f^{\prime\prime}(x)$ changes from positive to negative at $-3$, the graph changes from concave upward to concave downward at that point, so $(-3 , f(-3)$ is inflection point. Also, 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. Now ,we find $f( -3), f(1)$as follows: $$f(-3)=-3^{2}+8 \ln |-3+1|=9+8 \ln |-2|=9+8 \ln 2,\\ f(1)=(1)^{2}+8 \ln |1+1|=2+8 \ln |2|=2+8 \ln 2,$$ Finally, we have inflection points at $$(-3,9+8 \ln 2 ), \quad (1, 2+8 \ln 2 ).$$