Answer
$f''(x) = \frac{2((x^{10}+2x^8−7x^7+x^6−14x^5+x^4−7x^3+2x^2+1)tan^{-1}(x)−2x^9−x^7−x^6+x^4+x^3+2x)}{(x^2+1)^2~(x^3+1)^3}$
$f$ is concave down on the intervals $(-1.0, 0.0)\cup (0.7, 2.5)$
$f$ is concave up on the intervals $(-\infty, -1.0)\cup (0.0, 0.7) \cup (2.5, \infty)$
We can see a sketch of $f''(x)$ below.
Work Step by Step
A computer algebra system returns the following function as the second derivative:
$f''(x) = \frac{2((x^{10}+2x^8−7x^7+x^6−14x^5+x^4−7x^3+2x^2+1)tan^{-1}(x)−2x^9−x^7−x^6+x^4+x^3+2x)}{(x^2+1)^2~(x^3+1)^3}$
We can graph this function to find the values of $x$ such that $f''(x) = 0$
$f''(x) = 0$ when $x = 0.0, x = 0.7$ and $x = 2.5$
Also note that $f''(x)$ is undefined at $x = -1.0$
When $-1.0 \lt x \lt 0.0$ or $0.7 \lt x \lt 2.5$, then $f''(x) \lt 0$
$f$ is concave down on the intervals $(-1.0, 0.0)\cup (0.7, 2.5)$
When $x \lt -1.0$ or $0.0 \lt x \lt 0.7$ or $ x \gt 2.5$, then $f''(x) \gt 0$
$f$ is concave up on the intervals $(-\infty, -1.0)\cup (0.0, 0.7) \cup (2.5, \infty)$
We can see a sketch of $f''(x)$ below.
