Answer
$f''(x) = \frac{24x^6+64x^5+115x^4+115x^3+92x^2+32x−1}{
4(x^2+x+1)^{5/2}}$
$f$ is concave up on the intervals $(-\infty, -0.6)\cup (0.0, \infty)$
$f$ is concave down on the interval $(-0.6, 0.0)$
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:
$f''(x) = \frac{24x^6+64x^5+115x^4+115x^3+92x^2+32x−1}{
4(x^2+x+1)^{5/2}}$
We can graph this function to find the values of $x$ such that $f''(x) = 0$
$f''(x) = 0$ when $x = -0.6$ and $x = 0.0$
When $x \lt -0.6$ or $x \gt 0.0$, then $f''(x) \gt 0$
$f$ is concave up on the intervals $(-\infty, -0.6)\cup (0.0, \infty)$
When $-0.6 \lt x \lt 0.0$, then $f''(x) \lt 0$
$f$ is concave down on the interval $(-0.6, 0.0)$
We can see a sketch of $f''(x)$ below.
