Finding inflection points and concavity behavior

Write $f(x) = (1+x^2)^{-1}$. Then $f'(x) = -2x(1+x^2)^{-2}$ by the chain rule. Differentiating again using the product and chain rules gives $f''(x) = \dfrac{-2(1+x^2)^2 + (-2x)\cdot 2(1+x^2)(2x)}{(1+x^2)^4} = \dfrac{-2(1+x^2) + 8x^2}{(1+x^2)^3} = \dfrac{6x^2 - 2}{(1+x^2)^3}$.

Set the numerator equal to zero: $6x^2 - 2 = 0 \Rightarrow x^2 = \tfrac{1}{3} \Rightarrow x = \pm\tfrac{1}{\sqrt{3}}$. Since $(1+x^2)^3 > 0$ always, the sign of $f''(x)$ matches the sign of $6x^2-2$: positive for $|x| > 1/\sqrt{3}$ and negative for $|x| < 1/\sqrt{3}$. Thus $f''$ changes sign at both points, confirming they are genuine inflection points, and $f$ is concave up on $(-\infty, -1/\sqrt3)\cup(1/\sqrt3,\infty)$ and concave down on $(-1/\sqrt3, 1/\sqrt3)$.

The peak occurs at x equals zero, which lies between the two inflection points at minus one over root three and one over root three. On that middle interval, the solution showed f prime prime is negative, meaning the function is concave down there. So at the peak, the graph is concave down.