Second derivative computation

Using the quotient rule on $f(x) = \dfrac{x^2-1}{x^2+1}$, we get $f'(x) = \dfrac{(2x)(x^2+1) - (x^2-1)(2x)}{(x^2+1)^2} = \dfrac{4x}{(x^2+1)^2}$.

Applying the quotient rule again to $f'(x) = \dfrac{4x}{(x^2+1)^2}$: $f''(x) = \dfrac{4(x^2+1)^2 - 4x \cdot 2(x^2+1)(2x)}{(x^2+1)^4}$. Factor out $(x^2+1)$ from the numerator: $f''(x) = \dfrac{(x^2+1)[4(x^2+1) - 16x^2]}{(x^2+1)^4} = \dfrac{4(x^2+1) - 16x^2}{(x^2+1)^3} = \dfrac{4x^2 + 4 - 16x^2}{(x^2+1)^3} = \dfrac{4 - 12x^2}{(x^2+1)^3}$.

This simplifies to $f''(x) = \dfrac{4(1-3x^2)}{(x^2+1)^3}$, which is the fully simplified second derivative.

Since $(x^2+1)^3 > 0$ for all $x$, the sign of $f''(x)$ matches the sign of $1 - 3x^2$. Setting $1 - 3x^2 = 0$ gives $x^2 = \dfrac{1}{3}$, so $x = \pm \dfrac{\sqrt{3}}{3}$.

For $|x| < \sqrt{3}/3$, we have $1 - 3x^2 > 0$, so $f''(x) > 0$ and the function is concave up there. For $|x| > \sqrt{3}/3$, $1 - 3x^2 < 0$, so $f''(x) < 0$ and the function is concave down.

Therefore $f$ is concave up on $\left(-\tfrac{\sqrt{3}}{3}, \tfrac{\sqrt{3}}{3}\right)$ and concave down on $\left(-\infty, -\tfrac{\sqrt{3}}{3}\right)$ and $\left(\tfrac{\sqrt{3}}{3}, \infty\right)$.

From part (b), $f''(x) = 0$ at $x = \pm \sqrt{3}/3$, and the sign of $f''$ actually changes at each of these points (from negative to positive at $x=-\sqrt{3}/3$, and positive to negative at $x=\sqrt{3}/3$), confirming both are genuine inflection points.

Evaluate $f$ at $x = \sqrt{3}/3$: Since $x = \sqrt{3}/3$, we have $x^2 = 1/3$, so $f(\sqrt{3}/3) = \dfrac{1/3 - 1}{1/3 + 1} = \dfrac{-2/3}{4/3} = -\dfrac{1}{2}$. By symmetry (since $f$ is even), $f(-\sqrt{3}/3) = -\dfrac{1}{2}$ as well.

Thus the inflection points are $\left(-\tfrac{\sqrt{3}}{3}, -\tfrac{1}{2}\right)$ and $\left(\tfrac{\sqrt{3}}{3}, -\tfrac{1}{2}\right)$.