Concavity and inflection point of arctan(x)

Starting from $f'(x) = \frac{1}{1+x^2} = (1+x^2)^{-1}$, apply the chain rule: $f''(x) = -1\cdot(1+x^2)^{-2}\cdot 2x = \frac{-2x}{(1+x^2)^2}$. The denominator $(1+x^2)^2$ is always positive, so the sign of $f''(x)$ is determined entirely by the sign of $-2x$, i.e., by the sign of $x$.

For $x<0$, $-2x>0$ so $f''(x)>0$, meaning $f$ is concave up on $(-\infty, 0)$. For $x>0$, $-2x<0$ so $f''(x)<0$, meaning $f$ is concave down on $(0,\infty)$. Since $f''$ changes sign at $x=0$ (and $f''(0)=0$), $x=0$ is an inflection point, corresponding to the point $(0,0)$ on the graph.

The function is concave up for negative x values. This follows because the second derivative equals negative two x divided by a quantity that is always positive. When x is negative, negative two x becomes positive, making the second derivative positive, which signals concave up behavior.