Monotonicity of z(x)

Differentiate term by term: $z'(x) = \frac{1}{1+x^2} + 1$. Since $1+x^2 > 0$ for all real $x$, the term $\frac{1}{1+x^2}$ is always positive, so $z'(x) > 0$ for every real $x$.

Because $z'(x) > 0$ on the entire domain, $z(x)$ has no critical points and no local extrema. The function is therefore increasing on the whole real line, i.e., on $(-\infty, \infty)$.

Differentiate $z'(x) = (1+x^2)^{-1} + 1$: using the chain rule, $z''(x) = -2x(1+x^2)^{-2} = \frac{-2x}{(1+x^2)^2}$. The denominator is always positive, so the sign of $z''(x)$ matches the sign of $-2x$.

For $x < 0$, $-2x > 0$ so $z''(x) > 0$ (concave up); for $x > 0$, $-2x < 0$ so $z''(x) < 0$ (concave down). The concavity switches at $x=0$, and since $z''(x)=0$ there with a sign change, $x=0$ is an inflection point, giving the point $(0,0)$.