= Stationary points under an increasing outer function
{title2=$\nabla(F\circ h)=F'(h)\nabla h$}
Suppose $h$ is differentiable and $F'>0$ on its range. The <multivariable chain rule> shows that $F\circ h$ has exactly the same <stationary points> as $h$. Since $F$ is strictly increasing, the type of every <local maximum>, <local minimum>, and <saddle point> is preserved. If both functions are twice differentiable, then at a stationary point
$$
\operatorname{Hess}(F\circ h)=F'(h)\operatorname{Hess}h,
$$
so the <Hessian matrix> signature is unchanged. The hypothesis $F'>0$ matters for the converse about stationary points: a strictly increasing function such as $F(s)=s^3$ can have a zero <derivative> and create new stationary points in a composition.
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