Let $g:\mathbb{R}\to\mathbb{R}$ be a scritly increasing function. Let $f:\mathbb{R}^n\to\mathbb{R}$. Prove that minimizing $f(x)$ is equivalent to minimizing $g(f(x))$
My attempt:
$\min g(f(x))$ is a $x_0$ such that $g(f(x_0))\le g(f(x))$ for all $x$ close to $x_0$. Since $g$ is strictly increasing, $g(f(x_0))\le g(f(x))\implies f(x_0)\le f(x)$ for every $x$ in the ball. Therefore $x_0$ is also a minimum of $f$