Q:

Tricky elementary proof with limits

Let $f$ be a function from $(0,\infty) \to \mathbb{R}$ that is differentiable and satisfies the following:
\begin{align}
\lim_{x \to 0^+} f(x) &= \infty\\
\lim_{x \to \infty} f(x) &= -2. 
\end{align}
Prove that there is a point $x_0$ of the domain so that $f'(x_0)=-1$.

A:

Consider $g(x)=x+f(x)$. The statement is the same as saying there is $x_0$ with $g'(x_0)=0$.
Since
$$
\lim_{x\to0^+}g(x)=\infty=\lim_{x\to\infty}g(x)
$$
the function has an absolute minimum.  What's the derivative there?