Prove the following function must have a local min.

Here's a problem that should be intuitive, yet I cannot seem to get at a proof.

Let $f(x)$ be a function $\in C^\infty$ such that $f'(a)<0$ and $f'(b)>0$ for $a < b$. Show that f must have a local minimum.

Clearly, by the IVT, I can find a point $x\in[a,b]$ where $f'(x)=0$.

My idea is something along the lines of:

If $f''(x)< 0$, then we're done

If $f''(x)\geq 0$, then there will have to be another point, $x_1$ where $f'(x_1)=0$ and if $f''(x_1)<0$ we're done and if not then... and that this cannot go on forever.

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If $h$ is small enough positive, then $f(a+h)<f(a)$, else we could not have $f'(a)<0$. Similarly, if $k$ is small enough positive, then $f(b-k)<f(b)$. The function $f$ attains an absolute minimum at some $c$ in $[a,b]$, and by the above $c$ is not $a$ or $b$. So we have an absolute minimum and hence a local minimum at $c$.
$[a,b]$ is compact, so there is a minimum $f(x_0)=\operatorname{min}(f([a,b]))$. $x_0$ cannot be an endpoint because $f'(a)<0$ and $f'(b)>0$.