# First derivative test for absolute extrema proof.

Prove that if $I \subset \mathbb{R}$ is an open interval and $f: I \to \mathbb{R}$ differentiable, and $f$ has only one critical point $x_0$ and this critical point is a local minimum, then $x_0$ is also the absolute min of $f$, using Rolle's and IVT.

To me this seems "obvious", because if $f$ has only one critical point, then there can be no other point where $f' = 0$, meaning over its image $f'(x_0)$ is the only point where $f' = 0$. If this is the case then since $x_0$ is a local min, it is smaller than $f(x)$ for all points in a local interval, but since there are no other points $x$ in the entire image where $f'(x) = 0$, it is decreasing in the entire interval before it, and increasing in the entire interval after it. How do I go about formalizing this logic using IVT and Rolle's? Thanks in advance.

(edit) Description of problem also found on wikipedia: "For example, if a bounded differentiable function f defined on a closed interval in the real line has a single critical point, which is a local minimum, then it is also a global minimum (use the intermediate value theorem and Rolle's theorem to prove this by reductio ad absurdum)"

Let $x_1\in I$. Assume $f(x_1)<f(x_0)$ and wlog. $x_1>x_0$. Since $x_0$ is a local minimum, there is an $\epsilon>0$ such that $f(x)\ge f(x_0)$ for $x_0<x<x_0+\epsilon$. Select $x_2$ with $x_0<x_2<\min\{x_0+\epsilon,x_1\}$. Then $f(x_2)\ge f(x_0)>f(x_1)$ implies that there is $x_3\in[x_2,x_1)$ with $f(x_3)=f(x_0)$ and finally Rolle implies that there is $x_4\in(x_0,x_3)$ such that $f'(x_4)=0$, i.e. $x_4$ is a critical point different from $x_0$. Therefore the assumption $f(x_1)<f(x_0)$ must be wrong.
Well, assume that there's an $x_1$ with $f(x_1) < f(x_0)$. Since we know that $f(x_0\pm\epsilon) > f(x_0)$ for some $\epsilon > 0$, by the mean value theorem there's some point $x_2$ between $x_0\pm\epsilon$ and $x_1$ with $f(x_2) = f(x_0)$. Thus, by Rolle's theorem, there's an $x_3$ in the open(!) interval between $x_2$ and $x_0$ with $f'(x_3) = 0$. This contradicts the assumption that $f(x_0)$ is the only critical point.
(The proof uses $\pm\epsilon$, and avoids the usual interval notation $(a,b)$ since it'd otherwise have to distinguish between $x_1 < x_0$ and $x_1 > x_0$)
• Assume $x_1 < x_0$. Then $a = f(x_1) < f(x_0)$ and $f(x_0) < f(x_0-\epsilon) = b$. Since $a < f(x_0) < b$, by the mean value theorem there's an $x_2$ with $x_1 < x_2 < x_0-\epsilon$ and $f(x_2) = f(x_0)$. The case $x_2 > x_0$ works similarly. – fgp Oct 20 '12 at 21:10