A question about the second derivative test. Suppose we're given a function $f : X \rightarrow \mathbb{R}$ with $X \subseteq \mathbb{R}.$ Then by the second derivative test, we have that for all points $x \in X$ such that $f$ is twice differentiable at $x$, it holds that if $f'(x)=0$ and $f''(x)>0,$ then $x$ corresponds to a local minimum of $f$ and not a local maximum.
The italicized text is my own - its not officially part of the theorem. But, is this additional conclusion valid?
The problem, of course, is that just because $x$ corresponds to a local minimum, this doesn't rule out the possibility that it also corresponds to a local maximum. For instance, $f$ could be constant on some neighborhood of $x$. However, I'm guessing this is the only way that $x$ can correspond to both a local maximum and a local minimum, so I'm guessing the italicized text is valid extension of the theorem.
True? If so, I'd like to see a proof.
 A: If $f$ is a constant in some neighborhood of $x$ then it cannot be the case that $f''(x)>0$. Therefore, your concern is irrelevant and yes you can claim that the point, $x$, is not a local maximum provided $f'(x)=0$ and $f''(x) > 0$.
A: 
Let $f : X \rightarrow \mathbb{R}$ with $X \subseteq \mathbb{R}.$ Then by the second derivative test, we have that for all points $x \in X$ such that $f$ is twice differentiable at $x$, it holds that if $f'(x)=0$ and $f''(x)>0,$ then $x$ corresponds to a local minimum of $f$.

*

*The problem, of course, is that just because $x$ corresponds to a local minimum, this doesn't rule out the possibility that it also corresponds to a local maximum. For instance, $f$ could be constant on some neighborhood of $x$.


There is no problem actually, given the hypotheses of the Theorem. That is, if $f$ is indeed constant in some neighborhood of $x$, this would contradict the fact hypothesis $f''(x)>0$. So it would be redundant to tag on to the theorem the provision "...and not a local maximum."
Of course, it is valid to assert that "$x$, is not a local maximum provided $f'(x)=0$ and $f''(x) > 0$." But that's simply reiterating the theorem.
A: Assume that $x_0$ is a local minimum and a local maximum at the same time. This means that there exists a neighborhood $I$ of $x_0$ such that 
$$ f(x_0)\le f(x)\le f(x_0),\qquad \forall x \in I.$$
Therefore, $f$ is constant in $I$.
