If $f'(c)=0$ and $f''(c)\gt0$, then $f$ has a local minimum at $c$ Question: Let $f$ is differentiable on $I$. For $c\in I$, if $f'(c)=0$ and $\exists f''(c)\gt0$, then show that $f$ has a local minimum at $c$.
As you know, this is regarded as a fundamental theorem and is useful when graphing the function. But, I had some trouble because I only learned in the case that it holds if $f''$ is continuous near c.
When $f''$ is continuous near $c$, then there exists $\delta$ such that $\forall x\in(c-\delta, c+\delta)\implies f''(x)\gt0$, or $f'$ is increasing. So, $\forall x\in (c-\delta, c), f'(x)<f'(c)=0$ and $\forall x\in (c, c+\delta), f'(x)>f'(c)=0$, which in turn we conclude that $f$ has a local minimum at $c$.
But, what can we imply if there's no such condition? More basically, does the statement hold even though the condition($f''$ is continuous near $c$) is absent? Thanks a lot.
 A: Yes, even if $f$ is just twice differentiable at $c$, this condition will hold. Look for the definition of $f''$ at c
$$f''(c) = \lim_{h \to 0} \frac{f'(c+h) - f'(c)}{h} \\
 = \lim_{h \to 0} \frac{f'(c+h)}{h} \gt 0$$
Since $f''$ exists at c, that means that in a small neighbourhood around $c$, $f'(c+h)$ is positive, and hence $f(c+h) > f(c)$, which means that $f(c)$ is a local minimum
A: We don't need to assume that $f''$ is continuous, essentially because it's already defined as a limit. Specifically, we know that
$$
f''(c)=\lim_{x \to c} \frac{f'(x)-f'(c)}{x-c}=\lim_{x \to c}\frac{f'(x)}{x-c}
$$
Let $\varepsilon = \frac{1}{2}f''(c)$. Then there is some $\delta$ such that, whenever $0 < |x-c| < \delta$, we have
$$\left|f''(c)-\dfrac{f'(x)}{x-c}\right|<\varepsilon
$$
It follows that
$$
\frac{f'(x)}{x-c} > f''(c)-\varepsilon = \frac{1}{2}f''(c)
$$
and so $\dfrac{f'(x)}{x-c}$ is positive when $0 < |x-c| < \delta$. That is, $f'(x)$ is negative when $x \in (c-\delta,c)$ and positive when $x \in (c, c+\delta)$, which as you've noted is what we need.
