$f$ twice continuously differentiable on $[a, b], f^\prime(c) = 0$ for some $c \in[a, b], f''(c) > 0$ implies various statements Assume $f$ is twice continuously differentiable on $[a, b]$ and there's some $c \in [a, b]$ such that $f'(c) = 0$. Suppose $f''(c) &gt; 0$. Prove that:


there exists $\delta &gt; 0$ such that $f''(x) &gt; 0$ for all $x \in (c - \delta, c + \delta)$.


(I was able to prove this already, but I'm leaving it here in case it's helpful for 2. and 3., which I haven't had any luck with.)


$f'(x) &lt; 0$ for $x \in (c - \delta, c)$ and $f'(x) &gt; 0$ for $x \in (c, c + \delta)$.


What I'm confused about here is finding a way to apply $\epsilon-\delta$ continuity (if that is even the right way to approach this part).


$f(x) &gt; f(c)$ for $x \in (c - \delta, c + \delta)$ such that $x \neq c$.


Thanks in advance.
For part c, I first used the Mean Value Theorem that: for every interval $(c-\delta, c)$, we can find an $x_1$ and (likewise and $x_2$ for the interval $(c, c+\delta)$), such that:
$$f'(x_1) = (f(c) - f(c-\delta))/\delta$$ and $$f'(x_2) = (f(c+\delta) - f(c))/\delta$$

But we know from part b that $f'(x_1) &lt;0 \implies f(c-\delta) &gt; f(c)$ and $f'(x_2) &gt;0 \implies f(c+\delta) &gt; f(c)$ .

(*) Now we can choose smaller intervals within given interval, e.g. ($c-\delta_i, c+\delta_i$), where $\delta_i &lt; \delta $, ($\delta_i&gt;0$), and attain all possible values of f(x) = $f(c-\delta_i)&gt;f(c)$ or $ f(c+\delta_i) &gt;f(c)$ $ \implies f(x) &gt;$ f(c).

I'm a little unsure whether my argument (*) works. I'm saying that for all x in the given interval, $f(x)&gt; f(c)$ because any x in that interval can be found to equal a $c+\delta_i$ or $c-\delta_i$, which we found from part b is greater than f(c).
There is a counter example to 1) 
Take $f(x)= x^2sin(1/x)+ \frac{x}{2} \forall x\in(-\infty,0)\cup(0,+\infty)$
and $f(x)=0$ for $x=0$.
Take a look