So to start off with, fix $\displaystyle \epsilon > 0$. Then since f is $\displaystyle C^2$ there exists $\displaystyle \delta > 0$ such thatSuppose that $\displaystyle f : \mathbb{R} \rightarrow \mathbb{R}$ is $\displaystyle C^2$ everywhere, and suppose that $\displaystyle c \in R $ satisfies $\displaystyle f'(c) = 0$ and $\displaystyle f''(c) > 0$.

(a) Prove that there exists $\displaystyle \delta > 0$ such that $\displaystyle |x-c| < \delta \Rightarrow f''(x) > 0$

$\displaystyle 0 < |x-c| < \delta \Rightarrow |f''(x) - f''(c)| < \epsilon$

I'm not sure where I have to go from here. I know I have to show something like:

$\displaystyle 0 < |x-c| < \delta \Rightarrow |f''(x) - 0| < \epsilon$