Property of a function and its second derivative If $f(x)$ is a function and it's second derivative $f''(x)$ is continuous in a region around  $x=c$ and  $f''(c)<0$ then can we assume  that $f''(x)<0$ is also true in some open region, say $(a,b)$ around $x=c$ i.e $a<c<b$.
If the above statement is true how can I prove that.I need better explanation. 
 A: Yes, it is true. Let $$f''(c) = L.$$ Then, $$V =\left(\frac{1}{2} L, \frac{3}{2} L\right)$$ is a neighborhood about $L$ of radius $\frac{1}{2} L$. By continuity, there exists a neighborhood $U$ of $c$ that $$f''(U) \subseteq V.$$ If $x \in U$, then, $f''(x) \in V$. But, $L<0$ so that $f''(x) <0$.
A: It is true if $f''$ is continuous at and around $c$, not just around $c$:
recall
$$\mathrm{sgn}(x)=\begin{cases}
-1 & x<0,\\
0 & x=0,\\
1 & x>0.
\end{cases}$$
Now let $$g(t)=\int_{\beta}^{t}\mathrm{(sgn}(s^{2})-0.5)ds,$$
and let $$f(x)=\int_{\alpha}^{x}g(t)dt,$$
so, by the fundamental theorem of calculus, $$f'(x)=g(x)=\int_{\beta}^{x}\mathrm{(sgn}(s^{2})-0.5)ds,$$
which means $$f''(x)=g'(x)=\mathrm{sgn}(x^{2})-0.5.$$
We have that $f(x)$ is continuous everywhere and $f''(x)$ is continuous everywhere except at $x=0$. Note further that $f''(0)=-0.5<0$. We see that, for all $x\in\mathbb{R}\backslash  \{0\}$, we have $f''(x)>0$. So there is no open neighborhood $S=(a,b)$ around $0$ such that $f(y)<0$ whenever $y\in S$, since $f''(x)>0$ for all $x\in\mathbb{R}\backslash \{0\}$. Notice that the integrals all exist because the set of discontinuities of the $\mathrm{sgn}$ function on any compact set is of measure zero (there is only one discontinuity).
And so we require continuity at $c$ as well as around $c$.
If the other, more experienced members here at MSE could point out a flaw in my argument, please do, and I will [eventually] delete my answer.
