Suppose $f(0)=f'(0)=0$ and $f''(x)>0$ everywhere, prove $f(x)>0$ for all non-zero $x$. Any help would be appreciated. Not quite sure how to prove this: 
Suppose $f: \mathbb{R}\rightarrow \mathbb{R}$,  $f(0)=f'(0) = 0$ and $f''$ is everywhere positive. Prove $f(x)>0$ for all nonzero $x$ without referring to concavity. 
 A: $f'$ is strictly increasing since $f''$ is strictly positive. So for $a >0$, $f'(a)>0$ and for $a<0$, $f'(a)<0$. Let $x>0$. By the mean value theorem, there exists $t_{x}$ such that $0<t_{x}<x$, with $f(x)-f(0)=f'(t_{x})(x-0)$, hence $f(x)=f'(t_{x})x>0$. Now, let $x<0$. Then there exists $b_{x} \in ]x,0[$ such that $f(0)-f(x)=f'(b_{x})(0-x)$, which means that $f(x)=f'(b_{x})x$. $x<0$ and so is $f'(b_{x})$, then $f(x)>0$.
A: Since $f''(x) > 0$ for all $x$, it follows that $f'(x)$ is strictly increasing and clearly $f'(0) = 0$, it follows that $f'(x) < 0$ if $x < 0$ and $f'(x) > 0$ if $x > 0$. Thus $f$ is strictly decreasing on $(-\infty, 0]$ and strictly increasing in $[0, \infty)$. Since $f(0) = 0$, it now follows that $f(x) > 0$ for all $x \neq 0$.
A: Proceed by contradiction:
Suppose there exists $x \in (0,\infty)$ such that $f(x) \leq 0$. Then by MVT, there exists $c \in (0,x)$ such that $f^{\prime}(c)=\frac{f(c)-f(x)}{c-x}=\frac{f(c)}{c}$. Since $c>0$, $f^{\prime}(c) \leq 0$.
Can you take it from here?
