Prove that $f$ must have an inflection point at $0$ 
Suppose $f$ is a function such that $f'(x)>0$ $\forall x\not=0$ and $f'(0)=0$. Also, $f''$ is a continuous $\text{one to one}$ function on some open interval containing $0$. Prove that $f$ must have an inflection point at $0$.

Pf:
Let a,b be arbitrary values in the open interval containing $0$ such that $a<0<b$
Since $f''$ is a continuous $1-1$ on [a,b], then either $f'''>0$ $\forall x \in [a,b]$ or $f'''<0$ $\forall x \in [a,b]$. So, -$f''
$ is continuous on $[a,b]$ and $f''$ is differentiable on $(a,b)$
By the mean value theorem, there exists $c$ with, $a<c<b$ such that $$f''(c)=\frac{f'(b)-f'(a)}{b-a}$$
Assuming I am even going the right direction with this, I do not know how to show that $f''(0)=0$ and that $f''$ changes sign at $0$. Most of what I am reading speaks in regards to a $\delta$ neighborhood of $0$ where I may be able to prove a sign change. 
Any helpful hints or guidance is greatly appreciated. Preferably more leaning towards hints and NOT full blown spoilers on this question. :)
EDIT:
$f''$ is not known to be differentiable, $f'$ is known to be differentiable and we are applying the mean value theorem to it. 
 A: Since $f''$ is continuous and injective it is either increasing or decreasing.
Consider a sequence of points $x_k \searrow 0$. For each $k$ there exists $y_k \in (0,x_k)$ satisfying $$f''(y_k) = \frac{f'(x_k) - f'(0)}{x_k - 0} = \frac{f'(x_k)}{x_k} > 0.$$
Since $f''$ is continuous you can then show $f''(0) \ge 0$. Since $f''$ attains positive values along a sequence decreasing to $0$, it follows that $f''$ cannot be decreasing - it is thus increasing.
Can you proceed from there?
A: Let's work in an interval $(-c,c)$ where $f''$ is $1$-$1$ and continuous.
If $d\in(0,c)$, you have
$$
f''(y_+)=\frac{f'(d)-f(0)}{d-0}=\frac{f'(d)}{d}>0
$$
for some $y_+\in(0,d)$. Similarly you have
$$
f''(y_-)=\frac{f'(-d)-f(0)}{-d-0}=\frac{f'(-d)}{-d}<0
$$
for some $y_-\in(-d,0)$. By continuity of $f''$, there is $\xi\in(y_-,y_+)\subset(-d,d)$ such that $f''(\xi)=0$. By injectivity, $\xi$ is uniquely determined and belongs to $(-d,d)$, for every $0<d<c$.

 Therefore $\xi=0$.

Now finish up

 by observing that $f''(x)<0$ for $-c<x<0$ and $f''(x)>0$ for $0<x<c$.

A: Let's first prove that $f''(0) = 0$. Clearly if $f''(0) > 0$ then $f'$ is strictly increasing at $0$ and since $f'(0) = 0$ the derivative $f'(x)$ must be negative for all sufficiently small negative values of $x$. This contradicts that $f'(x) > 0$ for all $x \neq 0$. Similarly we can show that $f''(0)$ can not be negative.
Apart from this we must prove that there is some interval $[-h, h]$ around $0$ such that $f'$ is decreasing in $[-h, 0]$ and increasing in $[0, h]$. We are given that $f''$ is continuous and one-one on some interval containing $0$. We can take $h$ small enough so that $f''$ is continuous and one-one on $[-h, h]$. Then $f''$ is monotone (you should try to prove that continuous one-one functions are strictly monotone). Since $f''(0) = 0$ it follows that $f''(x) < 0$ for all $x \in [-h, 0)$ and $f''(x) > 0$ for all $x \in (0, h]$. This means that $f'$ is strictly decreasing in $[-h, 0]$ and strictly increasing in $[0, h]$ and therefore $f$ has an inflection point at $0$.
