Proving $f(x)$ is strictly increasing if $f'''(x) \gt 0$ 
Question:
Let f: X $\rightarrow \mathbb R$ be a continuous function and suppose it is differentiable and that $f'(x_0)= f''(x_0) = 0,$ and $f'''(x_0)\gt0$, Prove that $f$ is strictly increasing at $x_0$.

My attempt
I feel this has something to do with L'hopitals rule. So I began by defining $$g(x) = \frac{f(x)}{x^3}$$
By L'hopitals rule $$g'(x_0) = \frac{f'''(x_0)}{6} \gt 0$$
Therefore $g(x)$ is strictly increasing at $x_0$.
This must imply that $f(x)$ is also strictly increasing no?
Is any of this correct? Thank you
 A: Here's an idea. Put $\;g(x):=f'(x)\;$ , then we're given
$$\begin{cases}f''(x_0)=g'(x_0)=0\\{}\\f'''(x_0)=g''(x_0)>0\end{cases}\;\;\implies (x_0,g(x_0))\;\;\text{is a local minimum of}\;\;f'(x)=g(x)$$
Since $\;f'(x_0)=g(x_0)=0\;$ , this means there's a neighborhood $\;I_\epsilon:=(x_0-\epsilon,\,x_0+\epsilon)\;$ for which $\;f'(x)=g(x)\ge0\;\;\forall\,x\in I_\epsilon\;$ , and this means $\;f(x)\;$ monotonic non-decreasing in $\;I_\epsilon\;$ .
A: I am replacing $x_{0}$ by $a$ to save some typing effort. Now $f'''(a) > 0$ implies that $f''$ is strictly increasing at $a$. This means that there is a neighborhood $I$ of $a$ such that if $x \in I, x < a$ then $f''(x) < f''(a) = 0$ and if $x \in I, x > a$ then $f''(x) > f''(a) = 0$.
Next if $x \in I$ then $$f'(x) = f'(x) - f'(a) = (x - a)f''(c)$$ for some $c$ between $a$ and $x$ and by what we have established in the previous paragraph it follows that $f'(x) > 0$ for all $x \in I$ except $x = a$. And therefore $f$ is strictly increasing in interval $I$. This is far more stronger than saying that $f$ is strictly increasing at $a$.
