Let $g: \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function satisfying $g'(x) > 0$ for all $x \neq 0$. Prove that $g$ is one-to-one. Let $g: \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function satisfying $g'(x) > 0$ for all $x \neq 0$.  Prove that $g$ is one-to-one.
Proof:  Case 1: Consider $x_{2} > x_{1}>0$. Then by the Mean Value Theorem, there exists a point $x \in \mathbb{R}$ such that $g'(x) = \frac{g(x_{2}) - g(x_{1})}{x_{2}- x_{1}}$ which means that $g(x_{2}) > g_(x_{1})$ since $g'(x)> 0$ . Thus $g$ is $1-1$ in this case.    
Case 2: Similarily if $x_{1}<x_{2}<0$, then $g(x_{1})<g(x_{2})$. 
Case 3: The only case left to consider is that if $x_{1} \leq 0< x_{2}$
I am not sure how to approach case 3.  Do I use the MVT again or something else.  Also, am I missing any circumstances?

The MVT shows $f$ is strictly increasing on both $(-\infty,0]$ and $[0,\infty).$ Therefore $f$ is strictly increasing on $\mathbb {R}.$
