$f(x)=2x-e^x<0, \forall x \in \mathbb{R}$ The question is quite simple, but I'm finding some trouble doing it...  
Prove that the function $f(x)=2x-e^x$ is negative, i.e., $f(x)<0, \forall x \in \mathbb{R}$. 
Thanks for the help.

It's clearly true for negative numbers. For non-negative numbers, considering the Taylor expansion of $e^x$, we get $2x-e^x=2x-1-x-\frac{x^2}{2}-\frac{x^3}{3!}-\dots=-1+x-\frac{x^2}{2}-\dots=-(\frac{x}{2}-1)^2-\frac{x^2}{4}-\frac{x^3}{3!}-\dots<0.$
