$f(x)=2x-e^x<0, \forall x \in \mathbb{R}$ 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.

The function $e^x$ is strictly increasing and $y = e\cdot x$ is a tangent to $y = e^x$ at $x = 1$, so $e^x \geq e\cdot x > 2x$ for positive $x$. For nonpositive $x$: $e^x > 0 \geq 2x$. Finally, $e^x > 2x$
