Prove, using the mean value theorem, that $x+1 \lt e^x \lt xe^x+1$ for $x \gt 0$ Prove, using the mean value theorem, that $x+1 \lt e^x \lt xe^x+1$ for $x \gt 0$. 
What I tried so far: Let there be a function $f(x)=xe^x$. So $f'(x)=xe^x+e^x$. Let there be arbitrary interval $(0,b)$. So by the mean value theorem there is $c \in (0,b)$ so that:
$f'(c)= \frac{f(b)-f(0)}{b-0}$, thus $ce^c+e^c=\frac{be^b}{b}=e^b$.
So $e^c(c+1)=e^b$. So $c+1 \lt e^b$, but I don't really know how to continue.
Any hints will be greatly appreciated. Thank you! 

Hint: $$x+1 \lt e^x \lt xe^x+1 \iff 1 \lt \frac{e^x-e^0}{x - 0} \lt e^x$$
Now can you apply the Mean-Value Theorem?
