If $0So the problem is asking me to supose $0<x<\ln\ 2$ and then prove that $x+1<e^x<2x+1$. However, per the problem, I need to use the MVT. So far, I have noticed that if you exponentiate by e your hypothesis you get $e^0=1<e^x<2=e^{\ln 2}$. This happens to coincide with the derivative of your result. I thought about defining $f(x)=(2x+1)-(x+1)$ but that just results in $f(x)=x$ which I can't seem to then use for the MVT becuause the result will always be 1. So my question would be how to use the MVT to get from $1<e^x<2$ to   $x+1<e^x<2x+1$.

Let $f(x)=e^x$ and apply the MVT for the endpoints $0$ and $x$,
$$\frac{e^x-1}x=e^c$$
where $0< c < x$. Given $0<x<\ln 2$, we have
$$1< e^c < 2$$
and, in turn,
$$1<\frac{e^x-1}x<2$$
Thus,
$$x+1<e^x<2x+1$$
