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 $f(t)=e^t+1$ Apply LMVT to this on the interval $(0,x)$ where $0<x<\ln 2$.
Then $$\frac{(e^x+1)-2}{x-0}=f'(c)=e^c,~~0 <c <\ln 2.~~~~(1)$$
Next $$ 0<c<\ln 2 \implies 1<e^c <2~~~~(2)$$
Using (2) in (1) we get
$$1<\frac{e^x-1}{x-0}<2 \implies x+1 <e^x < 2x+1.$$
