I have the following task in my homework:
Let $a,b \in \mathbb{R}, a < b $ and $f:[a,b] \to \mathbb{R} $ be a continuous function that is differentiable on $(a,b)$. Show that if $f'(x)=f(x)$ for all $x \in (a,b)$, then there is a $c \in \mathbb{R}$ with $f(x)=ce^x$ for all $x \in [a,b]$. Hint: Consider the function $x\mapsto f(x)e^{-x} $
I know that the two solutions here would be $c=1$ and $c=0$, but apparently, that has to be shown with the mean value theorem, and that hint just confuses me more. Any help would be appreciated!
If $g (x)=f (x)e^{-x} $, then $$g'(x)=f (x)e^{-x}-f (x)e^{-x}=0. $$ The role of the Mean Value Theorem is to show that a function with zero derivative is constant.
As mentioned by Ryan, the above shows that $f(x)=ce^x$ on $(a,b)$. As $f$ is assumed continuous on $[a,b]$, the equality extends to $[a,b]$.