Application of Rolle's theorem? Establish existence of $c\in(a,b)$ such that $f(c)+f'(c)=f(c)f'(c)$
Solution 1:
Let $$ g(x) = f(x)e^{x-f(x)}. $$ Then $g(a) = g(b) = 0$ since $f(a)=f(b)=0$, so $g'(c) = 0$ for some $c\in(a,b)$. However, $$ g'(x) = f'(x)e^{x-f(x)} + f(x)e^{x-f(x)}(1-f'(x)) = e^{x-f(x)}(f'(x)+f(x)-f(x)f'(x)),$$ so $g'(c) = 0\implies f'(c)+f(c)-f(c)f'(c) = 0$, i.e. $f'(c)+f(c) = f(c)f'(c)$.