Let $f:[0,1]\to\mathbb{R}$ be twice differentiable function such that $f(0)=f(1)=0$ and $f''(x)-2f'(x)+f(x)\geq e^x$ then prove that $f’(x)\cdot f(x)$ has at least one root in $(0,1)$.
My thought:
$$f''(1)-2f'(1)\geq e$$ and $$f''(0)-2f'(0)\geq 1$$ If we assume $g(x)=f(x)-x$ and the function $g'(x)$ is $0$ at least once in the domain of $f$ such that $x\in(0,1)$. Also if we use the given equation we get $e>f''(a)-f(a)>1$ for any $a$ in $(0,1)$ where $g'(x)$ may be zero at least once in $(0,1)$.
After this I am not able to think how to proceed further. Please explain in simpler terms as I as newbie in this zone of mathematics. :-)
I think this is just an application of Rolle's theorem. $f(0)=f(1)$, therefore there exist $x\in (0,1)$ with $f'(x)=0$.