Prove that there exists $\eta \in (a,b)$ such that $f''(\eta)=f(\eta).$ Problem
Assume that $f(x)$ is  differentiable over $[a,b]$, twice differentiable over $(a,b)$, and satisfies that $f(a)=f(b)=0$, $f'_{+}(a)f'_{-}(b)>0.$ Prove that

*

*there exists $\xi \in (a,b)$ such that $f(\xi)=0$;

*there exists $\eta \in (a,b)$ such that $f''(\eta)=f(\eta).$
Proof
The first one is so easy,so we directly turn to deal with the second one.
Denote
$$F(x)=e^{-x}f(x).$$
Thus
$$F'(x)=e^{-x}[f'(x)-f(x)].$$
Since
$$F(a)=F(\xi)=F(b)=0,$$
where $\xi$ is defined in $(1)$, then according to Rolle's theorem,
$$\exists \xi_1 \in(a,\xi),\xi_2 \in (\xi,b):F'(\xi_1)=F'(\xi_2)=0,$$
namely
$$f'(\xi_1)=f(\xi_1),f'(\xi_2)=f(\xi_2).$$
Denote
$$G(x)=e^x[f'(x)-f(x)].$$
Then
$$G'(x)=e^x[f''(x)-f(x)].$$
Since $$G(\xi_1)=G(\xi_2)=0.$$
Applying Rolle's theorem again, we obtain that
$$\exists \eta \in(\xi_1,\xi_2) \subset(a,b):G'(\eta)=0,$$
namely
$$e^{\eta}[f''(\eta)-f(\eta)]=0,$$
which implies $$f''(\eta)=f(\eta).$$
The proof is completed by now.
Question
Any more elegant proof? Especially, can we apply Taylor's formula here?
 A: Here is another approach. Let the function $g:[a, b] \to\mathbb{R} $ be defined by  $$g(x) =f(x) \sinh x-f'(x) \cosh x$$ so that $$g'(x) =\{f(x)-f''(x) \} \cosh x$$ Also note that $\cosh x>0$ for all $x$ so that $g(a) g(b) >0$. And our job is done if we show that $g' $ vanishes somewhere in $(a, b) $. One way to ensure this is to prove that $g$ has a local extremum in $(a, b) $.
Let's observe that there is a point $c\in(a, b) $ with $f(c) =0$ (this is what you mention as so easy) and hence there are two points $c_1,c_2$ with $c_1\in(a,c),c_2\in(c,b)$ such that $f'(c_1)=f'(c_2)=0$. Moreover it can be proved that $c_1,c_2$ can be chosen such that $f(c_1)f(c_2)<0$ (prove this, it is related to the so easy part mentioned earlier) and therefore $g(c_1)g(c_2)\leq 0$. It follows that $g$ vanishes somewhere in $[c_1,c_2]$. Now we can note that $g(a), g(b) $ are of same sign and $g$ vanishes somewhere in $(a, b) $ therefore $g$ has a local extemeum in $(a, b) $ and our proof is complete.
Note: Although it is not mentioned the proof above requires the continuity of $f'$ on $[a, b] $ to ensure that $g$ is continuous on $[a, b] $. 
