TITLE: Show that there is $\eta\in(0,1)$ with $f''(\eta)=f(\eta)$
QUESTION [5 upvotes]: Suppose $f:[0,1]\to\mathbb{R}$ is twice differentiable with
$$\lim_{x\to0^+}\frac{f(x)}{x}=1,\lim_{x\to 1^-}\frac{f(x)}{x-1}=2$$
Show that there is $\eta\in(0,1)$ s.t. $f''(\eta)=f(\eta)$.
It is equivalent to prove that $f''(x)-f(x)$ has a root in $(0,1)$. Then I tried to construct function. First I noted that the ODE
$$y''-y=0$$
has solution of the form $\alpha e^x+\beta e^{-x}$, then I consider the following function :
$$(f-f')e^{x}$$
and apply mean value theorem to is. But thing does not behave that nice, and I was stuck.

REPLY [3 votes]: The basic idea of the following is essentially the same argument as the previous answer, but it's a little shorter so I'll post it anyway.
First, I claim that the function $f : [0,1] \to \mathbb{R}$ has a strictly positive global maximum and a strictly negative global minimum, and that each is attained in an interior point of $[0,1]$.
Assuming the claim, we finish the proof. Let $a,b \in (0,1)$ be points at which the maximum and minimum of $f$ are attained. Now, $f(a) > 0 >  f(b)$, and $f''(a) \leq 0, f''(b) \geq 0$, so that
$$
f(a) - f''(a) >0 >  f(b) - f''(b)
$$
It follows by the continuity of $f''$ and the intermediate value theroem that there is a point $\eta$ between $a,b$ for which $f(\eta) = f''(\eta)$.
We now prove the claim. This is not hard: $f(0+) = f(1-) = 0$, and from the left and right hand derivative limits, we know that $f$ is strictly positive in a neighborhood of $0$ and strictly negative in a neighborhood of $0$.