TITLE: How to prove there exists $c$ such $f(c)f'(c)+f''(c)=0$
QUESTION [22 upvotes]: Nice Question:

let $f(x)$ have two derivative on $[0,1]$,and such $$f(0)=2,f'(0)=-2,f(1)=1$$
  show that:
there exist $c\in(0,1)$,such
  $$f(c)f'(c)+f''(c)=0$$

my try: since $$f(0)=2,f'(0)=-2,f(1)=1$$
so we easy 
$$f(x)=x^2-2x+2$$ such  this condition,But we can't use this function prove  this problem.
and other  idea: we can find this ODE
$$yy'+y''=0?$$
Thank you for you help！

REPLY [2 votes]: I don't know, probably I do not understand the problem but it seems that one can do the following:
Consider the differential equation 
$$
f''(x) + f(x) f'(x) =0
$$
with boundary conditions $f(0) = 2, f'(0) = -2, f(1) = 1$.
Let's integrate it first time, thus we get:
$$
f'(x) + \frac{1}{2} f^2(x) = C_1
$$
$C_1$ we will find by applying the conditions for $x=0$:
$$
f'(0)+\frac{1}{2} f^2(0) = -2 + 2 = C_1
$$
Therefore, the new equation is:
$$
f'(x) + \frac{1}{2} f^2(x) = 0
$$
And it can easily be integrated:
$$
\frac{2}{f(x)} = x + C_2
$$
Applying the last condition which is $f(1)=1$, one can obtain:
$$
f(x) = \frac{2}{x+1}
$$
Seems like this function satisfies the equation $f'' + f f' = 0$ in any point of $(0,1)$, which means that one can always find a point $c$ such that $f(c)f'(c)+f''(c)=0$.