How to prove there exists $c$ such $f(c)f'(c)+f''(c)=0$ 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！
 A: 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 $f(c)f'(c)+f''(c)=0$.
A: Following Jyrki Lahtonen's observation, let
$$g(x) = \frac{f(x)^2}{2} + f'(x). $$
Since $g'(x) = f(x)f'(x) + f''(x)$, it is sufficient to prove that $g$ has a stationary point. If this is false, then $g$ is either increasing or decreasing on $(0,1)$.
Suppose first of all that $g(x)$ is increasing. Let $d = 1$ if $f(x) > 0$ for all $x \in [0,1]$, and otherwise let $d$ be smallest such that $f(d) = 0$. For $x \in [0,d)$ we have that $-f'(x) \le f(x)^2/2$. Dividing through by the strictly positive number $f(x)^2$ we get $(\star$):
$$ \frac{-f'(x)}{f(x)^2} = \Bigl( \frac{1}{f(x)} \Bigr)' \le \frac{1}{2} \quad\text{for $x\in [0,d)$.}  $$
Since $f(0) = 2$, integrating gives $1/f(x) \le 1/2 + x/2$, 
or equivalently,
$$ f(x) \ge \frac{2}{1+x} \quad\text{for $x \in [0,d)$.} $$
It follows that $d=1$. Moreover, using the continuity of $f'$, we see
that
$$f(1) > \frac{2}{1+1}$$
unless equality holds in ($\star$)
for all $x \in (0,1)$.
Hence $f(x) = \frac{2}{1+x}$ and $f(x)f'(x) + f''(x) = 0$ for all $x \in (0,1)$.
Now suppose that $g(x)$ is decreasing. 
Observe that if $f(d) = 0$ for some $d \in (0,1)$ then, since $f(1) = 1$, there exists $e \in (d,1)$ such that $f'(e) > 0$. Hence $g(e) > 0$. Since $g(0) = 0$ this is impossible. Therefore the same argument as before shows that
$$ f(x) \le \frac{2}{1+x} $$
for $x \in [0,1]$ and as before we get $f(x) = \frac{2}{1+x}$ for all $x \in [0,1]$.
