$f:\Bbb{R}\mapsto\Bbb{R}$ and $f$ is twice differentiable. $f:\Bbb{R}\mapsto\Bbb{R}$ and $f$ is twice differentiable such that $f(0)=2, f(1)=1, f'(0)=-2$. Prove that $\exists$ a $\varepsilon \in (0,1)$ such that $f(\varepsilon)f'(\varepsilon)+f''(\varepsilon)=0$.
Suppose $g(x) = \frac{2}{f(x)}-1-x$. Then $g(0)=g(1)=0\implies \exists$ $c \in (0, 1)$ such that $g'(c)=0$ by Rolle's theorem. 
$g'(x) = \frac{-2f'(x)-f(x)^2}{f(x)^2}$. Then $g'(0)=g'(c)=0$ 
$2f'(x)+f(x)^2=0$ for $x=0, c$. Suppose $h(x)=2f'(x)+f(x)^2$. Then $h(0)=h(c)=0\implies \exists$ $k \in (0, c)$ such that $h'(k)=0$ by Rolle's theorem.
$h'(k) = 2f''(k)+2f(k)f'(k)=0 \implies f''(k)+f(k)f'(k)=0$ for $k \in (0, c) \subset (0, 1)$. Hence, Proved.
But what to do if $f(k)=0$ for some $k \in (0, 1)$. Either a new function is to be found out or we need to prove that $f(x)$ is non-zero in the interval. 
 A: Consider $h(x)=2f'(x)+f(x)^2$.
You might guess there is two cases.
Note that $h(0) = 0$.

First: $\displaystyle{\min_{x\in[0,1]}}f(x) = 0$
Let $f(c) = 0$. Then since $f$ attains minimum at $x=c$ and differentiable, $f'(c) = 0 (\,\,\Rightarrow h(c) = 0)$.
Then by Rolle's theorem, $\exists c'\in(0,c)$ such that $h'(c') = 0.$ 

Second: $\displaystyle{\min_{x\in[0,1]}}f(x) = -\alpha <0.$
let us set $f(c_2) = -\alpha.$
Also here exists $c_1 \in(0,c_2)\,$ such that $f(c_1)= 0$, $f'(c_1)\leq0$. $(*)$
(This follows from $f(0),f(1) > 0$ but $f(c_2)<0$)


*

*$f'(c_1)<0$
Then $h(c_1) = 2f'(c_1) <0 $ and $h(c_2) = {\alpha}^2 >0$.
Thus there exists $c_4 \in (c_1,c_2)$ such that $h(c_4) = 0$. (IVT)
Finally, we get $\exists c_5 \in (0,c_4)$ satisfies $h'(c_5)=0$ By Rolle's theorem.


*$f'(c_1) = 0$
In this case $h(c_1) = 0$,
we get $\exists c_5 \in (0,c_1)$ satisfies $h'(c_5)=0$ By Rolle's theorem.

Proof for $(*)$:
First, by IVT, $A = \{x\in(0,c_2) :f(x) = 0\}$ is nonempty.
EDIT: Previous answer was logically right, but not mathematically clean. So I modified my answer a bit again.
Assume $f'(w) >0 $ for all $w\in A$. (In another case ($\exists w\in A \text{ s.t. } f'(w)\leq 0$), such $c_1$ clearly exists.)
let $x_0 = \inf(A)$. Then $f(x_0) = 0$ by continuity of $f$. ($\Rightarrow x_0 \in A$)
Now $f'(x_0) > 0 $, there exists small $\epsilon >0 $ such that $f(x_0 - \epsilon) <0.$
Then by IVT between $[0,x_0-\epsilon]$ ,
 there is $x_{-1} \in (0,x_0-\epsilon)$ such that $f(x_{-1}) = 0$. This contradicts with $x_0 = \inf(A)$. 
Indeed, the case is impossible.
