The function $f : \mathbb{R} → \mathbb{R}$ is twice differentiable and satisfies $f(0) =2, f ' (0) = 2 ~~and~~ f(1) = 1$. The function $f : \mathbb{R} → \mathbb{R}$ is twice differentiable and satisfies $f(0) =2, f ' (0) = 2\text{ and }f(1) = 1$. Prove that there exists a real number $n∈ (0,1)$ for which $f(n)f ’ (n)+f''(n)=0$
 A: This problem amounts to showing that the function $g(x)=f(x)f'(x)+f''(x)=0$ for some value in the interval. Since $f$ is differentiable, $f'$ has the intermediate value property, and it attains all values between $f'(0)$ and $f'(1)$. By the mean value theorem, there exists a $\theta \in (0,1)$ such that $f(1)-f(0)=f'(\theta)(1-0)$ which gives us $f'(\theta)=-1$. Since $f'(0)=2$, there exists an $x_0 \in (0, \theta)$ such that $f'(x_0)=0$, and thus $f$ has a maximum since the derivative is positive before the zero and negative after. Since the function $f$ is concave down at a max, $f''(x) \leq 0$ for all x before $\theta$. Plugging in our max of $f$ into $g(x)$ we can see that it obtains a negative value. 
Since our function is defined on a compact set it attains a minimum, and there is some $x\in [0,1]$ where $f(x)$ is minimal. If this is not at $1$, then we merely plug this minimum into $g(x)$, which since the function will be concave up around the minimum we will have $f''(x)\geq 0$, and thus $g(x)\geq 0$ and by the IVT there exists an $x$ so that $g(x)=0$. But in order to determine where our minimum is we need more information about the graph between $\theta$ and $1$. Specifically we need to know whether $f'(1)$ is positive, negative, or zero. Without this information we cannot conclude anything.
A: The claim is not true, and we can find a counter example the following way.
Let $f_k(x)$ be the solution to the differential equation 
$$ f'(x)=4-kx - \frac{1}{2}f(x)^2, f(0) =2 $$.
This is a one parameter family of differential equations, with $f_k'(0)=2$.
For $k=0$ the solution will be increasing on $[0, 1]$, then when $k$ increases $f_k(1)$ will decrease and with $k \approx 6.38$ we will have $f_k(1) = 1$. This value can be found from numerical integration of the differential equation.
Taking the second derivative we find that 
$$f_k''(x)=-k-f_k(x)f_k'(x)$$ so  $$f_k(x)f_k'(x)+f_k''(x)=-k$$ have no zero on $(0, 1)$.
