TITLE: If $f : \mathbb R \to [-2 , 2]$ with $(f(0))^2 + (f'(0))^2 =85$, then there exists $x \in (-4 , 4)$ such that $f(x) +f''(x) = 0$ and $f'(x) \neq 0$.
QUESTION [7 upvotes]: For every twice differentiable function $f : \mathbb R \to [-2 , 2]$ with $(f(0))^2 + (f'(0))^2 =85$ there exists $x \in (-4 , 4)$ such that $f(x) +f''(x) = 0$ and $f'(x) \neq 0$.

I was trying to get the answer by constructing a function $g(x)  = f(x) ^2  + f'(x) ^2$. But I can not proceed much. Can anyone help me?

REPLY [3 votes]: Let $g(x)=f(x)^2+f'(x)^2$. Then $g'(x)=2f'(x)(f(x)+f''(x))$ and $g(0)=85$. We have $f'(0)^2=85-f(0)^2\geq 81$. Without loss of generality, we can assume that $f'(0)>0$ (otherwise replace $f(x)$ by $-f(x)$). It follows that $f'(0)\geq 9$. We first show that $f'(c) <1$ for some $c\in (0,4)$. Otherwise, one has $f'(x)\geq 1$ for $x\in [0,4)$, which implies that $f(4)=f(0)+\int_0^4 f'(x)dx > f(0)+4>2$, a contradiction. Similarly, there exists $b\in (-4,0)$ such that $f'(b)<1$. If necessary, by changing $b,c$, we can assume that $f'(x)>0$ on $[b,c]$ (this is possible, since we have $f'(0)>0$). 
Now, we have $g(b)=f(b)^2+f'(b)^2<5$ and similarly $g(c)<5$. The function $g$ attains an absolute maximum on $(b,c)$, say at $x$. Then $g'(x)=2f'(x)(f(x)+f''(x))=0$. Since $x\in [b,c]$, we have $f'(x)\neq 0$, which implies that $f(x)+f''(x)=0$.