TITLE: Prove that there exists some $c\in(-3,3)$ such that$ \ \ g(c) \cdot g''(c)<0$.
QUESTION [7 upvotes]: $f(x)$ is a differentiable function and $g(x)$ is a double differentiable function such that $|f(x)|\leqslant 1$ and $f'(x)=g(x)$. If $$f(0)^2+g(0)^2=9$$
then prove that there exists some $c\in(-3,3)$ such that$ \ \ g(c) \cdot g''(c)<0$.

Attempt:
Let us define a function $h(x) = g(x) g~'(x)$. Then
$$h'(x) =  g(x)g''(x) + \left( g'(x) \right)^2 \tag 1$$
If we prove that for some $c \in (-3,3), h~'(c) < 0,$ then $$g(c)g''(c) <0 \tag 2$$
Also, $$\left|f(0)\right| < 1 \implies f'(0) \in (-3,-2\sqrt 2 ) \cup (2\sqrt 2,3) $$
Could someone please advise me how do I move forward from here.
Thank you very much for your help in this regard.

REPLY [4 votes]: It is clear that $g(0)\in (-3,-2\sqrt{2}]\cup [2\sqrt{2},3)$.
Consider the interval $(0,3)$.
By $LMVT$ there exists $x=c_{1}$ where $c_{1}\in(0,3)$such that 
$|g(c_{1})|=|f'(c_{1})|=\left|\frac{f(3)-f(0)}{3-0}\right|\leq \frac{|f(3)|+|f(0)|}{3}\leq\frac{1+1}{3}=\frac{2}{3}$
Similarly, on the interval $(-3,0)$ By $LMVT$ there exists $x=c_{2}$ where $c_{2}\in(-3,0)$such that 
$|g(c_{2})|=|f'(c_{2})|=\left|\frac{f(0)-f(-3)}{0-(-3)}\right|\leq \frac{|f(0)|+|f(-3)|}{3}\leq\frac{1+1}{3}=\frac{2}{3}$.
WLOG,let us  assume that $g(x)>0$.
Now clearly, $c_{2}<0<c_{1}$.
The values of $g(c_{2})$ and $g(c_{1})$ are less than or equal to $\frac{2}{3}$ whereas value of $g(0)\in [2\sqrt{2},3)$.
Since, $g(x)$ is twice differentiable, the graph of $y=g(x)$ must attain a point of maximum between 
$c_{2}$ and $c_{1}$ and clearly at such a point of maximum ,say $x=c$ the value of $g''(c)<0$ and we are 
done.