If $f(x),g(x)$ are two differentiable functions such that $f'(x)=g(x)$, $|f(x)|<1$ and $f(0)^2+g(0)^2=9$

Asked Jan 26 '16 at 23:18

Active Jan 27 '16 at 06:21

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If $f(x),g(x)$ are two differentiable functions such that $f'(x)=g(x)$, $|f(x)|<1$ and $f(0)^2+g(0)^2=9$, prove that there exists a $c\in(-3,3)$ such that $g(c)\cdot g''(c)<0$

$$f(0)^2+g(0)^2=9$$ Using the above equation, $g(0)\in (-3,-2\sqrt{2})\cup(2\sqrt{2},3)$

That's how much I was able to solve. I only have information about the value of functions at $x=0$.

edited Jan 27 '16 at 06:21

asked Jan 26 '16 at 23:18

Aditya Dev

4,774

So, $g$ is twice differentiable? – Mark Viola Jan 26 '16 at 23:36
Possible hint: assume for contradiction that g(c)*g''(c) <= 0 for all c in (-3,3). In particular, this means g(c) and g''(c) have opposite signs (unless one of them happens to be 0 at a given point). Not sure if this will help, but it may. – Jan 27 '16 at 00:36
These sure look like scaled sin and cos. – marty cohen Jan 27 '16 at 00:56
1

see http://math.stackexchange.com/q/1305612/72031 – Paramanand Singh Jan 27 '16 at 04:39
are you sure the conclusion is $g(c)g''(c) > 0$ and not $g(c)g''(c) < 0$ as mentioned in linked question. – Paramanand Singh Jan 27 '16 at 04:46
So this question already has an answer. – Aditya Dev Jan 27 '16 at 06:22

If $f(x),g(x)$ are two differentiable functions such that $f'(x)=g(x)$, $|f(x)|<1$ and $f(0)^2+g(0)^2=9$

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