Matematické Fórum

stuart clark · 04. 03. 2016 07:55

If $f(x)$ is a twice differentiable function such that $f(x)+f''(x)=-x|\sin x|f'(x)$

Where $x\geq 0$ and $f(0) = -3$ and $f'(0)=4\;,$

Then maximum value of $f(x)$

stuart clark

I have tried like this way:: Let $f=f(x)$ and $f' = f'(x)$ and $f'' = f''(x)$

So Here $f+f'' = -x|\sin x|f'\Rightarrow \left(f^2+f'^2\right)^{'} = -x\underbrace{2|\sin x|(f')^2}{\bf{\geq 0}}$

$\bullet\;$ If $x>0\;,$ Then function $f^2+f'^2$ is an Increasing function

$\bullet\;$ If $x<0\;,$ Then function $f^2+f'^2$ is an Decreasing function

So function $f^2+f'^2$ $\max$ at $x=0$

So $f^2+f'^2=25$ at $x=0$ . So $f^2=25-f'^2$

So $f^2\leq 25\Rightarrow |f(x)|\leq 5$

Is it right or not, If not then how can we solve it

Thanks

Matematické Fórum

#1 04. 03. 2016 07:55

Function with maximum value

#2 16. 03. 2016 12:03 — Editoval stuart clark (16. 03. 2016 12:04)

Re: Function with maximum value

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