Matematické Fórum

stuart clark · 10. 02. 2016 20:41

If $f(x)$ is a differentiable real valued function satisfying $f''(x)-3f'(x)>3\;\forall x \geq 0$

and $f'''(x)>0\;\forall x\geq 0$ and $f'(0)=-1\;,$ Then $f(x)+x\;\forall x>0$ is

$\bf{Options}:$

$(a)\;$ decreasing function

$(b)\;$ Increasing function

$(b)\;$ Constant function

$(d)\;\;$ Periodic function

I have tried like that way $\displaystyle f''(x)-3f'(x)>3\;,$ Now Multiplied both side by $e^{-3x}$

We get $\displaystyle e^{-3x}f''(x)-3f'(x)e^{-3x}>3\Rightarrow \frac{d}{dx}\left(e^{-3x}f'(x)\right)>3e^{-3x}$

Now Integrate both side w r to $x\;,$ We get $\displaystyle \int \frac{d}{dx}\left(e^{-3x}f'(x)\right)dx>\int 3e^{-3x}dx$

So we get $\displaystyle e^{-3x}f'(x)>-e^{-3x}+c\Rightarrow f'(x)>-1+ce^{3x}$

Now again integrate both side w r to $x\;,$ Means $\displaystyle \int f'(x)dx>\int (-1+ce^{3x})dx$

So $\displaystyle f(x)>-x+\frac{c}{3}e^{3x}+D$

Now How can I Proceed after that, Thanks