MHB Nature of real valued function f(x)

[juantheron]

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

 

[Kiwi1]

I think you should be able to make simple arguments for why it is not three of the four possible answers. So by deduction it must be the remaining one (assuming that the question stipulates that there is one correct answer). Having done that you may be able to make come up with a more constructive argument.

Start by asking what is f"(0) and then what do you know about f"(x).