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If $f(x)$ is a differentiable real valued function satisfying $f”(x)-3f’(x)>3$ for all $x$ greater than or equal to zero , and $f’(0)=-1$ then comment on the monotonocity of $f(x)+x$ for all $x>0$

This is how I attempted the question

$f”(x)-3f’(x)>3$

Multiplying the equation by $e^{-3x}$ yields

$e^{-3x}f”(x) -3e^{-3x}f’(x)>3e^{-3x}$

As $3e^{-3x}>0$ for all $x$ We may conclude

$\frac{d}{dx}e^{-3x}f’(x)>0$ for all $x>0$

Now the function $e^{-3x}f’(x)$ is an increasing function

Therefore , for any $x_1>x_2 (>0)$

We will have $e^{-3x_1}f’(x_1)>e^{-3x_2}f’(x_2)$

As $e^{-3x}$ is always positive we may conclude that $f’(x_1)>f’(x_2)$ Therefore $f’(x)>-1$ for each $x>0$ $(1)$

Now coming to the original question we have the function $f(x)+x$ taking it’s derivative we get $f’(x)+1$ now from $(1)$ we may conclude that this will always be positive for $x>0$ hence $f(x)+x$ will always be increasing .

Do you think that my solution is correct and I've made the correct assumptions without loss of generality at all steps ? Thanks for helping !

Aditi
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1 Answers1

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Posted proof looks good.

For a slightly different approach, let $\,g(x)=f'(x/3)+1\,$, then the given relation translates to $\,g'(x) \gt g(x)\,$ with $\,g(0)=0\,$, which reduces the problem to the one answered at $f:\mathbb R \to \mathbb R$ be differentiable such that $f(0)=0$ and $f'(x)>f(x),\forall x \in \mathbb R$ ; then is $f(x)>0,\forall x>0$?.

dxiv
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