Let $f\colon\mathbb{R} \to \mathbb{R}$ be a differentiable function satisfying
$f(0) = 0$
$f'(x) > f(x)$ for all $x\in\mathbb R$.
Prove that $f(x) > 0$ for all $x > 0$.
I considered $f'(0) > f(0) = 0$ Let f(x) < 0 for all x > 0 How do I apply indeterminate property and Rolle's theorem to this?
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Solve a differential equation? – Idonknow Jan 02 '18 at 15:54
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3I took the time to format your post, please do so yourself and add your thoughts – operatorerror Jan 02 '18 at 15:56
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Writing "for all x belong to R" makes no sense. Rather than awkwardly trying to indicate symbols without using them, just say what you actually mean. – Andrés E. Caicedo Jan 02 '18 at 16:00
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Possible duplicate of https://math.stackexchange.com/q/248076 and https://math.stackexchange.com/q/1576874. – Martin R Jan 02 '18 at 16:28
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Solved by proving Supremum = infinity – User12345 Jan 02 '18 at 16:45
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Hint: consider the function $g(x) = e^{-x} f(x)$.
Rigel
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@Singh. This is not a special case. Answer is constructing a new function from your given $f$. Information from $g$ might tell you something about $f$. – Matthew Leingang Jan 02 '18 at 16:03
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1@Singh have you done anything to make you think you should be able to proceed from this hint? – operatorerror Jan 02 '18 at 16:09
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If supremum is infinity, then since f(x) is increasing function, therefore, f(x) > 0 for x > 0 – User12345 Jan 02 '18 at 17:08
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@Singh The only thing you need to prove is that $f(x)$ is increasing, if it's increasing then $f(x)>f(0)=0$. – kingW3 Jan 02 '18 at 17:14
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It can be solved by proving f(x) > 0 on an interval (0,m) and then taking m as supremum and proving it equal to infinity – User12345 Jan 02 '18 at 22:01