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It seems to be the case, but i don't have a proof.

Given the function $f$ such that $f(x) \geq e^x$, is it true that $f'(x) \geq f(x)$?!

I was experimenting with wolfram and it appears that $\frac{f'(x)}{f(x)} \geq 1$ whenever $f$ is bigger or equal to $\exp(x)$.

Note : as suggested in the comments, i meant that for all positive $x$ which means that $f(x) \geq e^x \space \forall x$ such that $x\ge 0$.

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  • $\begingroup$ Looking at the answers, the next question could be: what functions satisfy $f'(x)\geq f(x)$ for all $x\geq 0$? Or less generally, does $f'(x)\geq f(x)$ implies $f(x)\geq e^x$? $\endgroup$ – Surb Mar 29 '17 at 9:50
  • $\begingroup$ @Surb: the antilogarithms of the antiderivatives of all functions bounded below by $1$. (Among which $e^x-1$.) $\endgroup$ – Yves Daoust Mar 29 '17 at 13:19
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Let us consider the function $g(x):=\log(f(x))$.

Your question is recast as: if $g(x)\ge x$, is it true that $g'(x)\ge1$ ?*

This is obviously false, as a curve lying above $x$ can take any slope.

enter image description here


*$\log f(x)\ge \log e^x$ vs. $(\log f(x))'=\dfrac{f'(x)}{f(x)}\ge1$.

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No and heres why.

Assume u have a function $f(x) = e^x+1$ which is greater than $e^x$ for all $x$. However its derivative would be $f'(x) = e^x$ which is smaller. I think a more interesting question would be if the function was asymptotically greater than $e^x$

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    $\begingroup$ You're still allowed to have the function go up massively and then decrease, and then go up massively and then decrease, and so on. Concretely, $f(x)$ is a linear interpolation between $\exp(e^n)$ and $\exp\left(\exp\left(n+1\right)\right)$ for $n \leq x \leq n+\frac{1}{2}$, and then is constant on $n+\frac{1}{2} \leq x \leq n+1$. $\endgroup$ – Patrick Stevens Mar 29 '17 at 8:09
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This is not true. The $e^x$ is always positive, so it suffices to give a function that is decreasing on some interval and yet greater than $e^x$. It should be obvious that such functions exist.

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    $\begingroup$ For a concrete example, I was going to suggest the function which is $e$ for $x < 1$, and $e^x$ otherwise. $\endgroup$ – Patrick Stevens Mar 29 '17 at 7:58
  • $\begingroup$ I think the question is more interesting if $f(x) \geq e^x \forall x$ $\endgroup$ – mrnovice Mar 29 '17 at 7:58
  • $\begingroup$ @mrnovice That's the question I am responding to. See Patrick's example. I am inferring a $\forall x$ in both the premise and the conclusion. $\endgroup$ – Stella Biderman Mar 29 '17 at 7:59
  • $\begingroup$ @StellaBiderman Oh yes, sorry I confused myself there. $\endgroup$ – mrnovice Mar 29 '17 at 8:02
  • $\begingroup$ @StellaBiderman i edited my question into your answer, sorry. $\endgroup$ – Ahmad Mar 29 '17 at 8:03
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let $f(x) = e^x + 1 + \cos x \ge e^x + 1 - 1 = e^x$

$f'(x) = e^x - \sin x$.

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    $\begingroup$ I don't feel that this adds anything to the ongoing discussion, nor is it notably different from the other answers posted 10 minutes ago. $\endgroup$ – Stella Biderman Mar 29 '17 at 8:13
  • $\begingroup$ Maybe not. But it's a very straightforward example if the op wanted one. $\endgroup$ – fleablood Mar 29 '17 at 8:14
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If $f(x) \geq e^x \space \forall x$ then we can take the logarithm of both sides to give $\log f(x) \geq x \space \forall x$. It is tempting to take the derivative of both sides and conclude that $\frac{f'(x)}{f(x)} \geq 1 \space \forall x$ - but, as the examples above show, that is just not true.

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