The implication clearly does not hold for example when $x \in (0, 1)$ for $f(x) = x$ and $g(x) = 1$.
Is there a simple condition under which it does hold?
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7Does this answer your question? When can I say that $f(x) \gt g(x) \implies f'(x) \gt g'(x)$? (Found by searching for the implication using approach0.) – Jakob Streipel Sep 14 '22 at 14:37
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1I think the simplest condition is $f'(x)<g'(x)$. – Henrik supports the community Sep 14 '22 at 14:38
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2The equivalent question is: when does $f$ being negative imply that $f$ is strictly decreasing? The answer is: "Meh..." – Sassatelli Giulio Sep 14 '22 at 14:38
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@Henriksupportsthecommunity - I don't think that condition works. Let $f(x)=-e^{-x}$ and $g(x)=e^{-x}.$ Then $f(x) < g(x)$ for all reals, but $f'(x) > g'(x)$ for all reals. – Chris Leary Sep 14 '22 at 14:44
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@ChrisLeary: That just shows that those funcions doesn't satisfy my condition (which is exactly the right hand side, so naturally satifying that will make the condition true, it's also quite irrelevant, which was the real point. – Henrik supports the community Sep 14 '22 at 14:50
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The answer is that under no conditions is your title statement guaranteed to be in effect.
Pick any two functions you want on a closed interval $[a,b]$. Adding or removing a constant makes the two functions have a bigger or smaller relationship with one another but they leave the derivative unchanged.
Another way to frame this is by taking your original claim and making it equal to $g(x) - f(x) > 0 \Leftrightarrow h(x) > 0$ and asking when $h'(x) > 0$ which in no way is guaranteed to hold for every choice of functions $f,g$.
It's trivial that there is no one answer to it.
EDIT: My first claim is already answered in a post linked above.
algevristis
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