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I'm not completely certain if the statement is true, but intuitively I think it is. My sort of intuitive explanation would be if there is a point where $f$ and $g$ are equal, then the fact that $f'$ is greater than $g'$ would prevent them from intersecting again.

I was thinking of doing something with the mean value theorem and a proof by contradiction:

If we assume $f$ and $g$ intersect at two points, say $a$ and $b$, then $f(a)=g(a)$ and $f(b)=g(b)$. By the MVT, there must be points $c_f$ and $c_g$ such that $f'(c_f)=\frac{f(b)-f(a)}{b-a}$ and $f'(c_g)=\frac{g(b)-g(a)}{b-a}$, which implies $f'(c_f)=g'(c_g)$. But I'm not sure how much this would help since it would only lead to a contradiction if $c_f=c_g$.

Is there a way to continue the proof from here or some other way to prove this statement?

jazhang
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Okay, I think I got the answer.

Since $f'(x)>g'(x)$, that means $f'(x)-g'(x)>0$ or the derivative of $f(x)-g(x)$ is greater than $0$. Using Rolle's theorem, if $f(x)-g(x)=0$ at two or more points, then there's a point where $f'(x)-g'(x)=0$, which is a contradiction since $f'(x)-g'(x)>0$. This means that $f(x)-g(x)=0$ at at most one point. Or, $f$ and $g$ can intersect at most once.

jazhang
  • 257