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If I have some equality $f(x) < g(x)$ that I want to prove to be true over some bounded interval for $x$, can I take the derivative wrt $x$ on both sides? Then, if I can reduce that to the point where it's obviously true over the bounds in question of $x$, does that prove the original case? I know the bounds on $x$ for $f'(x) < g'(x)$ are not the same as for $f(x) , but are they always less restrictive in the derivative case, allowing you to make implications for the non-derivative case?

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If $[a, b]$ is some interval, $f(a) < g(a)$, and $f'(x) \le g'(x)$ on the interval, then $f(x) < g(x)$ on the interval (exercise). This implication cannot be reversed; consider $[a, b] = [0, 10], f(x) = 20 - x, g(x) = x$.

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    I don't have to check $f'(b) < g'(b)$ at all?2011-09-15
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    @Angada: No, you don't. Intuitively, if $f$ starts below $g$ (at $x=a$) *and* $f$ has a smaller slope than $g$, then $f$ can never catch up to $g$.2011-09-15
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    I've seen at least one calculus text call this the [racetrack principle](http://en.wikipedia.org/wiki/Racetrack_principle). If horse $f$ starts behind horse $g$, and horse $f$ is always slower than horse $g$, then horse $f$ will always be behind horse $g$.2011-09-15
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    @Ted -- but what if the derivative isn't constant, but is itself a function of x? how can you be sure that, even if $f'(a) the opposite isn't true for $f'(a+e) ?2011-09-15
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    @Angada: sorry, I misstated the result I intended.2011-09-15
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    OH! I checked the edits so I could see what you changed. Now I get it. And it all makes sense! Thank you!2011-09-15
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Consider the functions $f(x)=2x$ and $g(x)=3x+c$. Clearly $f'(x)=2<3=g'(x)$ everywhere. On the other hand, $g(x) if and only if $x, so if you want to have $g(x) on $[a,b]$, just take $c>b$. Thus, on every bounded interval there is an easy counterexample.