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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 $f(x) <g(x)$, but are they always less restrictive in the derivative case, allowing you to make implications for the non-derivative case?

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up vote 9 down vote accepted

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? – Angada Sep 15 '11 at 4:14
@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 $g$. – Ted Sep 15 '11 at 4:21
I've seen at least one calculus text call this the 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 $g$. – Mike Spivey Sep 15 '11 at 4:27
@Ted -- but what if the derivative isn't constant, but is itself a function of x? how can you be sure that, even if $f'(a) <g'(a)$ the opposite isn't true for $f'(a+e) <g'(a_e)$? – Angada Sep 15 '11 at 4:32
@Angada: sorry, I misstated the result I intended. – Qiaochu Yuan Sep 15 '11 at 4:39

Consider the functions $f(x)=2x$ and $g(x)=3x+c$. Clearly $f'(x)=2<3=g'(x)$ everywhere. On the other hand, $g(x)<f(x)$ if and only if $x<c$, so if you want to have $g(x)<f(x)$ on $[a,b]$, just take $c>b$. Thus, on every bounded interval there is an easy counterexample.

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