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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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 $[a, b] = [0, 10], f(x) = 20 - x, g(x) = x$.
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.