Saturday, 25 April 2015

real analysis - Show that if f is differentiable and f'(x) ≥ 0 on (a, b), then f is strictly increasing

Show that if f is differentiable and f'(x) ≥ 0 on (a, b), then f is strictly increasing provided there is no sub interval (c, d) with с < d on which f' is identically zero.



So so far I'm trying to do this by contradiction:




Suppose not, that is suppose we have function f where f(x)0 on (a,b) where f ' is not identically 0 for a sub interval of (a,b) and f is not strictly increasing. Since f is not strictly increasing this implies there exists x1 and x2 where a <x1 < x2 < b and f(x1) = f(x2). Then for all y [x1, x2], f(y) = f(x2) which means that f is constant and f '(y) = 0.



Since f'(y)=0 for all y [x1, x2] this means f ' is identically 0 which is a contradiction. Thus f is strictly increasing.



I'm not sure if there is a better way to do this but any help or comments would be appreciated!

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