Posted by **sarah** on Thursday, March 8, 2012 at 10:19pm.

Suppose that a<b are extended real numbers and that f is differentiable on (a,b). If f' is bound on (a,b) prove f is uniformly continuous.

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**Steve**, Friday, March 9, 2012 at 12:57pm
We know that (f(y)-f(x))/(y-x) = f'(a-x,y) for all x < y in the domain

where a is in (x,y), from the Mean Value Theorem.

So if we assume that |f'| is boudned by M, then

(f(y) - f(x))/(y-x) <= M for all x < y.

Or |f(x)-f(y)| <= M*|x-y|.

Now if we are given e>0, we can choose d = e/M, and if

|x-y| < d, then |f(x)-f(y)| < e, as required.

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