Posted by sarah on .
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.

math 
Steve,
We know that (f(y)f(x))/(yx) = f'(ax,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))/(yx) <= M for all x < y.
Or f(x)f(y) <= M*xy.
Now if we are given e>0, we can choose d = e/M, and if
xy < d, then f(x)f(y) < e, as required.