Question

Hey guys,

I have a true false question here that I need help understanding

If a function f is differentiable on (1,4) then f is uniformly continuous on [2,3].

Could you say that, as f is differentiable on (1,4), then it is also continuous on (1,4) and hence by the Heine-Bolzano Theorem, f is uniformly continuous on (1,4) and since [2,3] is contained in (1,4), then f is uniformly continuous on [2,3]?

Answer 1

That sounds reasonable. Are there any particular details that are raising doubts?

Answer 2

And by "Heine-Bolzano" are you meaning a combination of Heine-Borel and Bolzano-Weierstrass theorems?

Answer 3

It also depends on what you mean by "uniformly continuous."
Either way, I think there's an easier way to think about this.

Answer 4

f might not be uniform continuous on (1,4)

Answer 5

Heine Bolzano. If f[a,b]->R is continuous on [a,b], then f is UC on [a,b]

Uniform continuity. |x-y|<delta => |f(x)-f(y)| < epsilon

Answer 6

Is there a counter example you can think of @Zarkon?

Answer 7

yes

Answer 8

Best response