Show Uniform Continuity: Let f:R->R be Differentiable with |f'|<=15

[Treadstone 71]

"Let f:R->R be differentiable such that |f'|<= 15, show that f is uniformly continuous."

I can't solve it. I tried writing down the definition, but it got no where.

 

[math-chick_41]

have you thought about mean value thm?

 

[matt grime]

Try the mean value theorem. Obviously 15 is a red herring, any constant will do.

 

[vincebs]

How about this:

f is differentiable, and the derivative's absolute value is bounded by 15. Then | f(x) - f(y) / (x-y)| is bounded by 15 for all x, y, since otherwise the MVP would indicate that there's a pt c between x and y s.t. f'(c) > 15.

So |f(x) - f(y) / (x-y)| = |f(x) - f(y)| / |x-y| <15

hence |f(x) - f(y)| < 15 |x-y|

Let epsilon = 15 delta.

Then for every epsilon greater than zero, exists a delta greater than zero s.t. if 0 <= |x - y| < delta, then |f(x) - f(y)| <= 15 delta = epsilon.

 

[matt grime]

Yet again I find myself pointing out that we shouldn't just give the answers out; we're not trying to prove how clever we are at answering other people's questions. Wasn't this in the disclaimer you were made to accept when logging into this particular subforum for the first time? It is expected that the OP indicates what they've done in trying to solve the question.

 

[Treadstone 71]

I found the answer shortly after I posted. I can't believe I didn't see it before. Thankfully I didn't see the answer here before I solved it. Thanks everyone.