# Differentiation question

1. Nov 16, 2007

### asif zaidi

Problem statement

Show that if F:(a,b)->R is differentiable at every point of (a,b) and increasing then f'(x) >= 0 for all x belonging to (a,b)

Problem Solution

Definition of differentiation:
Assume sequence {an} -> a then
f(an) - f(x) / (an -x) converges to some value

Using this definition I am saying since it is known that f is differentiable at a,b then the following is true

Assume sequence an->a and bn->b
then f(bn) -f(an) / (bn-an) --- (eq1)

Since this is an increasing function both numerator and denominator will be always greater than 0 and thus the derivative by eq1 is always positive.

Am I on the right track?

If not plz advise.

Thanks

Asif

2. Nov 17, 2007

### EnumaElish

I think you meant [f(b) -f(bn)] / (b-bn) and alternatively [f(an) -f(a)] / (an-a). Even then this only addresses f'(a) > 0 and f'(b) > 0.

3. Nov 17, 2007

### asif zaidi

No I meant what I had written - however I did realise after posting that it was wrong. I worked through a specific f(x) and figured it was wrong.

If I use the definition of differentiability as you have done (f(b) - f(bn))/(b-bn), can I safely say that it is f'(b) >= 0;

4. Nov 17, 2007

### HallsofIvy

Staff Emeritus
Is that really the definition of "differentiation" your text gives?

It should be "f is differentiable at a if and only if for every sequence {an} that converges to a, (f(an)- f(a))/(an-a) converges to the same limit. If f is differentiable at a, then its derivative is that common limit."
(NOT "f(an) - f(x) / (an -x)" that makes no sense. You must mean (f(an)- f(a))/(an-a) with "a", not "x". )

The way you've written it, it sounds like just one such sequence would be sufficient to determine that a function is differentiable. Of course, here, since you are given that f is differentiable, then it really doesn't matter.

I see no reason for looking at two sequences {an} and {bn}. Take {an} to be a sequence converging to a with an> a for all n.

Last edited: Nov 17, 2007
5. Nov 17, 2007

### Kummer

Just use mean-value theorem.

6. Nov 17, 2007

### HallsofIvy

Staff Emeritus
Well, if you want to do things the easy way!! :)

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