# Differentiability of function at only 0

1. Nov 10, 2007

### helpmeppl

1. function is x^2 if x is rational 0 if x is irrational. I need to prove that function is only differentiable at 0.

2. f'(x)=lim(h->0)=(f(x+h)-f(x))/h

3. fruitless attempt-----> So f'(0)=lim(h->0) f(h)/h=lim(h->0)x since it's 0 when irrational and x when rational, 0 when irrational, x=0. thanks.

Last edited: Nov 10, 2007
2. Nov 10, 2007

### rocomath

attempt attempt! i'm only letting you know ahead of time b/c when someone who can actually help you reads your post and sees no attempt, they'll say the same thing i'm telling you right now. so i'm saving you time :-]

3. Nov 10, 2007

### arildno

This is, indeed, the expression for the function value of f' at some particular value x.
Insofar as if it exists.

How, in particular, does it look if x=0?

That is, how must f'(0) look like?

IF f'(0) exists, then, since f(0)=0, we must have:
$$f'(0)=\lim_{h\to{0}}\frac{f(h)}{h}$$

Now, to make some headway:
Consider a sequence of RATIONAL numbers h converging upon 0; what is the limit of f(h)/h then? Does such a limit exist?

Make a similar converging sequence consisting solely of irrationals h. What is now the limit of f(h)/h?

Can you conclude anything from this?

4. Nov 10, 2007

### helpmeppl

that proves that the function is differentiable when x=0, but how can i prove that that's the only x that satisfies f's differentiability. thanks.

5. Nov 10, 2007

### helpmeppl

for the proof that it's only differentiable at 0, can proving that the function is not continuous everywhere except zero sufficient?

6. Nov 10, 2007

### arildno

Certainly; a function cannot be differentiable at any point upon which it is discontinuous.

This follows from the fact that the numerator in the limit expression for the derivative MUST go to zero in order for the fraction not to diverge.

But that the numerator must go to zero, i.e $$\lim_{h\to{0}}f(x+h)-f(x)=0$$ is equivalent to requiring the function to be continuous at x.

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