1. Help with proving deffferentiable

Define h : R - R by
h(x) = {x^2, x rational
h(x) = {0, x irrational

1. Prove that h is di erentiable at 0.

2. Prove that h is not continuous at c not = 0. You may use the fact that any
interval in R contains both rational and irrational points.
(Hint: Split the proof into two cases: one for c rational, the other for c irrational.)

3. Prove that h' is a function whose domain is {0} and that h" does not
exist.

2. Originally Posted by 450081592
Define h : R - R by
h(x) = {x^2, x rational
h(x) = {0, x irrational

1. Prove that h is di erentiable at 0.

2. Prove that h is not continuous at c not = 0. You may use the fact that any
interval in R contains both rational and irrational points.
(Hint: Split the proof into two cases: one for c rational, the other for c irrational.)

3. Prove that h' is a function whose domain is {0} and that h" does not
exist.

Ok, what've you done so far? Did you try the definition of h'(0), taking limits when x --> 0 and all x's are rational, or all are irrational? What happened? Where are you stuck in the rest?

Tonio

3. ok, now I have proved part 1 then I know h is continuous and differentiable at 0, now hoe do I prove that h is not continuous at c not = 0, how do I define the interval in R? I know I need to prove that exist a |f(x) - L| > absola, how do I start this.

4. Originally Posted by 450081592
ok, now I have proved part 1 then I know h is continuous and differentiable at 0, now hoe do I prove that h is not continuous at c not = 0, how do I define the interval in R? I know I need to prove that exist a |f(x) - L| > absola, how do I start this.

Do as they tell you: split in tow cases, rational and irrational: if $\displaystyle c\neq 0$ then there exists a sequence of rational points that --> c, but ALSO a sequence of irrational points that --> c . If f were continuous at c then f(x) --> f(c) no matter how we choose to make x --> c, but over rationals and over irrationals we get two different results.

Tonio

5. Originally Posted by tonio
Do as they tell you: split in tow cases, rational and irrational: if $\displaystyle c\neq 0$ then there exists a sequence of rational points that --> c, but ALSO a sequence of irrational points that --> c . If f were continuous at c then f(x) --> f(c) no matter how we choose to make x --> c, but over rationals and over irrationals we get two different results.

Tonio
I still dont understand what that means, can you tell me how to do it please?

6. Tonio was assuming that you knew that a function is differentiable at x= 0 if and only if $\displaystyle \lim_{h\to 0}\frac{f(h)- f(0)}{h}$ exists. And that will be true if and only if $\displaystyle \lim_{n\to \infty}\frac{f(h_n)- f(0)}{h_n}$ exists and is the same for all sequences $\displaystyle \{h_n\}$ that converge to 0. Since a sequence will converge as long as subsequences converge, it is sufficient to consider sequences consistin only of rational numbers and only of irrational numbers.

7. Originally Posted by HallsofIvy
Tonio was assuming that you knew that a function is differentiable at x= 0 if and only if $\displaystyle \lim_{h\to 0}\frac{f(h)- f(0)}{h}$ exists. And that will be true if and only if $\displaystyle \lim_{n\to \infty}\frac{f(h_n)- f(0)}{h_n}$ exists and is the same for all sequences $\displaystyle \{h_n\}$ that converge to 0. Since a sequence will converge as long as subsequences converge, it is sufficient to consider sequences consistin only of rational numbers and only of irrational numbers.
so should I use the absola and delta argument to prove it, my friend hinted me to take absola = to c^2/2, will that work, how do I implement it?

Can you show me the solution please, I am really confused now