# Math Help - Analysis Problem dealing with derivatives and neighborhoods

1. ## Analysis Problem dealing with derivatives and neighborhoods

1. Let f be a function defined on some neighborhood of x=a. Prove f '(a)=0 iff a is in D(abs. value f)

show me a nice proof..

2. Originally Posted by antoniowilliams46
1. Let f be a function defined on some neighborhood of x=a. Prove f '(a)=0 iff a is in D(abs. value f)

show me a nice proof..

What is $D(|f|)$?

3. D means differentiable

4. Originally Posted by antoniowilliams46
D means differentiable
So you're question is show that $f'(a)=0$ if and only if $|f|$ is differentiable at $a$?

5. f'(a)=0 iff a, then element sign, D(abs. val. of f)

6. Originally Posted by antoniowilliams46
f'(a)=0 iff a, then element sign, D(abs. val. of f)

8. If you mean that ${\mathrm{d}f\over\mathrm{d}x}(a) =0$ iff ${\mathrm{d}|f|\over\mathrm{d}x}(a) = 0$, try using the backwards triangle inequality $\Big||a|-|b|\Big|\le|a-b|$.

9. how would i go about doing that

10. $\Big|{|f(a+h)|-|f(a)|\over h}\Big| \le \Big|{f(a+h)-f(a)\over h}\Big|$ so if f' is 0 at a |f|' is zero at a too. The other direction is false, i.e. if |f| is differentiable at a, f could fail to be differentiable at a. I'm still not sure what you mean by D though, so I'm not sure if this is what you want...

11. D(f)={x:f'(x) exists}

12. so what do we do now

13. Let f = 1 on irrationals and -1 on rationals. Then f' does not exist anywhere, but |f| is differentiable everywhere. So $a\in D(|f|)$ does not imply f'(a)=0.

14. so how would i prove the originial problem?

15. Well the original problem appears to be wrong. If you mean " $f'(a)=0$ iff |f| is differentiable at a" then it is false by the example I gave in post #13. If you mean "if $f'(a)=0$ then |f| is differentiable at a (with derivative 0)" then I see post #10.

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