1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Function is differentiable at x=0?

  1. Feb 13, 2012 #1
    1. The problem statement, all variables and given/known data
    define a function f:R->R is given by f(x)=(|x|^a)*(sin(1/x)) if x is not equal 0 , and f(x)=0 if x=o . and a>0. for which values of a is f differentiable at x=o

    2. The attempt at a solution
    obviously, a=2 is one of the solution , but how about other values? can someone can show me the prove for other value of a that this function is differentiable at x=0?
     
  2. jcsd
  3. Feb 13, 2012 #2

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Re: differentiability

    Nope! But we'll help you work through the problem.

    You seem to already understand a=2 case. How does the proof for that work? In particular, how is a used?
     
  4. Feb 14, 2012 #3
    Re: differentiability

    for the a=2 case, i use the definition to prove it ,which when cancel the h in both the numerator and denominator then still an h left s.t. hsin(1/h) the limit exists. should it be all the a>1 is the answer?
     
  5. Feb 14, 2012 #4

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Re: differentiability

    That would be my first conjecture.

    Just to check, have you actually proven the derivative doesn't exist for a=1 or a<1? Or did you stop once you realized the algebraic manipulation you wanted to do assumes a>1?
     
  6. Feb 14, 2012 #5
    Re: differentiability

    the question just considered the a>0 , for 0<a<1 we have (sin(1/h))/(|h|^(1-a)) for h>0 then the denominator is 0 not defined and the same for h<0, for a=1 it is not continuouse then for a>1 we have (|h|^(a-1))sin(1/h)=(|h|^a)(hsin(1/h)) then both (|h|^a) (hsin(1/h))are differentiable then the function is differentiable and the same as the negative one


    does it make sense??
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook