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Homework Help: Show h is differentiable

  1. Dec 16, 2008 #1
    1. The problem statement, all variables and given/known data
    Define h(x)=x^3sin(1/x) for x[tex]\neq[/tex]0. and h(0)=0. Show h is differentiable everywhere and that h is cont everywhere, but fails to have a derivative at one point.



    2. Relevant equations



    3. The attempt at a solution
    [h(x)-h(0)]/[x-0]=x^2sin(1/x)
    h is diff everywhere because the limit exists and we know x[tex]\equiv[/tex]0.
    h'(x)=3x^2sin(1/x)-xcos(1/x)
    We know f is diff at [tex]x_{0}[/tex], then f must be continuous at [tex]x_{0}[/tex].
    h has a derivative at x if x[tex]\neq[/tex]0.
     
  2. jcsd
  3. Dec 16, 2008 #2
    I'm just unsure how to deal with it at x=0
     
  4. Dec 16, 2008 #3
    Ok I see my problem. I just don't understand why we cant have h(x) with x=0. Is it because the limit wouldn't exist?
     
  5. Dec 16, 2008 #4
    i had a similar problem that i figured out. maybe you can apply it to yours.

    Problem: Let f(x)=x^2sin1/x if x does not = 0, and f(0)=0. Find f '(x) and show that the lim as x goes to 0 of f '(x) does not exist.

    My answer: f '(x)= 2xsin1/x - cos1/x
    Now the lim as x goes to 0 of xsin1/x = 0, but the lim as x goes to 0 of cos1/x DNE. Since the value oscillates between -1 and 1 as x gets smaller.

    ~Sorry didn't have a chance to look into yours, but knew i had one similar. hope it helps.
     
  6. Dec 16, 2008 #5

    HallsofIvy

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    Surely you meant "that h' is cont everywhere, but fails to have a derivative at one point."



     
  7. Dec 16, 2008 #6

    HallsofIvy

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    ?? h(0) is defined- it is 0. The reason for the requirement that [itex]x\ne 0[/itex] in the first formula is that, of course, 1/x does not exist at x= 0 so x2sin(1/x) does not exist at x= 0.
     
  8. Dec 18, 2008 #7
    The graph approaches zero but is undefined at that point there it isn't continuous at that point and shouldn't be differentiable right? I don't know
     
  9. Dec 18, 2008 #8

    HallsofIvy

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    No, the function is NOT "undefined" at x= 0. f(0)= 0. The function is continuous at 0 because [itex]\lim_{x\rightarrow 0} f(x)= \lim_{x\rightarrow 0} x^3sin(1/x)= 0= f(0)[/itex]
     
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