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Homework Help: Is |x|^3 differentiable?

  1. Dec 5, 2007 #1
    [SOLVED] Is |x|^3 differentiable?

    1. The problem statement, all variables and given/known data

    Is [tex] |x|^3 [/tex] differentiable?

    2. Relevant equations

    [tex] Def: \ Let \ f \ be \ defined \ (and \ real-valued) \ on [a,b]. \ \ For \ any \ x \in [a,b], \ form \ the \ quotient [/tex]

    [tex]\phi(t)=\frac{f(t)-f(x)}{t-x} \ \ \ \ (a<t<b, \ t\neqx), \\ [/tex]

    [tex] and \ define \\ [/tex]

    [tex] f^{'}(x)=\lim_{\substack{t\rightarrow x}} \phi(t) [/tex]

    3. The attempt at a solution

    Well, using the definition, I calculated that the right-hand limit and left hand limit are different. But I'm not sure if that means anything or what I can conclude here. Nor am I sure how I should define the left and right limits here.

    Any help would be great, thanks!
  2. jcsd
  3. Dec 5, 2007 #2


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    If the left hand limit and the right hand limit are different, then the limit that defines the derivative at that point does not exists: the function is not differentiable.

    However, I'm not sure it's not differentiable in this case... if there were a problem it would be at x = 0 and when I sketch a picture I don't immediately see a problem arising. Can you show us the calculation for the limit?
  4. Dec 5, 2007 #3

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    The left hand derivative and the right hand derivative at x=0 are both zero. It's differentiable.
  5. Dec 5, 2007 #4


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    How did you get that the right and left hand derivatives are different? If t> 0, |f(t)|= |t^3|= t^3 then at x= 0,
    [tex]\frac{f(t)-f(x)}{t- x}= \frac{|t^3|}{t}= \frac{t^3}{t}= t^2[/tex]
    and the limit of that, as t goes to 0, is 0.

    If t< 0, |f(t)|= |t^3|= -t^3. At x= 0,
    [tex]\frac{f(t)-f(x)}{t-x}= \frac{|t^3|}{t}= \frac{-t^3}{t}= -t^2[/tex]
    but the limit, as t goes to 0, is still 0. The function is differentiable at 0 and the derivative there is 0.

    Obviously, if x> 0, [itex]f(x)=|x^3|= x^3[/itex] which is differentiable and if x< 0, [itex]f(x)= |x^3|= -x^3[/itex] which is differentiable.
  6. Dec 5, 2007 #5
    Ah, I see what I did now. Totally forgot to cube my expression in my calculations, hahaha. Alrighty, thanks Ivy.
  7. Dec 6, 2007 #6
    [tex] |x|^3 [/tex] the same as [tex] |x^3| [\tex] ?
  8. Dec 6, 2007 #7

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    They are equal, not same. They are different functions.
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