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How to differentiate an absolut value, f(x)=│x^2-4│

  1. Apr 30, 2005 #1
    how do u differentiate f(x)=│x^2-4│....?
    i don't know how to do it with absolute values...
     
  2. jcsd
  3. Apr 30, 2005 #2

    cepheid

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    If I remember right, define it in a piecewise fashion. Can you see that:

    |x^2 - 4| = x^2 - 4 if x^2 - 4 > 0

    |x^2 - 4| = -(x^2 - 4) if x^2 - 4 < 0

    The first condition becomes: if x^2 > 4 ===> |x| > 2, i.e. if x > 2 OR x < -2

    The second becomes: if x^2 < 4 ===> |x| < 2, i.e. -2 < x < 2

    So you have two cases. For x > 2 and x < -2, the function is:

    f(x) = x^2 - 4, and you can differentiate it.

    For -2 < x < 2, the function is:

    f(x) = -x^2 + 4, and you can differentiate it.

    The function is differentiable at x = 2 and x = -2 if and only if the left and right hand derivatives exist at those points.
     
  4. Apr 30, 2005 #3

    arildno

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    It is quite instructive to use the DEFINITION of the derivative at the problem points 2 and -2.
    I'll take the "2"-case:
    In general, we have:
    [tex]f'(2)=\lim_{\bigtriangleup{x}\to{0}}\frac{f(2+\bigtriangleup{x})-f(2)}{\bigtriangleup{x}}[/tex]
    if it exists.
    In our case, [tex]f(x)=|x^{2}-4|[/tex]
    which implies [tex]f(2)=0,f(2+\bigtriangleup{x})=|(2+\bigtriangleup{x})^{2}-4|=|4\bigtriangleup{x}+(\bigtriangleup{x})^{2}|[/tex]
    Hence, we must have:
    [tex]f'(2)=\lim_{\bigtriangleup{x}\to0}\frac{|\bigtriangleup{x}|}{\bigtriangleup{x}}|4+\bigtriangleup{x}|[/tex]
    if it exists.
    Can it exist?
     
  5. Apr 30, 2005 #4
    i found the first reply easier to understand...
    f(x)=x^2-4
    f'(x)=2x
    f'(x) as x->2+ would equal 2(2)=4

    f(x)=-x^2+4
    f'(x)=-2x
    f'(x) as x->2- would equal -2(2)=-4
    so does not exist at x=2

    f(x)=x^2-4
    f'(x)=2x
    f'(x) as x->-2- would equal 2(-2)=-4

    f(x)=-x^2+4
    f'(x)=-2x
    f'(x) as x->-2+ would equal -2(-2)=4
    so does not exist at x=-2
    is that right?
     
  6. Apr 30, 2005 #5

    Pyrrhus

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    Yes that's right, because the limit does not exist therefore the derivative does not exist at those points.
     
  7. Apr 30, 2005 #6
    okay...i see...thanks guys...
     
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