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For f(x) = abs(x^3 - 9x), does f'(0) exist

  1. Jan 10, 2015 #1
    1. The problem statement, all variables and given/known data

    For f(x) = abs(x^3 - 9x), does f'(0) exist?

    3. The attempt at a solution

    The way I tried to solve this question was to find the right hand and left hand derivative at x = 0.

    Right hand derivative
    = (lim h--> 0+) f(h) - f(0) / h
    = (lim h--> 0+) abs(h^3 - 9h) / h
    = (lim h--> 0+) h^2 - 9
    = (lim h--> 0+) h^2 - 9 = -9

    Left hand derivative
    = (lim h--> 0-) f(h) - f(0) / h
    = (lim h--> 0-) abs(h^3 - 9h) / h
    = (lim h--> 0-) -(h^2 - 9)
    = (lim h--> 0-) -h^2 + 9 = 9

    However, when I plug in the equation into the graphing calculator, the magnitude is correct, by the positive and negative signs are switched.
     
  2. jcsd
  3. Jan 10, 2015 #2

    LCKurtz

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    Science Advisor
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    Gold Member

    Note that when ##h## is small positive that ##h^3 - 9h < 0## so the last step should have ##9-h^2##.
    Similar comment for the second one.
     
  4. Jan 10, 2015 #3
    Thanks a million! This made perfect sense
     
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