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F'(x) existent or not.

  1. Oct 5, 2006 #1
    It's a question I had on a quiz a few minutes ago.

    f(x) = sin(x) for x < or = 0 and
    f(x) = x for x > 0

    Question was...does f'(x) exist? what is it?


    First I proved the f(x) is continuous at 0 by stating limit as x->0 from left = limit as x->0 from right = f(0).

    Then...taking the derivative turned out to be a problem.
    the lim[f(x+h)-f(x)]/h was unclear because...if x=0 and you add some h to it, that'd mean you use as your f(x) the x...but point (0, 0) belongs to Sin(x) not to x.

    What i did is state that in x-values close to zero, from left and from right, the f'(x) = 1. Basically I ruled out situations like |x| where dy/dx = -1 and 1 for values around x=0.

    I assumed that if f'(-0.0001) and f'(0.0001) are equal, the function (who we know is continuous) must be differentiable at that point.


    But my question is...is my logic something that will make my teacher grab his hairs before giving me a nice zero? or a 'wise' way to look the issue?

    Also, is there a better, clearer way to solve this?

    I always experience uncertainty on what to do and to which expression when a function gets split according to Domain.

    Thank you.

  2. jcsd
  3. Oct 5, 2006 #2


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    Dearly Missed

    Split your argument in two:
    1. Find the right-hand derivative
    2. Find the left-hand derivative

    If they are equal, then your function is differentiable at 0.
  4. Oct 5, 2006 #3

    matt grime

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    f is trivially differentiable for x>0 and x<0. At x=0, one merely needs to verify if the left and right derivatives exist and agree. They trivially exist and as cos(x) tends to 1 as x tends to zero they agree.
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