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Inverse Function of sine

  1. Dec 3, 2011 #1
    1. The problem statement, all variables and given/known data
    Determine if the following function is invertible.
    If it is, find the inverse function.

    f (x) = -sin(-x)
    -∏ < x < ∏




    3. The attempt at a solution

    f(x) = -sin(-x)
    y = -sin(-x)
    -(sin y)^-1 = -x
    (sin y)^-1 = x

    when x = -∏, y = 0
    when x = ∏, y = 0

    f^-1(x) = (sin x)^-1
    0 = x = 0
     
  2. jcsd
  3. Dec 3, 2011 #2
    Actually, now I'm wondering, is the answer simply that the function is not monotonic and therefore it is not invertible?
     
  4. Dec 3, 2011 #3

    Mark44

    Staff: Mentor

    The interval doesn't include [itex]\pi[/itex] or [itex]-\pi[/itex]
    Have you sketched a graph of this function? If you know the graph of y = f(x), you can get the graph of y = -f(-x) by doing a couple of reflections. Having a graph should give you a good idea of whether your function has an inverse.

    What rule or theorem are you using to determine whether a function has an inverse?
     
  5. Dec 3, 2011 #4
    Yes, I wasn't sure what to do about that. On other questions I've tried it usually states, for example, -∏ <= x <= ∏, but in this example it didn't have an equals in the interval.

    I've just input it into http://rechneronline.de/function-graphs/ and it appears that it is not monotonic, and therefore not invertible.

    I wasn't really using any specific rule for determining if the function is monotonic, as so far the functions I've been trying aren't too difficult eg (X^2 - 5) or (3x + 3).

    Below is the method I've been using to determine the inverse functions.

    For example,

    f(x) = x^2 - 5 0 <= x < ∞
    y = x^2 - 5
    (y + 5)^0.5 = x
    when x = 0, y = -5

    therefore f^-1(x) = (x + 5)^0.5
    when x = 0, y = -5


    Is there a specific rule for determining if a function is monotonic?
     
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