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Math Help - [SOLVED] For what values of x will F be undefined?

  1. #1
    Newbie beardedoneder's Avatar
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    [SOLVED] For what values of x will F be undefined?

    Given the domain 0 \leq x \leq 2\pi in F(x) = x \cdot tan(x) - \frac{sin(x)}{x}, is it possible to for tan(x) to cause F to be undefined for x = \frac{\pi}{2} \pm 2\pi K, where K is a natural number? Given the domain restriction, it would seem that ONLY x = \{\frac{\pi}{2}, \frac{3\pi}{2}\} in tan(x) would cause F to be undefined. Divide by zero is trivially obvious and is not in question.


    The solution key says that x in tan(x) will cause F to be undefined for all x = \frac{\pi}{2} \pm 2\pi K.
     \frac{\pi}{2} + 2\pi 1 is \frac{5\pi}{2}, which is greater than the 2\pi limit of x, and \frac{\pi}{2} - 2\pi 1 is  - \frac{3\pi}{2}, which is less than the 0 limit of x. Their solution would also exclude x = \frac{3\pi}{2}. Furthermore, the only value of K that will keep x within the domain restriction is 0. So, why do they bring " \pm 2\pi K" into consideration at all?
    Last edited by beardedoneder; April 9th 2010 at 10:28 AM. Reason: Add the second to the last sentence.
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  2. #2
    MHF Contributor
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    Hi

    tan(x) is not defined for x = \frac{\pi}{2} \pm k\pi
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  3. #3
    Newbie beardedoneder's Avatar
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    Domain Restrictions?

    Quote Originally Posted by running-gag View Post
    Hi

    tan(x) is not defined for x = \frac{\pi}{2} \pm k\pi
    x = \frac{\pi}{2} + K\pi works when K = 1. That's \frac{3\pi}{2}. However, x = \frac{\pi}{2} + 2\pi is \frac{5\pi}{2} AND x = \frac{\pi}{2} - 1\pi is - \frac{\pi}{2}, both of which are outside of the domain restriction 0 \leq x \leq 2\pi . Is there something I'm not understanding about how domain restrictions work?
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  4. #4
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    When you restrict the domain of a function, you only care about the range on the interval of restriction. So if pi/2 and 3pi/2 are the only answers on the interval, those would be your answers.
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