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


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

    tan(x) is not defined for $\displaystyle x = \frac{\pi}{2} \pm k\pi$
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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 $\displaystyle x = \frac{\pi}{2} \pm k\pi$
    $\displaystyle x = \frac{\pi}{2} + K\pi $ works when $\displaystyle K$ = 1. That's $\displaystyle \frac{3\pi}{2}$. However, $\displaystyle x = \frac{\pi}{2} + 2\pi $ is $\displaystyle \frac{5\pi}{2}$ AND $\displaystyle x = \frac{\pi}{2} - 1\pi$ is $\displaystyle - \frac{\pi}{2}$, both of which are outside of the domain restriction $\displaystyle 0 \leq x \leq 2\pi$ . 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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