# Thread: [SOLVED] For what values of x will F be undefined?

1. ## [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?

2. Hi

tan(x) is not defined for $x = \frac{\pi}{2} \pm k\pi$

3. ## Domain Restrictions?

Originally Posted by running-gag
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?

4. 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.