I tried to find the domain of each separate function but I do not understand if this is the right process and how to combine the resulting domains.

In the picture, I tried to find the domain of each separate function but I do not understand if this is the right process and how to combine the resulting domains.The function is $(\frac{1-x}{x-3})^{1/2}+(cos{(πx)})^{1/2}$ .

  • $\begingroup$ You may use mathJax to format your question. $\endgroup$
    – dmtri
    Sep 16, 2018 at 1:52

1 Answer 1


I liked this question a lot. But you should spend a little time and learn mathjax.

I’m tutoring in a couple of high-school classes that are worrying over exactly this kind of question. First, I should say that as given, it’s not a function, it’s just a mathematical expression, which may or may not make sense. When you identify all numbers for which it does make sense, then you may use this identification to define a set, and proclaim a function whose domain is this set, and whose rule is given by your expression.

Now, there’s a serious difficulty with what you have presented us, because your manuscript says $\sqrt{1-x}\big/\sqrt{x-3}$, while your typescript says $\sqrt{\frac{1-x}{x-3}}$. You may be surprised to see that these two expressions do not say the same thing!

For your specific expression $$ \sqrt{\frac{1-x}{x-3}}+\cos(\pi x)\,, $$ the second term, the trigonometric, poses no problem, since it makes sense for all real $x$. It’s the radical part that makes you stop. First, everything under a radical has to be nonnegative, i.e. $\ge0$, and when you examine that fraction more closely, you see that it’s not possible to get both top and bottom simultaneously positive. But if they’re simultaneously negative, then the radicand (what’s under the square root sign) becomes positive all right. Remembering that the denominator can’t be zero, you get two inequalities $1-x\le0$ and $x-3<0$, to which you apply standard techniques to get $1\le x$ and $x<3$. They both have to be true. You can describe the interval on which the expression gives you a good function.

  • $\begingroup$ I am sorry but I forget to add a square root under the cos(πx) so it should be √cos(πx) $\endgroup$
    – Josh Teal
    Sep 16, 2018 at 3:32
  • $\begingroup$ Well, the domain for the expression $\sqrt{\cos(\pi x)}$ is a bunch of nonooverlapping closed intervals, each of length $1$. For $x$ to be good, it must be good with respect to both conditions. So you’ll have to take the intersection ($\cap$) of the two sets. $\endgroup$
    – Lubin
    Sep 16, 2018 at 17:49
  • $\begingroup$ How do I find the domain of √cos(πx) using algebra and could you please explain what is meant by intersection? $\endgroup$
    – Josh Teal
    Sep 16, 2018 at 18:06
  • $\begingroup$ The intersection of two sets is the common part: all the points that are in both. That’s easy. For the domain of $\sqrt{\cos(\pi x)}$, that’s easy, too: all the $x$ that make the cosine-expression nonnegative. In your case, $[-\frac12,\frac12]$ and its translates by even numbers. According to your manuscript, you seem to have understood this, even if perhaps a little dimly. (That’s the way way we approach understanding.) $\endgroup$
    – Lubin
    Sep 16, 2018 at 23:25
  • $\begingroup$ When I do the intersection of the two sets, I get the final answer of [3/2,5/2]. Is this correct? Just for more understanding, why can I take the domain of these two terms separately? $\endgroup$
    – Josh Teal
    Sep 17, 2018 at 5:07

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