# Domain of $f(x)=\arctan\left(\frac{x^2}{2-x^2}\right)$

I need to find the domain of the function $f(x)=\arctan\left(\frac{x^2}{2-x^2}\right)$. I used a graphing calculator and found that the domain is $(-\infty,-\sqrt 2) ∪ (-\sqrt 2, \sqrt 2)∪(\sqrt 2, \infty)$ but I'm unsure as to how to do it without a calculator.

My assumption is that, using what we know about the calculation of the domain of $\sin$ and $\cos$, we make the denominator equal to $0$. So $$2-x^2=0$$ $$2=x^2$$ $$\pm\sqrt 2=x$$

But how do I get from there to the domain stated above?

• How would you describe the set of real numbers not equal to $\pm\sqrt2$? – Lord Shark the Unknown Dec 19 '17 at 20:17
• $\arctan$ is defined everywhere, and $\frac{x^2}{2-x^2}$ is only undefined where you showed. What can't you put into $\arctan\left(\frac{x^2}{2-x^2}\right)$? – Austin Weaver Dec 19 '17 at 20:20
• But in the answer $(-\sqrt 2, \sqrt 2)$ is also a possible solution to the question. – Ski Mask Dec 19 '17 at 20:22

The function $\arctan$ is defined everywhere in $\mathbb{R}$, but in mathematics you mustn’t divide by zero. This means that the only condition you have to apply is: $2-x^2\neq 0$. And this means $x\neq \pm \sqrt 2$.

Then the domain is $(-\infty,-\sqrt 2) ∪ (-\sqrt 2, \sqrt 2)∪(\sqrt 2, \infty)$ that is the same of $\mathbb{R}-\{-\sqrt 2,\sqrt 2\}$.

As you can see from the graphic of the function, it as the asymptotes in $x=\pm \sqrt 2$:

The function $arctan(x)$ is continuous, so this on its own would have no discontinuities. The only place then that this function will be discontinuous is when there is a division by $0$, which you correctly found to be $\sqrt2$ and $-\sqrt2$, therefore the domain is all reals except these two values.

To be clear, you don't have a function unless you have its domain to begin with. Asking for the domain of a function is like asking for the age of a man whose age is given to be $35.$

What you have with $\arctan (x^2/(2-x^2))$ is an expression. The set of all $x$ for which an expression is defined is often called the "natural domain" of the expression.

Assuming the natural domain is what we're after here leads to two questions: For which $x$ is $x^2/(2-x^2)$ defined, and for which $y$ is $\arctan y$ defined. The answers are: for the first question, it's all $x\ne \pm \sqrt 2;$ for the second question, it's all $y.$ Thus the natural domain for the expression $\arctan (x^2/(2-x^2))$ is $\{x\in \mathbb R: x\ne \pm \sqrt 2\}.$