Find the domain of $\arcsin\left(\frac{3-x}{\sqrt{9-x^2}}\right)$

Find the domain of $$\arcsin\left(\dfrac{3-x}{\sqrt{9-x^2}}\right)$$

This question is given in my book. The answer is as follows:

For the function to be defined, $-1\le3-x\le1$ and $9-x^2>0$. Solve these inequalities and take their common region. $$2\le x\le4\tag1$$ $$-3 < x < 3\tag2$$

The answer is $2\le x<3$. My question is why have they not taken $-1\le\dfrac{3-x}{\sqrt{9-x^2}}\le 1$, in which case the answer would have been different. Please help.

• The book answer would be correct if the function was $\cfrac{\arcsin(3-x)}{\sqrt{9-x^2}}\,$ instead, perhaps written as $\arcsin(3-x) / \sqrt{9-x^2}\,$.
– dxiv
May 10, 2017 at 4:08

For $x\ne3,\dfrac{3-x}{\sqrt{9-x^2}}=\sqrt{\dfrac{3-x}{3+x}}$

Now we need $$-1\le\sqrt{\dfrac{3-x}{3+x}}\le1\iff0\le\dfrac{3-x}{3+x}\le1$$

If $\dfrac{3-x}{3+x}=0\implies x=3\ \ \ \ (1)$

$\dfrac{3-x}{3+x}>0\iff(3-x)(3+x)>0\iff(x-3)(x+3)<0\iff-3<x<3\ \ \ \ (2)$

$\dfrac{3-x}{3+x}\le1\iff0\ge\dfrac{3-x}{3+x}-1=\dfrac{3-x-(3+x)}{3+x}=-\dfrac{2x}{3+x}$

By $(2),x+3>0,$ so we need $-2x\le0\iff x\ge0\ \ \ \ (3)$

By $(1),(2),(3):0\le x<3$

• First line For x≠3 should rather be $x \le 3\,$, though that hasn't been established yet. Last line 0 < x ≤ 3 should rather be $0 \le x \lt 3\,$.
– dxiv
May 10, 2017 at 5:28
• @dxiv, we can safely start with $x\ne3$ and then establish the condition May 10, 2017 at 5:34
• Without the pre-condition $x \le 3$ all that can be said is $\,\dfrac{3-x}{\sqrt{9-x^2}}=\operatorname{sgn}(3-x) \cdot \sqrt{\dfrac{3-x}{3+x}}\,$. You are right that the next inequality is symmetric around $0\,$, so the sign doesn't matter in the end, but that wasn't all that obvious on a first quick parse.
– dxiv
May 10, 2017 at 5:41

You are right. For instance, $x=1$ is clearly in the domain of the function, but it isn't in the interval given in your book.

The correct domain should be :

$$x \in [0,3)$$

(For its proof, see Mr.Lab Bhattacharjee's answer)