# Am I missing anything something when trying to find the domain of this function?

My son got the following question on a test:

Find the domain of the function.

$$q\left(y\right)=\sqrt{25-y^2}$$

a. all real numbers

b. $$y\ge0$$

c. $$y\le-5$$ or $$y\ge5$$

d. $$-5\le y\le 5$$

e. $$y\le5$$

He chose d, but it was marked incorrect, and he's given partial credit for finding the right answer. He couldn't figure out any other answer, so I tried, and I found the same answer.

Here's my reasoning: The argument to the square root must be non-negative, therefore…

$$\begin{eqnarray} 25-y^{ 2 } & \ge & 0 \\ 25 & \ge & y^{ 2 } \\ \sqrt { 25 } & \ge & \sqrt { y^{ 2 } } \\ 5 & \ge & \left| y \right| \\ -5 & \le y\le & 5 \end{eqnarray}$$

My son went back to the teacher, an was informed that the correct answer is c ($$y\le-5$$ or $$y\ge5$$). But I don't think this can be right, as $$10\ge5$$, but plugging $$10$$ into the function results in $$\sqrt{-75}$$, and his class hasn't been introduced to complex numbers yet, and $$\mathbb{C}$$ wasn't an option anyway.

Just to make sure I'm actually reading the problem correctly, here's a photo of it:

My guess is that the answer key has a misprint or something, but am looking for confirmation of what I've done before suggesting that to the teacher. The other possibility is that we're both missing something in the question or in our logic, which is why I've included the photo.

• You are correct @Chuck. The teacher was wrong. – Guillemus Callelus Sep 5 '17 at 23:24
• I agree; your reasoning and your answer are correct. – user84413 Sep 5 '17 at 23:43