1. ## Domain.

Can someone please show me how to work out the domain for this function?

$\frac{2}{3} \sqrt{36-p^2}$

I know the answer is 0 < x < 4
But just can't worked out how they got to that.

Any help is appreciated!

2. Originally Posted by el123
Can someone please show me how to work out the domain for this function?

$\frac{2}{3} \sqrt{36-p^2}$

I know the answer is 0 < x < 4
But just can't worked out how they got to that.

Any help is appreciated!
The domain for this function comes from the square root expression involved. If the function is a real function, then the argument of a square root can never be negative.

So what you are searching for, really, is the values of p for which $36 - p^2 \geq 0$.

You can find these values by solving that inequality.

You have your answer wrong though. The domain is $-6 \leq p \leq 6$

3. Do you mean 0 < p < 4? (there isn't an x in the term)

I would think the answer is $|p| <= 6$.

I am assuming the question only wants real numbers? If that is so, then you cannot have a negative number in the radical. So 36 minus any number less than 36 (which will always be positive because of the square) is fine, and 36-36 is fine too, since you can have a square root of zero. But 36 - 37 would be bad, since that would be the square root of a negative number, which is nonreal.

4. and i keep coming up with 0<x<6 but its not the correct answer.

5. Im using an old exam answer sheet. It says the domain is 0<x<4.

I get why it can't be negative. But this answer sheet says thats what the answer is.

6. The answer sheet is wrong. You can simply use a calculator to confirm. Try x = 4, x=5, x=6 and x = 7. If the answer sheet is right, you should get an error for 5, 6, 7. But if you and I are correct, then you should only get an error for 7.

7. Originally Posted by el123
and i keep coming up with 0<x<6 but its not the correct answer.
The correct method and answer was given in post #2. From the subsequent replies it appears no-one read it.