Can someone please show me how to work out the domain for this function?
$\displaystyle \frac{2}{3} \sqrt{36-p^2}$
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 $\displaystyle 36 - p^2 \geq 0 $.
You can find these values by solving that inequality.
You have your answer wrong though. The domain is $\displaystyle -6 \leq p \leq 6 $
Do you mean 0 < p < 4? (there isn't an x in the term)
I would think the answer is $\displaystyle |p| <= 6$.
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.