Thread: determining the domain of a fuction

So Determine the domain of this fuction
y=√x^2-16

so first i factorised and get (x+4) and (x-4)

i m sure the domain has to be between -4 <x<4

what i am confused about is what happens to the square root , can some 1 how me how to do the proper workout for this problem ?

cheers

Originally Posted by arsenal12345

So Determine the domain of this fuction
y=√x^2-16

so first i factorised and get (x+4) and (x-4)

i m sure the domain has to be between -4 <x<4

what i am confused about is what happens to the square root , can some 1 how me how to do the proper workout for this problem ?

cheers

Nothing negative can go inside a square root. So

thanks buddy would it be wrong if i wrote down domain is present in the fuction between -4 <x<4 ??

Originally Posted by arsenal12345

thanks buddy would it be wrong if i wrote down domain is present in the fuction between -4 <x<4 ??

Of course it would be wrong. The domain is everything EXCEPT for that interval...

wait i thought domain is the interval < when is the doma everything except interval ?

is it like when the y is real the domain is the interrval and when its not the domain is every thing except the interval ??

if yes how do i workout y ?

You were told that in Prove It's first response to your post: the quantity inside the square root must be non-negative and that happens for and .

I don't know what you mean by "work out y". If you mean just calculate y, for a given x, do exactly what the formula says: square x, subtract 16, then take the square root. If you mean that you want to find the range of the function, the set of all possible values of y, then you can note that a square root is never negative but, since can be any positive number and can be 0, the range is the set of all non-negative numbers.