How do you find the domain and range of g(z) = 1 / sqrt(4-z^2)?

1 Answer
Aug 9, 2017

Here the function is g(z) = 1/(sqrt(4-z^2). But sqrt(4-z^2 can't be 0 because denominator cannot be 0. That is because dividing by 0 is not defined.

From this, we can say that z^2 can't be 4. That means z !=sqrt4 => z!=+-2.

And the another thing to be noticed is that we can't have a negative number inside a square root. So z must be in the range of 2 & -2.
Now we can say that Domain is all the real numbers between -2 & 2.

So Domain = -2<z<2, z in RR or (-2,2)

From this Domain set, we can easily find out Range of the Function.

The Range of the Function is in between 0 & 1/sqrt3 including 1/sqrt3.

So Range = (0,1/sqrt3]