what is the domain and range of y=(16-x^2)^(1/2)

Question

What is the domain and range of the following problem?

y=(16-x^2)^(1/2)

Answer 1

The range consists of the possible y values in a function. In the case of square root functions, they can never be negative. So they y values must be equal to or greater than zero.

 

(16 - x^2)^.5 >=0

 

16 - x^2 >=0

 

x^2 <= 16

 

x <= +/- 4 or -4 <= x <= 4

For those x values, the y at its lowest is zero & at its highest 4.

 

Range: 0 <= y <= 4