What is the domain and range of y=sqrt(4-x^2) ?

2 Answers
Jun 9, 2017

Domain: [-2, 2]

Explanation:

Start by solving the equation

4 - x^2 = 0

Then

(2 + x)(2 -x) = 0

x = +- 2

Now select a test point, let it be x =0. Then y = sqrt(4 - 0^2) = 2, so the function is defined on [-2, 2[.

Thus, the graph of y= sqrt(4 - x^2) is a semicircle with radius 2 and domain [-2, 2].

Hopefully this helps!

Jun 12, 2017

Range: 0lt=ylt=2

Explanation:

The domain has already been determined to be -2lt=xlt=2. To find the range, we should find any absolute extrema of y on this interval.

y=sqrt(4-x^2)=(4-x^2)^(1/2)

dy/dx=1/2(4-x^2)^(-1/2)d/dx(4-x^2)=1/2(4-x^2)^(-1/2)(-2x)=(-x)/sqrt(4-x^2)

dy/dx=0 when x=0 and is undefined when x=pm2.

y(-2)=0, y(2)=0 and y(0)=2.

Thus the range is 0lt=ylt=2.


We could also arrive at this conclusion by considering the graph of the function:

y^2=4-x^2

x^2+y^2=4

Which is a circle centered at (0,0) with radius 2.

Note that solving for y gives y=pmsqrt(4-x^2), which is a set of two functions, since a circle by itself does not pass the vertical line test, so a circle is not a function but can be described by a set of 2 functions.

Thus y=sqrt(4-x^2) is the top half of the circle, which starts at (-2,0), rises to (0,2), then descends to (2,0), showing its range of 0lt=ylt=2.