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

##### 2 Answers

#### Answer:

Domain:

#### 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

Thus, the graph of

Hopefully this helps!

#### Answer:

Range:

#### Explanation:

The domain has already been determined to be

#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)#

Thus the range is

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

Note that solving for *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

Thus