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

2 Answers
Jun 9, 2017

Answer:

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

Answer:

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#.