How do you find the domain and range for #y =sqrt(9-x^2)#?

1 Answer
Feb 18, 2016

Answer:

Domain: #x in [-3,+3]#
Range: #yin [0,3]#

Explanation:

I assume that we are dealing only with Real numbers.

For Real numbers #sqrt(a)# is only defined if #a>=0#
#rarr (9-x^2)>=0#
#rarr abs(x)<=3#
which gives us the Domain.

The maximum value of #sqrt(9-x^2)# occurs when #x=0# and
this maximum is #sqrt(9) = 3#
The minimum value of #sqrt(9-x^2)# occurs when #x=+-3# and
this minimum is #sqrt(9-9) =0#
which together give us the Range.