What is the domain and range of #f(x) = sqrt (9 - x^2)#?

1 Answer
Cri007
Aug 12, 2017

Domain: #[-3,3]#
Range: #[0,3]#

Explanation:

The value under a square root cannot be negative, or else the solution is imaginary.

So, we need #9-x^2\geq0#, or #9\geqx^2#, so #x\leq3# and #x\geq-3#, or #[-3.3]#.

As #x# takes on these values, we see that the smallest value of the range is #0#, or when #x=pm3# (so #sqrt(9-9)=sqrt(0)=0#), and a max when #x=0#, where #y=\sqrt(9-0)=sqrt(9)=3#

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