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

Aug 12, 2017

Domain: $\left[- 3 , 3\right]$
Range: $\left[0 , 3\right]$
So, we need $9 - {x}^{2} \setminus \ge q 0$, or $9 \setminus \ge q {x}^{2}$, so $x \setminus \le q 3$ and $x \setminus \ge q - 3$, or $\left[- 3.3\right]$.
As $x$ takes on these values, we see that the smallest value of the range is $0$, or when $x = \pm 3$ (so $\sqrt{9 - 9} = \sqrt{0} = 0$), and a max when $x = 0$, where $y = \setminus \sqrt{9 - 0} = \sqrt{9} = 3$