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

Sep 27, 2015

Domain $\left\{t : \mathbb{R} , - 3 \le t \le 3\right\}$

Range $\left\{f \left(t\right) : \mathbb{R} , 0 \le f \left(t\right) \le 3\right\}$

#### Explanation:

For real f(t), $9 \ge {t}^{2}$ or $3 \ge \pm t$. That is $- 3 \le t \le 3$ would be the domain of f(t).

For range, f(t) would be 0 for t =3 or -3, it would be maximum at t=0

hence range would be$\left\{f \left(t\right) : \mathbb{R} , 0 \le f \left(t\right) \le 3\right\}$