# What is the domain and range of g(x)=sqrt(16-x^2) + 1 ?

May 27, 2018

$- 4 \le x \le 4$ and $1 \le y \le 5$

#### Explanation:

Since the radicand has never to be negative we get
$- 4 \le x \le 4$
Then we get
$1 \le \sqrt{16 - {x}^{2}} + 1 \le 5$
Since we have
$\sqrt{16 - {x}^{2}} \ge 0$
and
$\sqrt{16 - {x}^{2}} \le 4$
since
${x}^{2} \ge 0$