Question

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

Answer 1

`-4<=x<=4` and `1<=y<=5`

Explanation:

Since the radicand has never to be negative we get
`-4<=x<=4`
Then we get
`1<=sqrt(16-x^2)+1<=5`
Since we have
`sqrt(16-x^2)>=0`
and
`sqrt(16-x^2)<=4`
since
`x^2>=0`