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

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

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Answer 1 · 2018-05-27T08:28:46

$-4<=x<=4$ 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
$x^2>=0$

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

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

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