How do you find the domain and range of $f(x) = 1 + sqrt(9 - x^2)$ ?

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Answer 1 · 2017-06-05T23:30:58

The domain of $f(x)$ is all the values for which

$9-x^2>=0$

$(3-x)*(3+x)>=0$

Which is valid for $x$ in $[-3,3]$

Hence the domain is $D_f=[-3,3]$

For the range of the function we have that

$f(-3)=f(3)=1$

and the maximum value of f(x) is achieved when

$9-x^2$ is maximized which happens for $x=0$

and that is $f(0)=4$

Hence the range of the function is

$R_f=[1,4]$

The graph of the function is

enter image source here

How do you find the domain and range of f(x) = 1 + sqrt(9 - x^2)f(x) = 1 + sqrt(9 - x^2)?