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

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Answer 1 · 2017-06-05T23:30:58

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

$9 - x^{2} \geq 0$ $9-x^2>=0$

$(3 - x) \cdot (3 + x) \geq 0$ $(3-x)*(3+x)>=0$

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

Hence the domain is $D_{f} = [- 3, 3]$ $D_f=[-3,3]$

For the range of the function we have that

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

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

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

and that is $f (0) = 4$ $f(0)=4$

Hence the range of the function is

$R_{f} = [1, 4]$ $R_f=[1,4]$

The graph of the function is

enter image source here

How do you find the domain and range of f(x)=1+√9−x2f(x) = 1 + sqrt(9 - x^2)?