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

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Answer 1 · 2015-09-20T21:37:19

Refer to explanation

For the domain we have that

$9 - x^{2} \geq 0 \Rightarrow (3 - x) \cdot (3 + x) \geq 0 \Rightarrow - 3 \leq x \leq 3$ $9-x^2>=0=>(3-x)*(3+x)>=0=>-3<=x<=3$ hence

$D (f) = [- 3, 3]$ $D(f)=[-3,3]$

and the range is

$R (f) = [0, 3]$ $R(f)=[0,3]$

The graph of the function is

graph{(9-x^2)^(1/2) [-10, 10, -5, 5]}

Actually the function represents the upper half of the circle with
equation

$x^{2} + y^{2} = 3^{2}$ $x^2+y^2=3^2$

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