The range of the function f(x) = sqrt{9 - x^2 | vedclass

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(B) Given the function $f(x) = \sqrt{9 - x^2}$.For the function to be defined,the expression inside the square root must be non-negative:$9 - x^2 \geq

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$ [0, 3] $

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(B) Given the function $f(x) = \sqrt{9 - x^2}$.For the function to be defined,the expression inside the square root must be non-negative:$9 - x^2 \geq 0 \Rightarrow x^2 \leq 9 \Rightarrow -3 \leq x \leq 3$.Thus,the domain of the function is $x \in [-3, 3]$.When $x = 0$,$f(0) = \sqrt{9 - 0} = 3$.When $x = \pm 3$,$f(\pm 3) = \sqrt{9 - 9} = 0$.Since the square root function always yields non-negative values,the minimum value is $0$ and the maximum value is $3$.Therefore,the range of the function is $[0, 3]$.

The range of the function $f(x) = \sqrt{9 - x^2}$ is

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