The range of the real valued function f(x) = | vedclass

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Question

(C) Given the function $f(x) = \sqrt{9-x^2}$.For $f(x)$ to be defined,the expression inside the square root must be non-negative:$9 - x^2 \geq 0$$x^2

Vedclass · Accepted Answer

$[0, 3]$

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(C) Given the function $f(x) = \sqrt{9-x^2}$.For $f(x)$ to be defined,the expression inside the square root must be non-negative:$9 - x^2 \geq 0$$x^2 \leq 9$$-3 \leq x \leq 3$Since $x^2$ ranges from $0$ to $9$,the expression $9 - x^2$ ranges from $9 - 9 = 0$ to $9 - 0 = 9$.Therefore,$\sqrt{9 - x^2}$ ranges from $\sqrt{0}$ to $\sqrt{9}$,which is $0$ to $3$.Thus,the range of the function is $[0, 3]$.

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

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