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The range of the function \( f(x)=\sqrt{9-x^{2}} \) is
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The correct answer is:
\( [0,3] \)
Given that, $f(x)=\sqrt{9-x^{2}}$
$x^{2}=9 \Rightarrow x \pm 3$
So, domain is given by $x \in[-.3,3]$
Therefore, range of $f(x) \in[0,3]$
$x^{2}=9 \Rightarrow x \pm 3$
So, domain is given by $x \in[-.3,3]$
Therefore, range of $f(x) \in[0,3]$
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