Question:
Mark (√) against the correct answer in the following:
Let $f(x)=\sqrt{9-x^{2}}$. Then, dom $(f)=?$
A. $[-3,3]$
B. $[-\infty,-3]$
C. $[3, \infty)$
D. $(-\infty,-3] \cup(4, \infty)$
Solution:
$F(x)=\sqrt{9-x^{2}}$
$\sqrt{9-\mathrm{x}^{2}}$ should be $\geq 0$
$\Rightarrow 9-x^{2} \geq 0$
$\Rightarrow x^{2} \leq 9$
$\Rightarrow-3 \leq x \leq 3$
$\therefore \operatorname{dom}(f)=[-3,3]$
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