Question
Find the domain and range of the real function $f(x)=\sqrt{9-x^2}$.
✓
Answer
It is clear that, $f ( x )=\sqrt{9-x^2}$ is not defined when $\left(9- x ^2\right)<0$, i.e.
When $x^2>9 i$,e when $x>3$ or $x<-3$
$\begin{array}{l}\operatorname{dom}( f )=|x \in R:-3 \leq x \leq 3| \\
\text { Also, } y =\sqrt{9-x^2} \Rightarrow y^2=\left(9-x^2\right) \\
\Rightarrow x=\sqrt{9-y^2}\end{array}$
$\begin{array}{l}\text { clearly, } x \text { is not defined when }\left(9-y^2\right)<0 \\
\text { but }\left(9-y^2\right)<0 \Rightarrow y^2>9 \\ \Rightarrow y>3 \text { or } y<-3 \\
\text { range }(f)=\{y \in R:-3 \leq y \leq 3\}\end{array}$
When $x^2>9 i$,e when $x>3$ or $x<-3$
$\begin{array}{l}\operatorname{dom}( f )=|x \in R:-3 \leq x \leq 3| \\
\text { Also, } y =\sqrt{9-x^2} \Rightarrow y^2=\left(9-x^2\right) \\
\Rightarrow x=\sqrt{9-y^2}\end{array}$
$\begin{array}{l}\text { clearly, } x \text { is not defined when }\left(9-y^2\right)<0 \\
\text { but }\left(9-y^2\right)<0 \Rightarrow y^2>9 \\ \Rightarrow y>3 \text { or } y<-3 \\
\text { range }(f)=\{y \in R:-3 \leq y \leq 3\}\end{array}$
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