The domain of the function
\[f\left( x \right) = {{{{\sin }^{ - 1}}\left( {x - 3} \right)} \over {\sqrt {9 - {x^2}} }}\]
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Official solution
\[f\left( x \right) = {{{{\sin }^{ - 1}}\left( {x - 3} \right)} \over {\sqrt {9 - {x^2}} }}\] is defined
if (i) \[\,\,\, - 1 \le x - 3 \le 1 \Rightarrow 2 \le x \le 4\]
and (ii) \[9 - {x^2} > 0 \Rightarrow - 3 < x < 3\]
Taking common solution of \[\left( i \right)\] and \[\left( {ii} \right),\]
we get \[2 \le x < 3\]
\[\therefore\] Domain \[ = \left[ {2,\left. 3 \right)} \right.\]