Question
Find the domain and range of the function $f(x)=\frac{x^2-9}{x-3}$

Answer

Here $f(x)=\frac{x^2-9}{x-3}$
$f(x)$ assume real values for all real values of $x$ except for $x - 3 = 0$
i.e $x = 3$
Thus domain of $f (x) = R - {3}$
Let $f (x) = y$
$\therefore y=\frac{x^2-9}{x-3}$
$=\frac{(x+3)(x-3)}{(x-3)}$
$\Rightarrow y=x+3$
$y$ takes all real values except $6$ as domain $=R-{3}$
Thus range of $f(x) = R - {6}.$

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