Find the domain and the range of each of the following real

Given: $f(x)=\frac{x^{2}-9}{x-3}$

Need to find: Where the functions are defined.

To find the domain of the function $f(x)$ we need to equate the denominator of the function to 0 .

Therefore

$x-3=0$

$\Rightarrow x=3$

It means that the denominator is zero when $x=3$

So, the domain of the function is the set of all the real numbers except 3 .

The domain of the function, $\mathrm{D}_{\mathrm{f}(\mathrm{x})}=(-\infty, 3) \cup(3, \infty)$.

Now if we put any value of $x$ from the domain set the output value will be either (-ve) or (+ve), but the value will never be 6

So, the range of the function is the set of all the real numbers except 6.

The range of the function, $\mathrm{R}_{\mathrm{f}(\mathrm{x})}=(-\infty, 6) \cup(6, \infty)$.