**Question:**

Find the domain and the range of each of the following real function: f(x)

$=\frac{a x-b}{c x-d}$

**Solution:**

Given: $\mathrm{f}(\mathrm{x})=\frac{a x-b}{c x-d}$

Need to find: Where the functions are defined.

Let, $f(x)=\frac{a x-b}{c x-d}=y$ ......(1)

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

Therefore,

$c x-d=0$

$\Rightarrow \mathrm{x}=\frac{d}{c}$

It means that the denominator is zero when $x=\frac{d}{c}$

So, the domain of the function is the set of all the real numbers except d/c

The domain of the function, $\left.\mathrm{D}_{\mathrm{f}(\mathrm{x})}=\left(-\infty, \frac{d}{c}\right) \cup \frac{d}{(c}, \infty\right)$

Now, to find the range of the function we need to interchange x and y in the equation no. (1)

So the equation becomes,

$\frac{a y-b}{c y-d}=x$

$\Rightarrow a y-b=c x y-d x$

$\Rightarrow a y-c x y=b-d x$

$\Rightarrow y=\frac{b-d x}{a-c x}$

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

Therefore

$a-c x=0$

$\Rightarrow \quad x=\frac{a}{c}$

It means that the denominator is zero when $x=\frac{a}{c}$

So, the range of the function is the set of all the real numbers except a/c.

The range of the function, $\operatorname{R}_{\mathrm{f}(x)}=\left(-\infty, \frac{a}{c}\right) \cup\left(\frac{a}{c}, \infty\right)$.