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

Solution:

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

Need to find: Where the functions are defined.

Let, $f(x)=\frac{x-3}{2-x}=y$  .......(1)

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

Therefore,

$2-x=0$

$\Rightarrow x=2$

It means that the denominator is zero when $x=2$

So, the domain of the function is the set of all the real numbers except 2 .

The domain of the function, $\operatorname{Df}(x)=(-\infty, 2) \cup(2, \infty)$.

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

So the equation becomes

$\frac{y-3}{2-y}=x$

$\Rightarrow y-3=2 x-x y$

$\Rightarrow y+x y=2 x+3$

$\Rightarrow y(1+x)=2 x+3$

$\Rightarrow y=\frac{2 x+3}{1+x}=f\left(x_{1}\right)$

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

Therefore

$x+1=0$

$\Rightarrow x=-1$

It means that the denominator is zero when $x=-1$

So, the range of the function is the set of all the real numbers except $-1$.

The range of the function, $\mathrm{R}_{\mathrm{f}(\mathrm{x})}=(-\infty,-1) \cup(-1, \infty)$.