Find the domain and the range of each of the following real
function: $f(x)=\frac{x-3}{2-x}$
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, $\mathrm{R}_{\mathrm{f}(\mathrm{x})}=(-\infty,-1) \cup(-1, \infty)$.
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