Question

find the domain and range of the real function f(x)=x-2/3-x.

Answer 1

domain

3-X ≠ 0
X≠3
so domain R-{3}
and
range
put f(x)=y
3y-yx=x-2
3y-x(y+1)=2
x=3y-2/y+1
so R-{1} is the range

Answer 2

Answer:

$D=(-\infty,3)\cup(3,\infty) [x|x\neq3]$

$R=(-\infty,-1)\cup(-1,\infty) [y|y\neq-1]$

Step-by-step explanation:

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

To find : The domain and range of the real function

Solution :

To find domain  : Equate the denominator to zero

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

Denominator (3-x)=0

x=3

This means at x=3 function is not defined

And by definition of domain - The domain is where the function is not defined.

Domain is $D=(-\infty,3)\cup(3,\infty) [x|x\neq3]$

Range

Put f(x)=y

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

$x=\frac{3y-2}{y+1}$

Range is the set of value that correspond to domain

Equate the denominator to zero

$x=\frac{3y-2}{y+1}$

Denominator (y+1)=0

y=-1

This means at y=-1 function is not defined

Range is $R=(-\infty,-1)\cup(-1,\infty) [y|y\neq-1]$