Question

Q28. Find the Domain and range of (x - 2)/(3 - x).​

Answer 1

Answer:

Domain : R - {3} or (-∞, 3) ∪ (3, ∞)

Range : R - {-1} or (-∞, -1) ∪ (-1, ∞)

Step-by-step explanation:

Given equation :

$\dfrac{x-2}{3-x}$

To find :

Domain and range of the given equation

Solution :

We know that, Domain is the set of all the elements which satisfy the given equation, without giving any indefinite value.

So, we must find the elements which give the indefinite value for the given question

As given equation is a rational polynomial, denominator should never be zero as it gives infinity as a value

So, 3 - x ≠ 0

=> x ≠ 3

Hence, the only number which does not satisfy the given equation is 3

${\therefore \textsf {\bf {Domain of \dfrac{x-2}{3-x} is \ \boxed {\textsf {R - \{3\}}} }}}$

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We know that, Range is the set of all the elements which gives the result of the given equation

So, whenever we need to find the range of any function, we must always assign a variable to the function and then find the domain of its inverse. The Domain of the inverse gives the range of the given equation

So, let the given function or equation be f(x) and it be assigned with variable y

Thus,

$f(x) = y \\\\=> f^{-1}(y) = x$

So,

$y = \dfrac{x-2}{3-x} \\\\=> y (3-x) = x-2 \\\\=> 3y - xy = x -2\\\\=> 3y + 2 = x + xy \\\\=> x (y+1) = 3y+2 \\\\=> x = \dfrac {3y+2}{y+1} \\\\=> f^{-1}(y) = \dfrac {3y+2}{y+1}$

Replacing y with x, we get

$f^{-1}(x) = \dfrac {3x+2}{x+1}$

Hence this is the inverse equation of given equation f(x)

So, finding the domain of f⁻¹(x) gives range

So, for finding the domain, following the same method. Denominator must not be zero

Thus, x + 1 ≠ 0

=> x ≠ -1

So, for the given equation we never get -1

${\therefore \textsf {\bf {Range of \dfrac{x-2}{3-x} is \ \boxed {\textsf {R - \{-1\}}} }}}$

_______________________________

❖ Extra information:

⟡ Graph of given equation $\dfrac{x-2}{3-x}$ is in the attachment.

Hope it helps!

Please mark as brainliest

Answer 2

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:\dfrac{x - 2}{3 - x}$

Let assume that

$\rm :\longmapsto\:f(x) = \dfrac{x - 2}{3 - x}$

We know,

Domain of a function f(x) is defined as set of those real values of x for which f(x) is well defined.

So,

$\rm :\longmapsto\:f(x) = \dfrac{x - 2}{3 - x} \: is \: defined \: when \: 3 - x \ne \: 0$

So, Domain of f(x) = R - {3}

or

$\boxed{ \bf \:Domain \: of \: f(x) \: is \: x \: \in \: ( - \infty ,3) \: \cup \: (3, \infty )}$

Now, we know

Range of a function f(x) is defined as set of those real values of f(x) which is taken for x from its domain.

To find the Range of a function following steps have to be followed :

Step :- 1 Put y = f(x)

Step :- 2 Solve this to get the value of x in terms of y.

Step :- 3 Now, find the values of y for which x is defined. The set of values of y provides the range of f(x).

Now, given function is

$\rm :\longmapsto\:f(x) = \dfrac{x - 2}{3 - x}$

So,

$\rm :\longmapsto\:y= \dfrac{x - 2}{3 - x}$

$\rm :\longmapsto\:3y - xy = x - 2$

$\rm :\longmapsto\:3y - 2= x + xy$

$\rm :\longmapsto\:3y - 2= x(1 + y)$

$\rm :\longmapsto\:x = \dfrac{3y - 2}{y + 1}$

So,

$\rm :\longmapsto\:x = \dfrac{3y - 2}{y + 1} \: is \: defined \: if \: y + 1 \ne \: 0$

$\boxed{ \bf \:Range \: of \: f(x) \: is \: y \: \in \: ( - \infty , - 1) \: \cup \: ( - 1, \infty )}$