f(x) =

4x+3

6x−4

Let f (x1) = f (x2)
⇒

4x1+3

6x1−4

=

4x2+3

6x2−4

⇒ 24 x1x2 - 16 x1 + 18 x2 - 12 = 24 x1x2 + 18 x1 - 16 x2 - 12
⇒ 18 x2 + 16 x2 = 18 x1 + 16 x1
⇒ 34 x2 = 34 x1
Since,

4x1+3

6x1−4

is a real number, therefore, for every y in the co-domain of f, there exists a number x in R - {

2

3

} such that f(x) = y =

4x1+3

6x1−4

Therefore, f(x) is onto.
Hence, f−1 exists.
Now, let y =

4x1+3

6x1−4

⇒ 6xy - 4y = 4x + 3
⇒ 6xy - 4xx = 4y + 3
⇒ x (6y - 4) = 4y+ 3
⇒ x =

4y+3

6y−4

⇒ y =

4x+3

6x−4

int exchanging the variables x and y
⇒ f−1 (x) =

4x+3

6x−4

[put y = f−1 x]