Show that the function f on A = R – {2/3}, defined as f(x) = (4x + 3

1.	Show that the function f on A = R – {2/3}, defined as f(x) = (4x + 3)/ (6x – 4) in one-one and onto. Hence, find f -1.
Answer» One-one f: Consider x₁ and x₂ ∈ dom (f) We know that f(x₁) = f(x₂) It can be written as (4x₁ + 3)/ (6x₁ – 4) = (4x₂ + 3)/ (6x₂ – 4) So we get (4x₁ + 3) (6x₂ – 4) = (4x₂ + 3) (6x₁ – 4) On further calculation 24 x₁ x₂ – 16x₁ + 18 x₂ – 12 = 24 x₁ x₂ – 16x₂ + 18x₁ – 12 We get 34 x₁ = 34 x₂where x₁ = x₂ f is one-one Onto f: Consider y ∈ co domain (f) We know that y = f(x) y = (4x + 3)/ (6x – 4) On further calculation 6xy – 4y = 4x + 3 So we get 6xy – 4x = 3 + 4y It can be written as x (6y – 4) = 3 + 4y So x = (3 + 4y)/ (6y – 4) ∈ domain Ɐ y ∈ co-domain f is an onto function Here, x = (3 + 4y)/ (6y – 4) where y ≠ 2/3 We get f ^-1 (y) = (3 + 4y)/ (6y – 4) where y ≠ 2/3

Show that the function f on A = R – {2/3}, defined as f(x) = (4x + 3)/ (6x – 4) in one-one and onto. Hence, find f -1.

Answer»

One-one f:

Consider x₁ and x₂ ∈ dom (f)

We know that

f(x₁) = f(x₂)

It can be written as

(4x₁ + 3)/ (6x₁ – 4) = (4x₂ + 3)/ (6x₂ – 4)

So we get

(4x₁ + 3) (6x₂ – 4) = (4x₂ + 3) (6x₁ – 4)

On further calculation

24 x₁ x₂ – 16x₁ + 18 x₂ – 12 = 24 x₁ x₂ – 16x₂ + 18x₁ – 12

We get

34 x₁ = 34 x₂where x₁ = x₂

f is one-one

Onto f:

Consider y ∈ co domain (f)

We know that y = f(x)

y = (4x + 3)/ (6x – 4)

On further calculation

6xy – 4y = 4x + 3

So we get

6xy – 4x = 3 + 4y

It can be written as

x (6y – 4) = 3 + 4y

So x = (3 + 4y)/ (6y – 4) ∈ domain Ɐ y ∈ co-domain

f is an onto function

Here, x = (3 + 4y)/ (6y – 4) where y ≠ 2/3

We get

f ^-1 (y) = (3 + 4y)/ (6y – 4) where y ≠ 2/3