Let f: R → R: f(x) = ½ (3x + 1). Show that f is invertible and find

1.	Let f: R → R: f(x) = ½ (3x + 1). Show that f is invertible and find f -1.
Answer» We know that f(x₁) = f(x₂) It can be written as ½ (3x₁ + 1) = ½ (3x₂ + 1) So we get 3x₁ + 1 = 3x₂ + 1 On further calculation 3x₁ = 3x₂ where x₁ = x₂ f is one-one. Take y = ½ (3x + 1) It can be written as 2y = 3x + 1 We get 2y – 1 = 3x So x = (2y – 1)/3 If y ∈ R there exists x = (2y – 1)/3 ∈ R We know that f(x) = f([2y – 1]/ 3) = ½ (3([2y – 1]/ 3) + 1) = y f is onto Here, f is one-one and invertible. Take y = f(x) So y = (3x + 1)/2 It can be written as x = (2y – 1)/ 3 Hence, we define f ^-1: R → R: f ^-1(y) = (2y – 1)/ 3 for all y ∈ R

Let f: R → R: f(x) = ½ (3x + 1). Show that f is invertible and find f -1.

Answer»

We know that

f(x₁) = f(x₂)

It can be written as

½ (3x₁ + 1) = ½ (3x₂ + 1)

So we get

3x₁ + 1 = 3x₂ + 1

On further calculation

3x₁ = 3x₂ where x₁ = x₂

f is one-one.

Take y = ½ (3x + 1)

It can be written as

2y = 3x + 1

We get

2y – 1 = 3x

So x = (2y – 1)/3

If y ∈ R there exists x = (2y – 1)/3 ∈ R

We know that

f(x) = f([2y – 1]/ 3) = ½ (3([2y – 1]/ 3) + 1) = y

f is onto

Here, f is one-one and invertible.

Take y = f(x)

So y = (3x + 1)/2

It can be written as

x = (2y – 1)/ 3

Hence, we define f ^-1: R → R: f ^-1(y) = (2y – 1)/ 3 for all y ∈ R