Find the inverse of the function f(x) = 5x – 8, where x ∈ R.

1.	Find the inverse of the function f(x) = 5x – 8, where x ∈ R.
Answer» Given: f (x) = 5x – 8, x ∈ R. For a function to be invertible, it should be a bijection, i.e., one-one onto function For all x₁, x₂ ∈ R, f (x₁) = f (x₂) ⇒ 5x₁ – 8 = 5x₂ – 8 ⇒ x₁ = x₂ ⇒ f (x) is one-one. Let y = f (x) = 5x – 8 ⇒ 5x = y + 8 ⇒ x = \(\frac{y+8}{5}\) Thus, for all y ∈ R, x = \(\frac{y+8}{5}\) ∈ R ⇒ f (x) is onto. f being one-one onto ⇒ f is invertible. Let y = f (x) = 5x – 8 ⇒ x = \(\frac{y+8}{5}\) ⇒ f^–1(y) = \(\frac{y+8}{5}\) ⇒ f^–1(x) = \(\frac{x+8}{5}\)

Answer»

Given:

f (x) = 5x – 8, x ∈ R.

For a function to be invertible, it should be a bijection, i.e., one-one onto function

For all x₁, x₂ ∈ R, f (x₁) = f (x₂) ⇒ 5x₁ – 8 = 5x₂ – 8 ⇒ x₁ = x₂ ⇒ f (x) is one-one.

Let y = f (x) = 5x – 8 ⇒ 5x = y + 8 ⇒ x = \(\frac{y+8}{5}\)

Thus, for all y ∈ R, x = \(\frac{y+8}{5}\) ∈ R

⇒ f (x) is onto.

f being one-one onto ⇒ f is invertible.

Let y = f (x) = 5x – 8

⇒ x = \(\frac{y+8}{5}\)

⇒ f^–1(y) = \(\frac{y+8}{5}\)

⇒ f^–1(x) = \(\frac{x+8}{5}\)

Discussion