Question

If f : R → R, such that f(x) = x³ – 3, then prove that f^-1 exists and find its formula. Thus, find f^-1(24) and f^-1(5).

Answer 1

Given, f : R → R, f(x) = x³ – 3

One-one/many-one:

Let a, b ∈ R

∴ f(a) = f(b)

⇒ a³ – 3 = b³ – 3

⇒ a³ = b³

⇒ a = b

Hence, . f(a) = f(b) ⇒ a = b

∴ f is a one-one function.

Onto/into:

Let y ∈ R (Co-domain)

f(x) = y

⇒ x³ – 3 = y

⇒ x= (y + 3)^1/3 ∈ R, ∀ y ∈ R

Here, for each value of y, x exists in domain R.

Thus, Range of F= co-domain of f.

So, ‘f’ is onto function.

It is clear that ‘f is one-one onto function.

Hence, f^-1: R → R exists.

f^-1(y) = x ⇒ f(x) = y

But f(x)= x³ – 3

∴ x³ = 3 = y

⇒ x³ = y + 3

⇒ x = (y + 3)^1/3

⇒ f^-1(y) = (x + 3)^1/3

⇒ f^-1(x) = (x + 3)^1/3, ∀ X ∈ R

For x = 24

∴ f^-1(24)= (24 + 3)^1/3

= (27)^1/3

= 3^{3 x 1/3} = 3

For x = 5

f^-1(5) = (5 + 3)^1/3

= (8)^1/3

= 2^{3 x 1/3} = 2