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
= 33 x 1/3 = 3
For x = 5
f-1(5) = (5 + 3)1/3
= (8)1/3
= 23 x 1/3 = 2