Question

If f: R → R be defined by f(x) = x³ − 3, then prove that f⁻¹ exists and find a formula for f⁻¹. Hence, find f⁻¹(24) and f⁻¹(5)

Answer 1

Given function f: R → R be defined by f(x) = x³ − 3

Let us prove that f⁻¹ exists

Injectivity of f:

Let x and y be two elements in domain (R),

Such that, x³ − 3 = y³ − 3            

⇒ x³ = y³        

⇒ x = y

Therefore, f is one-one.

Surjectivity of f:

Let y be in the co-domain (R)

Such that f(x) = y

⇒ x³ – 3 = y

⇒ x³ = y + 3 

⇒ x = ³√(y + 3) in R

⇒ f is onto.

Therefore, f is a bijection and, hence, it is invertible.

Find f ^-1

Let f^-1(x) = y ...(1)

⇒ x = f(y)

⇒ x = y³ − 3

⇒ x + 3 = y³

⇒ y = ³√(x + 3) = f^-1(x) [from (1)]

Therefore, f^-1(x) = ³√(x + 3)

Now, f^-1(24) = ³√(24 + 3)

= ³√27

= ³√3³

= 3

And f^-1(5) = ³√(5 + 3)

= ³√8

= ³√2³

= 2