1.

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»

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


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