Question -
Answer -
Given thatf: R → R is defined as f(x) = x3 +4
Injectivity of f:
Let x and y be two elements of domain (R),
Such thatf (x) = f (y)
⇒ x3 + 4 = y3 + 4
⇒ x3 = y3
⇒ x = y
So, f isone-one.
Surjectivityof f:
Let y be in the co-domain (R),
Such that f(x) =y.
⇒ x3 +4 = y
⇒ x3 =y – 4
⇒ x = ∛ (y – 4) in R (domain)
⇒ f is onto.
So, f is a bijection and, hence, it is invertible.
Finding f-1:
Let f−1 (x) = y……(1)
⇒ x = f (y)
⇒ x = y3 + 4
⇒ x − 4 = y3
⇒ y =∛ (x-4)
So, f-1(x) =∛ (x-4) [from (1)]
f-1 (3)= ∛(3 – 4)
= ∛-1
= -1