Question -
Answer -
Givenf: R → R be defined by f(x) = x3 −3
Now we have to prove thatf−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
So, f isone-one.
Surjectivityof f:
Let y be in the co-domain (R)
Such that f(x) =y
⇒ x3 –3 = y
⇒ x3 =y + 3
⇒ x = ∛(y+3) in R
⇒ 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 −3
⇒ x + 3 = y3
⇒ y = ∛(x + 3) = f-1(x) [from (1)]
So, f-1(x)= ∛(x + 3)
Now, f-1(24)= ∛ (24 + 3)
= ∛27
= ∛33
= 3
And f-1(5)=∛ (5 + 3)
= ∛8
= ∛23
= 2