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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).
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Solution
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 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 = 3sqrt(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 = 3sqrt(x+3) = f^-1 (x)` [from (1)]
`So, f^-1 (x) = 3sqrt(x+3)`
Now`, f^1 (24) = 3sqrt(24+3) = 3sqrt27 = 3sqrt3^3 =3`
and `f^-1 (5) = 3sqrt(5+3) = 3 sqrt8 = 3sqrt2^3 =2`
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