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Question
A function f : R → R is defined as f(x) = x3 + 4. Is it a bijection or not? In case it is a bijection, find f−1 (3).
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Solution
Injectivity of f:
Let x and y be two elements of domain (R), such that
f (x) = f (y)
⇒ x3 + 4 = y3 + 4
⇒ 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.
⇒ x2 + 4 = y
⇒ x3 = y - 4
⇒ `x = 3sqrt( y-4) in R` (domain)
⇒ f is onto.
So, f is a bijection and, hence, is invertible.
Finding f -1:
Let f−1 (x) = y ...(1)
⇒ x = f (y)
⇒ x = y3+ 4
⇒ x − 4 = y3
⇒ `y = 3sqrt(x-4)`
So, f−1(x) = `3sqrt(x-4)` [from (1)]
`f^-1(3) = 3sqrt(3-4) = 3sqrt-1 = -1`
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