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Question
If f : R → R be the function defined by f(x) = 4x3 + 7, show that f is a bijection.
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Solution
Injectivity:
Let x and y be any two elements in the domain (R), such that f(x) = f(y)
⇒ 4x3+7 = 4y3+ 7
⇒ 4x3= 4y3
⇒ x3= y3
⇒ x = y
So, f is one-one.
Surjectivity:
Let y be any element in the co-domain (R), such that f(x) = y for some element x in R (domain)
f(x) = y
⇒ 4x3+7 = y
⇒ 4x3= y −7
⇒ `x^3 = (y - 7)/4`
⇒ `x = 3sqrt(y-7)/4 in R`
So, for every element in the co-domain, there exists some pre-image in the domain. - f is onto.
Since, f is both one-to-one and onto, it is a bijection.
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