Functions-22

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Show that the function f : R → R defined by f(x) = 3x3 + 5 for all x ∈ R is a bijection.

Solution: Injectivity : Let x, y be any two elements of R(domain).

Then, f(x) = f(y)

=> 3x3 + 5 = 3y3 + 5

=> 3x3 = 3y3

=>x = y

Thus, f(x) = f(y) => x = y for all x,y ∈ R.

So, f is an injective map.

Surjectivity: Let y be an arbitrary element of R(co-domain). Then,

f(x) = y => 3x3 + 5 = y

=>x3 = (y - 5)/3

=> x = [(y - 5)/3]1/3

Thus, we find that for all y ∈ R(co-domain) there exists x = [(y - 5)/3]1/3 ∈ R (domain) such that

f(x) = f{(y – 5)/3]1/3} = 3{{[(y – 5)/3]1/3}3} + 5 = y – 5 + 5 = y

This shows that every element in the co-domain has its pre-image in the domain So, f is a surjection. Hence, f is a bijection.

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