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Show that the function f: R → R: f(x) = x5 is one-one and onto.

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Injectivity:

Consider x, y ∈ R where f(x) = f(y)

We get

x5 = y5

It can be written as

x5 – y5 = 0

By multiplying and dividing by 2 on both sides

(x5/2)2 – (y5/2)2 = 0

We know that

(x5/2 + y5/2) (x5/2 – y5/2) = 0

So we get

x5/2 – y5/2 = 0

where x = y

Hence, f(x) = f(y) is x = y for all x, y ∈ R

f is injective.

Surjectivity:

Consider y as an arbitrary element of R

We know that

f(x) = y

It can be written as

x= y

So we get

x5 – y = 0

Odd degree equation has one real root

Thus, for every real value of y

x5 – y = 0 has real root α where

α 5 – y = 0

So α 5 = y

Thus, f(α) = y

For every y ∈ R there exists α ∈ R where f(α) = y

f is surjective

Thus, f: R → R is bijective.

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