**Injectivity:**

**Consider x, y ∈ R where f(x) = f(y)**

We get

x^{5} = y^{5}

It can be written as

x^{5} – y^{5} = 0

**By multiplying and dividing by 2 on both sides**

(x^{5/2})^{2} – (y^{5/2})^{2} = 0

We know that

(x^{5/2} + y^{5/2}) (x^{5/2} – y^{5/2}) = 0

So we get

x^{5/2} – y^{5/2} = 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^{5 }= y

So we get

x^{5} – y = 0

Odd degree equation has one real root

**Thus, for every real value of y**

x^{5} – 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.**