# Functions-22

### From Homeworkwiki

**Show that the function f : R → R defined by f(x) = 3x ^{3} + 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)

=> 3x^{3} + 5 = 3y^{3} + 5

=> 3x^{3} = 3y^{3}

=>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 => 3x^{3} + 5 = y

=>x^{3} = (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.