If f : R → R be the function defined by

Solution:

Injectivity:
Let x and y be any two elements in the domain (R), such that f(x) = f(y)

$\Rightarrow 4 x^{3}+7=4 y^{3}+7$

$\Rightarrow 4 x^{3}=4 y^{3}$

$\Rightarrow x^{3}=y^{3}$

$\Rightarrow x=y$

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).

$f(x)=y$

$\Rightarrow 4 x^{3}+7=y$

$\Rightarrow 4 x^{3}=y-7$

$\Rightarrow x^{3}=\frac{y-7}{4}$

$\Rightarrow x=\sqrt[3]{\frac{y-7}{4}} \in R$

So, for every element in the co-domain, there exists some pre-image in the domain.

$\Rightarrow f$ is onto.

Since, $f$ is both one-to-one and onto, it is a bijection.