Question

If f : R → R be the function defined by f(x) = 4x³ + 7, show that f is a bijection.

Answer 1

Injectivity:
Let x and y be any two elements in the domain (R), such that f(x) = f(y)
$$\begin{gathered} \Rightarrow 4x^3+7=4y^3+7 \\ \Rightarrow 4x^3=4y^3 \\ \Rightarrow x^3=y^3 \\ \Rightarrow x=y \end{gathered}$$
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
$$\begin{gathered} \Rightarrow 4x^3+7=y \\ \Rightarrow 4x^3=y-7 \\ \Rightarrow x^3=\frac{y-7}{4} \\ \Rightarrow x=\sqrt[3]{\frac{y-7}{4}}\in R \end{gathered}$$
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.