Question

A function f : R → R is defined as f(x) = x³ + 4. Is it a bijection or not? In case it is a bijection, find f⁻¹ (3).

Answer 1

Injectivity of f:
Let x and y be two elements of domain (R), such that
$$\begin{gathered} f\left(x\right)=f\left(y\right) \\ \Rightarrow x^3+4=y^3+4 \\ \Rightarrow x^3=y^3 \\ \Rightarrow x=y \end{gathered}$$
So, f is one-one.

Surjectivity of f:
Let y be in the co-domain (R), such that f(x) = y.

$$\begin{gathered} \Rightarrow x^3+4=y \\ \Rightarrow x^3=y-4 \\ \Rightarrow x=\sqrt[3]{y-4}\in R\left(\text{domain}\right) \end{gathered}$$

$\Rightarrow$ f is onto.
So, f is a bijection and, hence, is invertible.

Finding f ^-1:
$$\begin{gathered} \text{Let}{f}^{-1}\left(x\right)=y...\left(1\right) \\ \Rightarrow x=f\left(y\right) \\ \Rightarrow x=y^3+4 \\ \Rightarrow x-4=y^3 \\ \Rightarrow y=\sqrt[3]{x-4} \\ \text{So,}{f}^{-1}\left(x\right)=\sqrt[3]{x-4}[\text{from}\left(1\right)] \\ {f}^{-1}\left(3\right)=\sqrt[3]{3-4}=\sqrt[3]{-1}=-1 \end{gathered}$$