Question

If f : R → R be defined by f(x) = x³ −3, then prove that f⁻¹ exists and find a formula for f⁻¹. Hence, find f⁻¹ (24) and f⁻¹ (5).

Answer 1

Injectivity of f :
Let x and y be two elements in domain (R),

$$\begin{gathered} \mathrm{such}\mathrm{that},x^3-3=y^3-3 \\ \Rightarrow x^3=y^3 \\ \Rightarrow x=y \end{gathered}$$
So, f is one-one.

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

$$\begin{gathered} \Rightarrow x^3-3=y \\ \Rightarrow x^3=y+3 \\ \Rightarrow x=\sqrt[3]{y+3}\in R \end{gathered}$$

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

Finding f ^-1:
$$\begin{gathered} \text{Let}{f}^{-1}\left(x\right)=y...\left(1\right) \\ \Rightarrow x=f\left(y\right) \\ \Rightarrow x=y^3-3 \\ \Rightarrow x+3=y^3 \\ \Rightarrow y=\sqrt[3]{x+3}=f^{-1}\left(x\right)[\text{from}\left(1\right)] \\ \text{So,}{f}^{-1}\left(x\right)=\sqrt[3]{x+3} \\ \text{Now,}{f}^{-1}\left(24\right)=\sqrt[3]{24+3}=\sqrt[3]{27}=\sqrt[3]{3^3}=3 \\ \mathrm{and}{f}^{-1}\left(5\right)=\sqrt[3]{5+3}=\sqrt[3]{8}=\sqrt[3]{2^3}=2 \end{gathered}$$