Question

If f : C → C is defined by f(x) = 8x³, then f^–1(8) = . _________.

Answer 1

Given: f(x) = 8x³

$$\begin{gathered} f\left(x\right)=8x^3 \\ \Rightarrow y=8x^3 \\ \Rightarrow x^3=\frac{y}{8} \\ \Rightarrow x={\left(\frac{y}{8}\right)}^{\frac{1}{3}} \\ \mathrm{Thus},f^{-1}\left(x\right)={\left(\frac{x}{8}\right)}^{\frac{1}{3}} \\ {f}^{-1}\left(8\right)={\left(\frac{8}{8}\right)}^{\frac{1}{3}} \\ =1^{\frac{1}{3}} \\ =1,\omega ,{\omega }^2\left(\because f:C\to C\right) \\ \mathrm{where},\omega \mathrm{is}\mathrm{the}\mathrm{cube}\mathrm{root}\mathrm{of}\mathrm{unity}. \end{gathered}$$

Hence, if f : C → C is defined by f(x) = 8x³, then f⁻¹(8) = $\underline{1,\omega ,{\omega }^2}$.