Question

If f : R → R is defined by f(x) = 6 – (x – 9)³, then f^–1(x) = ___________.

Answer 1

Given: A function f : R → R is defined by f(x) = 6 – (x – 9)³

$$\begin{gathered} f\left(x\right)=6-{\left(x-9\right)}^3 \\ \Rightarrow y=6-{\left(x-9\right)}^3 \\ \Rightarrow {\left(x-9\right)}^3=6-y \\ \Rightarrow x-9={\left(6-y\right)}^{\frac{1}{3}} \\ \Rightarrow x=9+{\left(6-y\right)}^{\frac{1}{3}} \\ \mathrm{Thus},f^{-1}\left(x\right)=9+{\left(6-x\right)}^{\frac{1}{3}} \end{gathered}$$

Hence, if f : R → R is defined by f(x) = 6 – (x – 9)³, then f⁻¹(x) = $\underline{9+{\left(6-x\right)}^{\frac{1}{3}}}$.