12) To answer this question, we will set the equation y=f(x), solve for x, and then exchange x and y.
Setting y=f(x) we get:
[tex]y=9+2\sqrt[3]{x}.[/tex]
Subtracting 9 from the above equation we get:
[tex]\begin{gathered} y-9=9+2\sqrt[3]{x}-9, \\ y-9=2\sqrt[3]{x}. \end{gathered}[/tex]
Dividing the above equation by 2 we get:
[tex]\begin{gathered} \frac{y-9}{2}=\frac{2\sqrt[3]{x}}{2}, \\ \frac{y-9}{2}=\sqrt[3]{x}. \end{gathered}[/tex]
Taking the above equation to the power of 3 we get:
[tex]\begin{gathered} (\frac{y-9}{2})^3=(\sqrt[3]{x})^3, \\ (\frac{y-9}{2})^3=x. \end{gathered}[/tex]
Exchanging x and y we get:
[tex](\frac{x-9}{2})^3=y.[/tex]
Therefore:
[tex]f^{-1}(x)=(\frac{x-9}{2})^3.[/tex]
Answer:
[tex]f^{-1}(x)=(\frac{x-9}{2})^3.[/tex]
