replace f(x) with y switch x and y solve for y replace y with f⁻¹(x)
so I'm not sure if it is [tex]f(x)=\sqrt[3]{\frac{x}{7}}-9[/tex] or [tex]f(x)=\sqrt[3]{\frac{x}{7}-9}[/tex]
first one [tex]f(x)=\sqrt[3]{\frac{x}{7}}-9[/tex] replace f(x) with y [tex]y=\sqrt[3]{\frac{x}{7}}-9[/tex] switch x and y [tex]x=\sqrt[3]{\frac{y}{7}}-9[/tex] solve for y [tex]x+9=\sqrt[3]{\frac{y}{7}}[/tex] [tex](x+9)^3=\frac{y}{7}[/tex] [tex]7(x+9)^3=y[/tex] replace y with f⁻¹(x) [tex]f^{-1}(x)=7(x+9)^3[/tex]
2nd one [tex]f(x)=\sqrt[3]{\frac{x}{7}-9}[/tex] replce f(x) with y [tex]y=\sqrt[3]{\frac{x}{7}-9}[/tex] switch x and y [tex]x=\sqrt[3]{\frac{y}{7}-9}[/tex] solve for y [tex]x^3=\frac{y}{7}-9[/tex] [tex]x^3+9=\frac{y}{7}[/tex] [tex]7(x^3+9)=y[/tex] replace y with f⁻¹(x) [tex]f^{-1}(x)=7(x^3+9)[/tex]
if you meant [tex]f(x)=\sqrt[3]{\frac{x}{7}}-9[/tex] then the inverse is [tex]f^{-1}(x)=7(x+9)^3[/tex]
if you meant [tex]f(x)=\sqrt[3]{\frac{x}{7}-9}[/tex] then the inverse is [tex]f^{-1}(x)=7(x^3+9)[/tex]
