to solve
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]
