Find the Inverse f(x)=7(x-3)^(1/5)

Rewrite the equation as ${\displaystyle 7{(y-3)}^{\frac{1}{5}}=x}$.

Divide each term in ${\displaystyle 7{(y-3)}^{\frac{1}{5}}=x}$ by ${\displaystyle 7}$ and simplify.

Raise each side of the equation to the power of ${\displaystyle 5}$ to eliminate the fractional exponent on the left side.

Simplify the exponent.

Add ${\displaystyle 3}$ to both sides of the equation.

To verify the inverse, check if ${\displaystyle f^{-1}\left(f\left(x\right)\right)=x}$ and ${\displaystyle f\left(f^{-1}\left(x\right)\right)=x}$.

Evaluate ${\displaystyle f^{-1}\left(f\left(x\right)\right)}$.

Evaluate ${\displaystyle f\left(f^{-1}\left(x\right)\right)}$.

Since ${\displaystyle f^{-1}\left(f\left(x\right)\right)=x}$ and ${\displaystyle f\left(f^{-1}\left(x\right)\right)=x}$, then ${\displaystyle f^{-1}\left(x\right)=\frac{x^5}{16807}+3}$ is the inverse of ${\displaystyle f\left(x\right)=7{(x-3)}^{\frac{1}{5}}}$.