Finding inverse functions (ACT). Game on.

Set $y=x^{3}-5$. Swap: $x=y^{3}-5$. Add $5$: $x+5=y^{3}$. Take the cube root: $y=\sqrt[3]{x+5}$. So $f^{-1}(x)=\sqrt[3]{x+5}\textsf{.}$

Set $y=\sqrt[3]{4x+1}$. Swap: $x=\sqrt[3]{4y+1}$. Cube both sides: $x^{3}=4y+1$. Subtract $1$ and divide by $4$: $y=\dfrac{x^{3}-1}{4}$. So $f^{-1}(x)=\dfrac{x^{3}-1}{4}\textsf{.}$

Set $y=\sqrt{x+2}$. Swap: $x=\sqrt{y+2}$. Square both sides: $x^{2}=y+2$. Subtract $2$: $y=x^{2}-2$. So $f^{-1}(x)=x^{2}-2\textsf{.}$

Set $y=\dfrac{3}{x-1}$. Swap: $x=\dfrac{3}{y-1}$. Multiply: $x(y-1)=3$. Divide by $x$: $y-1=\dfrac{3}{x}$. Add $1$: $y=\dfrac{3}{x}+1$. So $f^{-1}(x)=\dfrac{3}{x}+1\textsf{.}$

Set $y=2x^{3}+1$. Swap: $x=2y^{3}+1$. Subtract $1$: $x-1=2y^{3}$. Divide by $2$: $\dfrac{x-1}{2}=y^{3}$. Take the cube root: $y=\sqrt[3]{\dfrac{x-1}{2}}$. So $f^{-1}(x)=\sqrt[3]{\dfrac{x-1}{2}}\textsf{.}$

Set $y=7-x$. Swap: $x=7-y$. Solve for $y$: $y=7-x$. So $f^{-1}(x)=7-x\textsf{.}$

Set $y=\dfrac{2x+3}{5}$. Swap: $x=\dfrac{2y+3}{5}$. Multiply by $5$: $5x=2y+3$. Subtract $3$: $5x-3=2y$. Divide by $2$: $y=\dfrac{5x-3}{2}$. So $f^{-1}(x)=\dfrac{5x-3}{2}\textsf{.}$

Set $y=(x-4)^{3}$. Swap: $x=(y-4)^{3}$. Take the cube root: $\sqrt[3]{x}=y-4$. Add $4$: $y=\sqrt[3]{x}+4$. So $f^{-1}(x)=\sqrt[3]{x}+4\textsf{.}$