Finding inverse functions (ACT). Game on.

Set $y = 4x + 12\textsf{.}$ Swap $x$ and $y\textsf{:}$ $x = 4y + 12\textsf{.}$ Subtract $12$ and divide by $4\textsf{:}$ $y = \dfrac{x - 12}{4}\textsf{.}$ So $f^{-1}(x) = \dfrac{x - 12}{4}\textsf{.}$

Set $y = \dfrac{x}{2} - 5\textsf{.}$ Swap: $x = \dfrac{y}{2} - 5\textsf{.}$ Add $5$ and multiply by $2\textsf{:}$ $y = 2x + 10\textsf{.}$ So $f^{-1}(x) = 2x + 10\textsf{.}$

Set $y = 1 - 3x\textsf{.}$ Swap: $x = 1 - 3y\textsf{.}$ Then $3y = 1 - x\textsf{,}$ so $y = \dfrac{1 - x}{3}\textsf{.}$ So $f^{-1}(x) = \dfrac{1 - x}{3}\textsf{.}$

Set $y = \dfrac{2x + 6}{5}\textsf{.}$ Swap: $x = \dfrac{2y + 6}{5}\textsf{.}$ Multiply by $5\textsf{,}$ subtract $6\textsf{,}$ and divide by $2\textsf{:}$ $y = \dfrac{5x - 6}{2}\textsf{.}$ So $f^{-1}(x) = \dfrac{5x - 6}{2}\textsf{.}$

Set $y = 5x - 1\textsf{.}$ Swap: $x = 5y - 1\textsf{.}$ Add $1$ and divide by $5\textsf{:}$ $y = \dfrac{x + 1}{5}\textsf{.}$ So $f^{-1}(x) = \dfrac{x + 1}{5}\textsf{.}$

Set $y = \dfrac{x - 4}{3}\textsf{.}$ Swap: $x = \dfrac{y - 4}{3}\textsf{.}$ Multiply by $3$ and add $4\textsf{:}$ $y = 3x + 4\textsf{.}$ So $f^{-1}(x) = 3x + 4\textsf{.}$

Set $y = 6 - x\textsf{.}$ Swap: $x = 6 - y\textsf{.}$ Then $y = 6 - x\textsf{.}$ So $f^{-1}(x) = 6 - x\textsf{.}$

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 $f^{-1}(x)=\sqrt[3]{x+5}\textsf{.}$