Is there a general rule for the inverse of the function $ g(x) = a f(x) $, where $a$ is a constant, assuming $f^{-1}(x)$ is defined?
Follow up: $g(x) = a f(x) + b$. Is the following correct, given $f^{-1}(x)$ is defined?
$g(x)=y$
$g^{-1}(x) = f^{-1}( \frac{y-b}{a} )$ $\:\:\:\: \leftarrow$ correct?
$$ \begin{align*} g(x) = y & \Leftrightarrow {} af(x) + b = y \\[3mm] &\Leftrightarrow a f(x) = y - b \\[2mm] &\Leftrightarrow f(x) = \frac{y-b}{a} \\[2mm] &\Leftrightarrow x = f^{-1} \Big( \frac{y-b}{a} \Big) \\ \end{align*} $$
So, $\displaystyle g^{-1}(y) = f^{-1} \Big( \frac{y-b}{a} \Big)$.
Another point of view : let $\varphi \, : \, x \, \longmapsto \, ax+b$ an affine function. If $a \neq 0$, $\varphi$ is a bijection. Therefore, it admits an inverse $\varphi^{-1}$, which is : $\displaystyle \varphi^{-1}(y) = \frac{y-b}{a}$. Now, let's assume that $a \neq 0$ and let $g = \varphi \circ f$ (the composition of $\varphi$ and $f$). Since both $f$ and $\varphi$ are bijections, $g$ is also a bijection and its inverse is : $g^{-1} = (\varphi \circ f)^{-1} = f^{-1} \circ \varphi^{-1}$. So,
$$ g^{-1}(y) = f^{-1} \Big( \varphi^{-1}(y) \Big) = f^{-1} \Big( \frac{y-b}{a} \Big) $$