Relationship between the inverses of two functions Suppose we are given two functions $g(x)$ and $f(x)$ where $f(x)$ is monotonous, such that
$$g(x) = af(x) + b,$$
with $a,b$ constant. I was wondering if there is any relationship between their inverses.
 A: Letting $y = g(x)$
$$y = af(g^{-1}(y))+b$$
$$f^{-1}\left(\frac{y-b}{a}\right) = g^{-1}(y)$$
and so
$$f^{-1}(y) = g^{-1}(ay+b)$$
A: We have
$$
af(x)+b=g(x)=y \iff x=g^{-1}(y),
$$
and
$$
f(x)=\dfrac{g(x)-b}{a}=z \iff x=f^{-1}(z)
$$
Hence
$$
g^{-1}(y)=f^{-1}\left(\dfrac{g(x)-b}{a}\right)=f^{-1}\left(\dfrac{y-b}{a}\right).
$$
The relationship between $f^{-1}$ and $g^{-1}$ is therefore
$$
g^{-1}(x)=f^{-1}\left(\dfrac{x-b}{a}\right)
$$
or equivalently
$$
f^{-1}(x)=g^{-1}(ax+b)
$$
A: Yes. Your function $g$ is the composition $h \circ f$, where $h(x) = ax + b$ is a linear (or affine if you're picky) function. 
It just so happens that for invertible functions $f, h$, we have $(h \circ f)^{-1} = f^{-1} \circ h^{-1}$. In this case, $h^{-1}(x) = \frac{x - b}{a}$, so $g^{-1}(x) = f^{-1}\left(\frac{x-b}{a}\right)$.
But you can arrive at the same conclusion using the usual procedure for finding the inverse of a function. If we have $y = g(x) = af(x) + b$, switch $x$ and $y$ then solve for $y$:
\begin{align*}
x &= af(y) + b \\
x - b &= af(y) \\
\frac{x-b}{a} &= f(y) \\
f^{-1}\Big(\frac{x - b}{a}\Big) &= y = g^{-1}(x)
\end{align*} 
where the last line is the result of applying $f^{-1}$ to both sides of the preceding equation.
