Relationship between the inverses of two functions

Question

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.

Answer 1

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 $$ f^{-1}(x)=g^{-1}(ax+b) $$

Answer 2

Letting $y = g(x)$ $$y = af(g^{-1}(y))+b$$ $$f^{-1}\left(\frac{y-b}{a}\right) = g^{-1}(y)$$ and so $$f^{-1}(y) = g^{-1}(ay+b)$$

Answer 3

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.