Find the inverse of the function $r(x)=1-2f(3-4x)$ in terms of $f^{-1}$

Question

I have absolutely no idea to inverse functions containing different functions. Apparently this is a one-to-one function with inverse $f^{-1}$ and I'm asked to calculate the inverses of the given functions in terms of $f^{-1}$

Answer 1

Let $y=r^{-1}(x)$

Then $r(y)=x$

So $1-2f(3-4y)=x$

Rearrange to make $y$ the subject...

$1-x=2f(3-4y)$

$\frac{1-x}2=f(3-4y)$

$f^{-1} \left (\frac{1-x}2 \right )=3-4y$

$4y=3-f^{-1} \left (\frac{1-x}2 \right )$

$y=\frac 34- \frac 14 f^{-1} \left (\frac{1-x}2 \right )$

$r^{-1}(x)=\frac 34- \frac 14 f^{-1} \left (\frac{1-x}2 \right )$

Answer 2

Let $g(x)=3-4x$ and $h(x)=1-2x$

Note that both $g$ and $h$ are one to one and invertible over the real numbers.

Note further that $r(x)= (h\circ f\circ g)(x)$

By properties of composition of invertible functions, $r^{-1}=(h\circ f\circ g)^{-1}=g^{-1}\circ f^{-1}\circ h^{-1}$

We can find $h^{-1}$ and $g^{-1}$ and simplify leaving an expression for $r^{-1}$ that depends only on $x$ and $f^{-1}$