One-to-one function's inverse

Question

I've been trying to solve this question for a while and couldn't find the correct way. We're looking for the inverse of the given function $r$ in terms of $f^{-1}$, where $r$ is defined by: $$r(x) = 1 - 2f(3-4x). $$

I've tried setting $r(x)$ equal to $h(f(3-4x))$ but couldn't solve it. If you have any solution please let me know.

Answer 1

We have the functions:

• $x\mapsto 3-4x$ with inverse $x\mapsto \frac34-\frac14x$. Let's denote this function by $u$.
• $x\mapsto f(x)$ with inverse $x\mapsto f^{-1}(x)$
• $x\mapsto 1-2x$ with inverse $x\mapsto\frac12-\frac12x$. Let's denote this function by $v$.

Then $r=v\circ f\circ u$ so that $r^{-1}=u^{-1}\circ f^{-1}\circ v^{-1}$.

Work out the RHS.

Answer 2

So you want to solve $$ y = r(x) = 1 - 2f(3-4x) $$ for $x$. We start by isolating the $f$-term, we have $$ y = 1 - 2f(3-4x) \iff \frac{1-y}2 = f(3-4x) $$ Now apply $f^{-1}$ to both sides $$ \frac{1-y}2 = f(3-4x) \iff 3-4x = f^{-1}\left(\frac{1-y}2\right) $$ and isolate $x$, $$ 3- 4x = f^{-1}\left(\frac{1-y}2\right) \iff x =\frac 34 - \frac 14 f^{-1}\left(\frac{1-y}2\right) $$ so $$ r^{-1}(y) = \frac 34 - \frac 14 f^{-1}\left(\frac{1-y}2\right) $$

Answer 3

$$ r^{-1}(y)=\frac{3-f^{-1}(\tfrac{y-1}{2}{}))}{4} $$

As a hint, substitute $y=r(x)$ and find $x$ as a function of $y$

Answer 4

To find the inverse of $r$ we let $r(x)=y$ then switch $x$ and $y$. This gives \begin{align*} x&=-1-2f(3-4y)\\ -\frac{x+1}{2}&=f(3-4y)\\ f^{-1}\left(-\frac{x+1}{2}\right)&=3-4y\\ y&=\frac{3-f^{-1}\left(-\frac{x+1}{2}\right)}{4}. \end{align*} so that $$r^{-1}(x)=\frac{3-f^{-1}\left(-\frac{x+1}{2}\right)}{4}.$$

Answer 5

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

Its inverse function is

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

Just as

$$ y = x^2/2 \rightarrow x = y^2/2. $$

Swapped re-symbolization can follow.