# Problem of the Week Problem E and Solution Fine Line

## Problem

Suppose that $$f(x) = ax + b$$ and $$g(x) = f^{-1}(x)$$ for all values of $$x$$. That is, $$g$$ is the inverse of the function $$f$$.

If $$f(x) - g(x) = 2022$$ for all values of $$x$$, determine all possible values for $$a$$ and $$b$$.

## Solution

Since $$f(x) = ax+b$$, we can determine an expression for $$g(x) = f^{-1}(x)$$ by letting $$y = f(x)$$ to obtain $$y = ax + b$$. We then interchange $$x$$ and $$y$$ to obtain $$x = ay + b$$, which we solve for $$y$$ to obtain $$ay = x - b$$ or $$y = \dfrac{x}{a} - \dfrac{b}{a}$$.

Therefore, $$f^{-1}(x) = \dfrac{x}{a} - \dfrac{b}{a}$$. Note that $$a \neq 0$$. (This makes sense since the function $$f(x) = b$$ has a graph which is a horizontal line, and so cannot be invertible.)

Therefore, the equation $$f(x) - g(x) = 2022$$ becomes \begin{aligned} (ax + b) - \left( \dfrac{x}{a} - \dfrac{b}{a} \right) &= 2022\\ \left( a - \dfrac{1}{a} \right) x + \left(b + \dfrac{b}{a} \right)& = 2022 \end{aligned}

This is true for all $$x$$.

From here, we will present two approaches for determining the possible values for $$a$$ and $$b$$.

• Approach 1: Comparing coefficients

Since the equation $\left( a - \dfrac{1}{a} \right) x + \left(b + \dfrac{b}{a} \right) = 2022 = 0x + 2022$ is true for all $$x$$, then the coefficients of the linear expression on the left side must match the coefficients of the linear expression on the right side.

Therefore, $$a - \dfrac{1}{a} = 0$$ and $$b + \dfrac{b}{a} = 2022$$.

From the first of these equations, we obtain $$a = \dfrac{1}{a}$$ or $$a^2 = 1$$, which gives $$a = 1$$ or $$a = -1$$.

If $$a = 1$$, the equation $$b + \dfrac{b}{a} = 2022$$ becomes $$b + b = 2022$$, which gives $$b = 1011$$.

If $$a = -1$$ the equation $$b + \dfrac{b}{a} = 2022$$ becomes $$b - b = 2022$$ which is not possible.

Therefore, we must have $$a=1$$ and $$b = 1011$$, and so $$f(x) = x + 1011$$.

• Approach 2: Trying specific values for $$x$$

Since the equation $\left( a - \dfrac{1}{a} \right) x + \left(b + \dfrac{b}{a} \right) = 2022$ is true for all values of $$x$$, then it must be true for any specific values of $$x$$ that we choose.

Choosing $$x = b$$, we obtain \begin{align*} \left( a - \dfrac{1}{a} \right) b + \left(b + \dfrac{b}{a} \right) &= 2022\\ ab + b &= 2022 \tag{1} \end{align*}

Choosing $$x = 0$$, we obtain \begin{aligned} 0 + b + \dfrac{b}{a} &=2022\\ b + \dfrac{b}{a} &= 2022\\ \dfrac{ab+b}{a} &= 2022 \end{aligned} Then, substituting $$ab+b=2022$$ from equation $$(1)$$, we obtain \begin{aligned} \dfrac{ab + b}{a} &= 2022\\ \dfrac{2022}{a} &= 2022\\ a &=1\end{aligned}

Since $$a = 1$$, then $$ab + b = 2022$$ gives $$2b = 2022$$ or $$b = 1011$$.

Thus, $$f(x) = x + 1011$$.

In summary, the only possible values for $$a$$ and $$b$$ for which the given equation is true for all $$x$$ are $$a=1$$ and $$b=1011$$.