Problem E and Solution

Fine Line

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 \(b\).

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 coefficientsSince 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 \(b=1011\).