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Problem of the Week
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 \(b\).

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