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Homework Statement
If ##f\left(x\right) = x^2 + 2x + 2##, find two functions ##g## for which ##\left(f \circ g\right)\left(x\right) = x^2 - 4x + 5##.
Homework Equations
If ##f\left(x\right) = x^2 + 2x + 2##, then ##\left(f \circ g\right)\left(x\right) = g\left(x\right)^2 + 2g\left(x\right) + 2##, so ##g\left(x\right)^2 + 2g\left(x\right) + 2 = x^2 - 4x + 5##
The Attempt at a Solution
By solving for ##g\left(x\right)##,
\begin{align*}
g\left(x\right)^2 + 2g\left(x\right) + 2 - 2&= x^2 - 4x + 5 - 2 \\
g\left(x\right)^2 + 2g\left(x\right) &= x^2 - 4x + 3 \\
g\left(x\right)\left[g\left(x\right) + 2\right] &= \left(x - 3\right)\left(x - 1\right)
\end{align*}
From here ##\left(x - 3\right)## is a solution, but ##\left(x - 1\right)## is not. Now, according to the answers provided, another function ##g## is ##1 - x##, which is the negative of ##x - 1##. How do you arrive at this solution?
I know that it is possible to let ##\left(x - 3\right)\left(x - 1\right) = -\left(x - 3\right)\left(1 - x\right)##, but what is the intuition behind it?