Curiosity about why this is not a Function composition

[ecastro]

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?

 

[RPinPA]

Well, you have $g(x) \left [ g(x) + 2 \right ] = (x - 3)(x - 1)$ but you could flip the signs and also write $g(x) \left [ g(x) + 2 \right ] = (3 - x)(1 - x)$ and since $(3 - x)$ = $(1 - x) + 2$ you can then identify $g(x)$ with $(1 - x)$ and $g(x) + 2$ with $(3 - x)$

 

[SammyS]

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 - \color{red}2 & = x^2 - 4x + 5 - \color{red}2 \\

\end{align*}

That's a good start.

Rather than subtracting 2, subtract 1. That gives the following.

$\left(g(x)\right)^2 + 2g\left(x\right) + 1 = x^2 - 4x + 4$​

The expression on each side of the equation is a perfect square.

 

[ecastro]

Well, you have g(x)[g(x)+2]=(x−3)(x−1)g(x)[g(x)+2]=(x−3)(x−1)g(x) \left [ g(x) + 2 \right ] = (x - 3)(x - 1) but you could flip the signs and also write g(x)[g(x)+2]=(3−x)(1−x)g(x)[g(x)+2]=(3−x)(1−x)g(x) \left [ g(x) + 2 \right ] = (3 - x)(1 - x) and since (3−x)(3−x)(3 - x) = (1−x)+2(1−x)+2(1 - x) + 2 you can then identify g(x)g(x)g(x) with (1−x)(1−x)(1 - x) and g(x)+2g(x)+2g(x) + 2 with (3−x)

Is there any particular reason why we should flip signs?

That's a good start.

Rather than subtracting 2, subtract 1. That gives the following.

$\left(g(x)\right)^2 + 2g\left(x\right) + 1 = x^2 - 4x + 4$​

The expression on each side of the equation is a perfect square.

Oh, I never thought of this. Thanks!

 

[RPinPA]

Is there any particular reason why we should flip signs?

Because it gives another solution. It's just a thing to try, an equivalent equation with potentially a different solution. Which turns out to be the case.

 

[WWGD]

Sorry if this is too far of an aside, but I always wondered how to determine the number of solution to functional equations, or a method that exhausts all solutions. What if we, e.g., have the equation f(x)^3 =Id, aka, find all f with fofof(x) =x (where o means function composition).

 

[SammyS]

Is there any particular reason why we should flip signs?

Because it gives another solution. It's just a thing to try, an equivalent equation with potentially a different solution. Which turns out to be the case.

To further answer OP's question:

Rather than saying "flip the signs", consider the following.

You can write $\ (x-3)\$ as $\ -(-x+3)\$.

Similarly write $\ (x-1)\$ as $\ -(-x+1)\$.

So, $\ (x-3)(x-1) =(-(-x+3))(-(-x+1)) = (-x+3)(-x+1) \$.

To finish things off:
Write this as $(-x+1)(-x+3) = ((-x+1))((-x+1)+2) \$.
Combine this with what you have done in the OP.

 

[mfb]

Sorry if this is too far of an aside, but I always wondered how to determine the number of solution to functional equations, or a method that exhausts all solutions. What if we, e.g., have the equation f(x)^3 =Id, aka, find all f with fofof(x) =x (where o means function composition).

This is complicated and way beyond the scope of this thread if treated properly.