Vtyjkyuk

In Spivaks Calculus there is problem 8 in chapter 3:

For which numbers $a$, $b$, $c$, and $d$ will the function
$$
f(x) = fracax + bcx + d
$$
satisfy $f(f(x)) = x$ for all $x$?

Now, the solution I found suggests to expand out $f(f(x))$ and then simplify and arrive at the solution.

8. If
$$
x
= f(f(x))
= fraca left( fracax+bcx+d right) + b
c left( fracax+bcx+d right) + d
$$
for all $x$ then
$$
(ac+cd)x^2 + (d^2 - a^2)x - ab - bd = 0
qquad
textfor all $x$.
$$

(Original image here.)

I understand how generally expanding out works, but:

• I wonder why it makes sense to do that for this problem.


• What the solution in the end tells me/how it helps me to find numbers $a$, $b$, $c$, and $d$?

1. It's sensible to do this because you're given information about how the composition $(fcirc f)(x) = fbig(f(x)big)$ should be defined, namely that
   $$
   fbig(f(x)big) = x.
   $$
   Since you're given that $f(x)$ has the fractional form, you substitute it into itself for the composition and rearrange to get the equation that is quadratic in $x$. The reason for doing so is answered next.



2. 
   

   You have a quadratic in $x$ that equals zero, namely the quadratic
   $$
   (ac + cd)x^2 + (d^2 - a^2)x -(ab + bd) = 0.
   $$
   For this to be true for every value of $x$, I hope that you can see that we must have
   $$
   ac + cd = 0,qquad d^2 - a^2 = 0,qquad ab + bd = 0.
   $$
   These can be rewritten as
   $$
   c(a + d) = 0,qquad (d - a)(d + a) = 0,qquad b(a + d) = 0.
   $$
   Thus

   
   
   
   1. the first equation is zero if either $c = 0$ or $a = -d$;
   
   
   2. the second equation is zero if either $a = d$ or $a = -d$;
   
   
   3. the third equation is zero if either $b = 0$ or $a = -d$.
   
   

This actually gives two nice cases to consider. First, if $a = -d$ then all equations are satisfied so that
$$
f(x) = fracax + bcx - a.
$$
You should check that $(fcirc f)(x) = fbig(f(x)big) = x$ for this $f$.

The second case is to suppose that $a = d$ so that $c = 0$ and $b = 0$. This gives
$$
f(x) = fracaxa = x,
$$
which clearly satisfies $(fcirc f)(x) = fbig(f(x)big) = x$.

We note that if both $a = d$ and $a = -d$ are satisfied, then $a = d = 0$ so that
$$
f(x) = fracbcx.
$$
This also satisfies $(fcirc f)(x) = fbig(f(x)big) = x$ which you should check. This is just a special case of the first case.