Take what they give you and see where it leads. For instance:
Suppose f(f(x)) = x. Then you have:
. . . . .f(ax + b|x|) = a(ax + b|x|) + b|ax + b|x|| = x
. . . . .a<sup>2</sup>x + ab|x| + |a|b|x| + b|bx| = x
Let x = 1. Then:
. . . . .a<sup>2</sup> + ab + |a|b + b|b| = 1
Let x = -1. Then:
. . . . .-a<sup>2</sup> + ab + |a|b + b|b| = -1
Subtracting, we get:
. . . . .2a<sup>2</sup> = 2
. . . . .a<sup>2</sup> = 1
. . . . .a = ±1
If a = 1, then:
. . . . .f(f(-1)) = -1 = -1 + b + b + b|b|
. . . . .-1 = -1 + 2b + b|b|
. . . . .0 = 2b + b|b|
If b > 0, then:
. . . . .0 = 2b + b<sup>2</sup>
. . . . .0 = b(2 + b)
. . . . .b = 0 or b = -2
By assumption, b can't equal =2. So let b = 0. Then we have:
. . . . .f(x) = x
This is trivial, and g(x) would then also equal x, so g(g(x)) = x for all x.
If b < 0, then:
. . . . .0 = 2b - b<sup>2</sup>
. . . . .0 = b(2 - b)
. . . . .b = 0 or b = 2
By assumption, b can't equal either of 2 or 0, so b cannot be negative, if a = 1.
Try a = -1. Then:
...and so forth.
Eliz.
