Question

for which real numbers a, b, c will the function f(x)=(ax+b)/(cx+d) satisfy f(f(x))=x for all x

Answer 1

Expand the f(f(x)) and try to find something

Answer 2

What I got is a=cx and b=dx

Answer 3

Keep the quotient form, you will have\[\frac{ a \frac{ ax+b }{ cx+d } +b }{ c \frac{ ax+b }{ cx+d }+ d } =x\]

Answer 4

i know,then i get\[-x^{2}(ac+bc)-x(db-a^{2})+b(a+b)=0\], but what i can do next?

Answer 5

no

Answer 6

\[(a-cx)\frac{ ax+b }{ cx+d }+(b-dx)=0\]

Answer 7

then a=cx and b=dx is a solution to this equation

Answer 8

I believe you should multiply both sides by (cx + d) and then simplify first.

Answer 9

What I end up with is:\[c(a+d)x^2+(d^2-a^2)x-b(a+d)=0\]

Answer 10

Since this has to hold for ANY x, then every term in this equation must be zero.

Answer 11

The simplest solution is therefore:\[a=-d\]

Answer 12

why every term in your equation must be zero?

Answer 13

because it has to hold for ANY x

Answer 14

i.e. for f(f(x)) = x for all x, we must have:\[c(a+d)x^2+(d^2-a^2)x-b(a+d)=0\]

Answer 15

for all x

Answer 16

got it

Answer 17

this therefore leads to:\[f(x)=\frac{ax+b}{cx-a}\]

Answer 18

for any a, b and c will give:\[f(f(x))=x\]

Answer 19

yeah, i got it, thx a lot

Answer 20

yw :)