Find Values of a, b, c, and d for f(f(x)) = x | Confusing Problem Solution

[nietzsche]

Homework Statement

For which numbers a, b, c, and d will the function

$$f(x) &= \frac{ax+b}{cx+d}$$

satisfy f(f(x)) = x for all x (for which this equation makes sense.

Homework Equations

The Attempt at a Solution

I'm not sure if the answer will consist of specific values, or just a general solution, but here is my attempt so far:

$$\begin{align*} f(f(x)) &= \frac{a\frac{ax+b}{cx+d} + b}{c\frac{ax+b}{cx+d} + d}\\ x &= \frac{a^2x + ab + bcx + bd}{cax + cb + dcx + d^2}\\ 0 &= cx^2 + (d-a)x - b \end{align*}$$

I skipped a bunch of steps in between, but that was the final equation I ended up with after simplification. Which was really weird, because I don't know what to do with that. If I had to guess, from the last equation, a, b, c, d must be values that satisfy the equation

$$x = \frac{ax+b}{cx+d}$$

but again, I'm really confused and am not sure how to proceed.

 

[nietzsche]

uh...

i can see that it will work when a=d and b=c=0, but that seems a bit trivial.

 

[Dick]

No, you won't get specific values for a,b,c and d. You'll get a condition from them to satisfy. But you are equating the wrong things. Your expression for f(f(x)) isn't equal to x. It's equal to f(x)=(ax+b)/(cx+d). Equate those. I.e.
$$\frac{a^2x + ab + bcx + bd}{cax + cb + dcx + d^2}=\frac{ax+b}{cx+d}$$.
If you clear out the denominators and move everything to one side, you'll get a quadratic in x^2 that must be 0 for all x. So the coefficients of x^2, x and the constant must all be equal to zero. See if you can figure out a condition on a,b,c and d that makes that true.

 

[nietzsche]

No, you won't get specific values for a,b,c and d. You'll get a condition from them to satisfy. But you are equating the wrong things. Your expression for f(f(x)) isn't equal to x. It's equal to f(x)=(ax+b)/(cx+d). Equate those. I.e.
$$\frac{a^2x + ab + bcx + bd}{cax + cb + dcx + d^2}=\frac{ax+b}{cx+d}$$.
If you clear out the denominators and move everything to one side, you'll get a quadratic in x^2 that must be 0 for all x. So the coefficients of x^2, x and the constant must all be equal to zero. See if you can figure out a condition on a,b,c and d that makes that true.

but the question says "for which a, b, c ,d will the f(f(x)) = x for all x"

i thought that means that f(f(x)) = x?

 

[Dick]

but the question says "for which a, b, c ,d will the f(f(x)) = x for all x"

i thought that means that f(f(x)) = x?

Indeed you are right. I was reading the problem wrong. Let me work it out again. But you still should turn the expression into a quadratic in x equalling 0 and then set the coefficients to zero.

 

[nietzsche]

$$\begin{align*} f(f(x)) &= \frac{a\frac{ax+b}{cx+d} + b}{c\frac{ax+b}{cx+d} + d}\\ x &= \frac{a^2x + ab + bcx + bd}{cax + cb + dcx + d^2}\\ 0 &= cx^2 + (d-a)x - b \end{align*}$$

that's the quadratic i got

 

[Dick]

$$\begin{align*} f(f(x)) &= \frac{a\frac{ax+b}{cx+d} + b}{c\frac{ax+b}{cx+d} + d}\\ x &= \frac{a^2x + ab + bcx + bd}{cax + cb + dcx + d^2}\\ 0 &= cx^2 + (d-a)x - b \end{align*}$$

that's the quadratic i got

That's great. You are way ahead of me. Now the only way that can be zero for all x is if each coefficient equals zero. What do you conclude?

 

[nietzsche]

d=a, c=0, b=0

i think...

thanks for your help today dick.

 

[Dick]

d=a, c=0, b=0

i think...

thanks for your help today dick.

And that corresponds to f(x)=x, the trivial solution. I'm noticing you dropped an overall factor in the quadratic. Shouldn't it be multiplied by (a+d)? That gives you another solution possibility. Careful!

 

[nietzsche]

oh...yeah, there was an a+d in there. does that mean that if a+d = 0, it is also true?

 

[Dick]

oh...yeah, there was an a+d in there. does that mean that if a+d = 0, it is also true?

It sure does! Put d=(-a) into your expression for f(f(x)).

 

[nietzsche]

Thanks very much!