Algebra problem that you have to assume certain criteria at the end.

Question

I was trying to solve this problem:

If $f(x)=\frac{ax+b}{cx+d}, abcd\neq0$ and $f(f(x))=x$ for all $x$ in the domain of $f$, what is the value of $a+d$?

I start off by just plugging in and simplifying: $$\frac{a(\frac{ax+b}{cx+d})+b}{c(\frac{ax+b}{cx+d})+d}=x$$$$\implies\frac{\frac{a^2x+ab}{cx+d}+\frac{bcx+bd}{cx+d}}{\frac{cax+cb}{cx+d}+\frac{dcx+d^2}{cx+d}}=x$$$$\implies\frac{\frac{(a^2+bc)x+b(a+d)}{cx+d}}{\frac{(ca+dc)x+cb+d^2}{cx+d}}=x$$$$(a^2+bc)x+b(a+d)=(ca+dc)x^2+(cb+d^2)x$$$$-(a+d)cx^2+(a+d)(a-d)x+(a+d)\cdot b=0$$$$(a+d)[-cx^2+(a-d)x+b]=0$$

Ok, so now, by prior knowledge from factoring and solving quadratics, I know that $(a+d)$ and/or $[-cx^2+(a-d)x+b]$ is equal to $0$. So that means that if $(a+d)=0$, then, well, $(a+d)=0$. If they are both equal to $0$, then $(a+d)$ still equals $0$. So it is apparent that $(a+d)$ is equal to $0$. However, what if $[-cx^2+(a-d)x+b]$ is equal to $0$? Would that have any effect on $(a+d)=0$?

Furthermore, the solution I have sets $x$ to $0$ in the equation $$(a+d)[-cx^2+(a-d)x+b]=0$$ I have derived. Then, it follows that $(a+d)\cdot b=0$, and due to $abcd\neq0$, $b$ can't equal $0$ and thus $(a+d)=0$. However, doesn't this "let $x$ be $0$" thing not work for all $x$? How is this right, if at all?

Thanks for your help!

Max0815

Answer 1

For the product $(a+d)[-cx^2+(a-d)x+b]=0$

The quadratic term $-cx^2 + (a-d)x+b$ cannot be 0 for all x in the domain.

This is because it is a function of x, and varies with the value of x. It cannot be 0 for ALL x in the domain unless ALL the coefficients of the function are 0

Then b, a coefficient of the function would have to be $0$ but this is impossible, as we have assumed $abcd\not=0$

Thus, the other term in the product is 0:

$a+d=0$

Answer 2

Alternatively, note the property of inverse function: $$f(f^{-1}(x))=f^{-1}(f(x))=x$$ Hence: $$f(f(x))=x \iff f(x)=f^{-1}(x)\\ \begin{align}f(x)&=\frac{ax+b}{cx+d} \Rightarrow y(cx+d)=ax+b \Rightarrow x=\frac{dy-b}{-cy+a} \Rightarrow \\ f^{-1}(x)&=\frac{dx-b}{-cx+a}\end{align}$$ Now: $$f(0)=\frac{b}{d}=-\frac ba=f^{-1}(0) \Rightarrow a=-d \Rightarrow a+d=0.$$