Fdtxjr

I was trying to solve this problem:

If $f(x)=fracax+bcx+d, abcdneq0$ 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:
$$fraca(fracax+bcx+d)+bc(fracax+bcx+d)+d=x$$$$impliesfracfraca^2x+abcx+d+fracbcx+bdcx+dfraccax+cbcx+d+fracdcx+d^2cx+d=x$$$$impliesfracfrac(a^2+bc)x+b(a+d)cx+dfrac(ca+dc)x+cb+d^2cx+d=x$$$$(a^2+bc)x+b(a+d)=(ca+dc)x^2+(cb+d^2)x$$$$-(a+d)cx^2+(a+d)(a-d)x+(a+d)cdot b=0$$$$(a+d)[-cx^2+(a-d)x+b]=0$$

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 $abcdneq0$, $b$ 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

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 $abcdnot=0$

Thus, the other term in the product is 0:

$a+d=0$

Alternatively, note the property of inverse function:
$$f(f^-1(x))=f^-1(f(x))=x$$
Hence:
$$f(f(x))=x iff f(x)=f^-1(x)\
beginalignf(x)&=fracax+bcx+d Rightarrow y(cx+d)=ax+b Rightarrow x=fracdy-b-cy+a Rightarrow \
f^-1(x)&=fracdx-b-cx+aendalign$$
Now:
$$f(0)=fracbd=-frac ba=f^-1(0) Rightarrow a=-d Rightarrow a+d=0.$$