question: Let $a,b,c,d$ be given constants with the property that $ad-bc\neq0$. If $f(x)=\frac{ax+b}{cx+d}$, show that there exists a function $g(x)=\frac{\alpha x+\beta}{\gamma x+\delta}$ such that $f(g(x))=x$. Also show that for these two functions it is true that $f(g(x))=g(f(x))$.
how can I get the answer $\alpha=\frac{d}{ad-bc}$, $\beta=\frac{-b}{ad-bc}$, $\gamma=\frac{-c}{ad-bc}$, $\delta=\frac{a}{ad-bc}$
my approach: first calculate the value of $f(g(x))$
$$f(g(x))=\frac{(a\alpha+b\gamma)x+(a\beta+b\delta)}{(c\alpha+d\gamma)x+(c\beta+d\delta)}$$
then, $f(g(x))=x$
$$
\begin{align}
\frac{(a\alpha+b\gamma)x+(a\beta+b\delta)}{(c\alpha+d\gamma)x+(c\beta+d\delta)}&=x\\
(a\alpha+b\gamma)x+(a\beta+b\delta)&=(c\alpha+d\gamma)x^2+(c\beta+d\delta)x
\end{align}
$$
and I stuck at this step. please help!
Best Answer
You're trying to find the inverse function of $f(x)$, so set the function equal to $y$ and solve for $x$:
$$y = \frac{ax+b}{cx+d}$$
$$ycx+yd = ax+b$$
$$ycx-ax = b-yd$$
$$x(yc-a) = b-yd$$
$$x = \frac{-dy+b}{-a+cy}=\frac{dy-b}{a-cy}.$$
Interchange $x$ and $y$ and you have
$$g(x) =\frac{dx-b}{a-cx}.$$
If you divide top and bottom by $ad-bc$, you'll have the answer you want.