Prove that if $a^2+bc \neq 0$, then the graph of $f(x)= \frac{ax+b}{cx-a}$ is symmetric about the line y=x

Question

Prove that if $a^2+bc \ne 0$, then the graph of $f(x)= \frac{ax+b}{cx-a}$ is symmetric about the line $y=x$.

Maybe this is a simple exercise, but I need help from you guys to understand how is the best way to solve this problem.

Reversing the coordinates of a point on the graph of $f(x)$ produce a point in the graph of $f^{-1}(x)$.

Answer 1

Just a different proof:

$$f(f(x))=\frac{af(x)+b}{cf(x)-a}=\frac{a\frac{ax+b}{cx-a}+b}{c\frac{ax+b}{cx-a}-a} = \frac{\frac{a^2x+ab+bcx-ab}{cx-a}}{\frac{cax+cb-acx+a^2}{cx-a}}= \frac{a^2x+bcx}{cb+a^2}=x$$

So $f(x)$ is its own inverse and so its graph is symmetric respect to $y=x$ (recall that the graph of a function and its inverse are symmetric respect to $y=x$)

Answer 2

A function is symmetric about $y = x$ if $f(\lambda) = \mu \iff f(\mu) = \lambda$ holds for all $\mu, \lambda \in \mathbb R$.

We have \begin{align} f(\lambda) = \mu & \iff \frac{a \mu + b}{c \mu - a} = \lambda \iff a \mu + b = \lambda c \mu - \lambda a \iff (a - \lambda c) \mu = - \lambda a - b \\ & \iff \mu = \frac{- \lambda a - b}{a - \lambda c} = \frac{\lambda a + b}{\lambda c - a} = f(\lambda). \end{align} The condition $-a^2 \ne bc$ ($\star$) is relevant as in this case we have $$ f(x) = \frac{a x + b}{c x - a} = \frac{b a x + b^2}{bc x - ab} \overset{(\star)}{=} \frac{b (a x + b)}{-a(ax + b)} = - \frac{b}{a}, $$ which is constant and thus not symmetrical about $y = x$.