Find all odd functions of the form $f(x) = \frac{ax + b}{x + c}$ I am working through a pure maths book and am stuck on odd and even functions.
Let $f(x) = \frac{ax + b}{x + c}$ where x, a, b, c are real and $x \ne \pm c$.  Show that if $f$ is an even function then $ac = b$. Deduce that if $f$ is an even function then $f(x)$ must reduce to the form $f(x) = k$, where k is constant. Find all odd functions of the form $\frac{ax + b}{x + c}$
I have solved the first part. I am fairly sure of the second part. But I cannot solve the third part.
My calculations are as follows:
If $f(x)$ is even $f(x) = f(-x)$
$\implies \frac{ax + b}{x + c} = \frac{-ax + b}{-x + c}$
$\implies (ax + b)(c-x)= (x+c)(b-ax)$
$ac = b$
$\implies \frac{ax + b}{x + c} = \frac{ax + ac}{x + c} = \frac{a(x + c)}{x + c} = a$
So I am assuming that the a is the k to which the question refers.
Now, if $\frac{ax + b}{x + c}$ is odd, $f(-x) =-f(x)$. So
$\frac{-ax + b}{-x + c} = \frac{-(ax + b)}{x + c}$
$\implies (b-ax)(c+x)= (c-x)(-ax-b)$
$\implies 2bc=2ax^2 \implies bc=ax^2 \implies x = \sqrt\frac{bc}{a}$
but this does not lead to the answer in the book, which is $f(x)=\frac{k}{x}$

You need $$ax^2 -bc =0 $$ for all $x$. That is, $ax^2-bc$ should be the zero polynomial, i.e. $a=0$ and $bc=0$. If $b=0$, then $$f(x) =0$$ If $c=0$, then $$f(x)=\frac bx $$ for any $b\in \mathbb R$. But setting $b=0$ gives the first case, so a general solution is $$f(x)=\frac kx$$
