Functional substitution

First we make $f\left(\dfrac {x-3}{x+1} \right) = f \left(\dfrac 12\right)$ , $\implies \dfrac {x-3}{x+1} = \dfrac 12$ or $x = 7$ . Then $f \left(\dfrac {3+x}{1-x} \right) = f \left(\dfrac {3+7}{1-7} \right) = f \left(-\dfrac 53 \right)$ . Therefore,

$f\left(\frac 12\right) + f\left(-\frac 53\right) = 7\quad ...(1)$

Putting $x = \dfrac 12 \implies f\left(\dfrac {\frac 12-3}{\frac 12 +1} \right) = f \left(-\dfrac 53 \right)$ and $f \left(\dfrac {3+\frac 12}{1-\frac 12} \right) = f \left(7\right)$ , then

$f\left(-\frac 53 \right) + f\left(7 \right) = \frac 12 \quad ...(2)$

Similarly $x = - \dfrac 53 \implies f\left(\dfrac {-\frac 53-3}{-\frac 53 +1} \right) = f \left(7\right)$ and $f \left(\dfrac {3-\frac 53}{1+\frac 53} \right) = f \left(\dfrac 12\right)$ , then

$f\left(7\right) + f\left(\frac 12 \right) = - \frac 53 \quad ...(3)$

From $(1)-(2)+(3)$ :

$2 f \left(\frac 12 \right) = 7 - \frac 12 + \frac 53 = \frac {29}6 \implies f \left(\frac 12 \right) = \frac {29}{12}$

Therefore $m+n = 29+12 = \boxed{41}$ .

Observe that if $g(x) = \frac{x-3}{x+1}$ , then $g^{-1}(x) = \frac{3+x}{1-x} = -g(-x)$ . Thus the identity becomes:

$f(g(x))+f(-g(-x)) = x$ ,

for all $x \notin \{-1, 1\}$ . Let

• $x = 7$ : $f(\frac{1}{2}) + f(\frac{-5}{3}) = 7$
• $x = \frac{1}{2}$ : $f(\frac{-5}{3}) + f(7) = \frac{1}{2}$
• $x = \frac{-5}{3}$ : $f(7) + f(\frac{1}{2}) = \frac{-5}{3}$ .

Thus, $f(\frac{-5}{3}) = 7 - f(\frac{1}{2}) = \frac{1}{2} + \frac{5}{3} + f(\frac{1}{2})$

which gives $f(\frac{1}{2}) = \frac{29}{12}$ .