A calculus problem by Alf-archie Sangkula

For this problem, since we need the value of f(-3), it is a good idea to substitute x = -3 and see what happens:

$\begin{aligned} f(-3) + f \left( \frac{1}{1-(-3)} \right) &= \frac{-3}{-3-1},&& \\ f(-3) + f\left( \frac{1}{4} \right) &= \frac{3}{4}. &&-(1) \end{aligned}$

From the equation above, we will know the value of f(-3) once we know the value of $f \left( \frac{1}{4} \right)$ . So let’s repeat the step above! (Repeating steps is a common problem solving technique…) Substituting $x = \frac{1}{4}$ into the original functional equation:

$\begin{aligned} f \left(\frac{1}{4}\right) + f \left( \frac{1}{1-\frac{1}{4}} \right) &= \frac{\frac{1}{4}}{\frac{1}{4}-1},&& \\ f \left( \frac{1}{4} \right) + f\left( \frac{4}{3} \right) &= -\frac{1}{3}. &&-(2) \end{aligned}$

Again! Substituting $x = \frac{4}{3}$ into the original functional equation:

$\begin{aligned} f \left(\frac{4}{3}\right) + f \left( \frac{1}{1-\frac{4}{3}} \right) &= \frac{\frac{4}{3}}{\frac{4}{3}-1},&& \\ f \left( \frac{4}{3} \right) + f\left( -3 \right) &= 4. &&-(3) \end{aligned}$

Our persistence has paid off! Taking (1) + (3) - (2):

$\begin{aligned} 2f(-3) &= \frac{3}{4} + 4 - \left( -\frac{1}{3} \right), \\ 24f(-3) &= 9 + 48 + 4 \\ &= \boldsymbol{61}. \end{aligned}$

The answer is 61.