An Addition-Formula Function

Problem

A function f(x)f(x) defined on (1,1)(-1, 1) satisfies:

  • for all x,y(1,1)x, y \in (-1, 1),  f(x)+f(y)=f(x+y1+xy)\ f(x) + f(y) = f\left(\dfrac{x + y}{1 + xy}\right);
  • f(x)>0f(x) > 0 whenever x>0x > 0.

Determine, with proof, which of the following are true:

1. f(x)f(x) is odd; 2. f(x)f(x) is decreasing; 3. if f(15)=12f\left(\dfrac{1}{5}\right) = \dfrac{1}{2}, then f(12)f(17)=1f\left(\dfrac{1}{2}\right) - f\left(\dfrac{1}{7}\right) = 1; 4. f(111)+f(119)+f(129)+f(141)<f(13)f\left(\dfrac{1}{11}\right) + f\left(\dfrac{1}{19}\right) + f\left(\dfrac{1}{29}\right) + f\left(\dfrac{1}{41}\right) < f\left(\dfrac{1}{3}\right).

Answer

Solution

Difficulty7/10
Topicsfunctions, Monotonicity, Telescoping, Functional Equations

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