Another Functional Equation Past Time

Algebra Level 3

Given that for x ∈ R , x ≠ 1 , − 1 x \in \mathbb{R} , x \neq 1, -1 ,

f ( x − 3 x + 1 ) + f ( 3 + x 1 − x ) = x f \left (\frac{x-3}{x+1} \right ) + f \left (\frac{3+x}{1-x} \right) = x ,

f ( 2 ) f(2) can be expressed as − m n -\frac{m}{n} where m , n m, n are relatively prime. Find m + n m+n .


The answer is 14.

Alan Yan by Alan Yan , 20, USA

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1 solution

Alan Yan
Sep 13, 2015

Solution 1:

x = 3 + x 1 − x ⟹ f ( x ) + f ( x − 3 x + 1 ) = 3 + x 1 − x x = \frac{3+x}{1-x} \implies f(x) + f(\frac{x-3}{x+1}) = \frac{3+x}{1-x} x = x − 3 x + 1 ⟹ f ( x ) + f ( 3 + x 1 − x ) = x − 3 x + 1 x = \frac{x-3}{x+1} \implies f(x) + f(\frac{3+x}{1-x}) = \frac{x-3}{x+1}

Adding the two equations yield that

2 f ( x ) = 3 + x 1 − x + x − 3 x + 1 − x ⟹ f ( 2 ) = − 11 3 ⟹ 11 + 3 = 14 2f(x) = \frac{3+x}{1-x} + \frac{x-3}{x+1} - x \implies f(2) = -\frac{11}{3} \implies 11 + 3 = \boxed{14}

Solution 2:

x = − 5 ⟹ f ( 2 ) + f ( − 1 3 ) = − 5 x = -5 \implies f(2) + f(-\frac{1}{3}) = -5 x = 2 ⟹ f ( − 1 3 ) + f ( − 5 ) = 2 x = 2 \implies f(-\frac{1}{3}) + f(-5) = 2 x = − 1 3 ⟹ f ( − 5 ) + f ( 2 ) = − 1 3 x = -\frac{1}{3} \implies f(-5) + f(2) = -\frac{1}{3} f ( 2 ) = − 11 3 f(2) = -\frac{11}{3} .

This problem is from CleverMath.

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