Skip to main content
AMC 12 focus

Functional Equations

Pinning down an unknown function from a few constraints, using targeted substitution instead of guesswork.

A functional equation gives you a rule that a function ff must satisfy for all inputs, and asks you to determine ff (or just a specific value of it). The technique is always the same: plug in cleverly chosen values to generate equations you can combine.

Substitutions worth trying first

  • Set a variable to 00 (or 11) — often reveals a simple starting value like f(0)f(0).
  • Set two variables equal to each other.
  • Swap the roles of the variables and compare with the original equation (exposes symmetry).
  • Set a variable equal to another expression already in the equation, to create a substitution chain.

Two named equations worth recognizing

Cauchy's equation, f(x+y)=f(x)+f(y)f(x+y) = f(x)+f(y), forces f(x)=cxf(x) = cx for some constantcc under mild regularity conditions (which AMC problems always implicitly assume). The multiplicative version, f(xy)=f(x)+f(y)f(xy) = f(x)+f(y) for x,y>0x,y>0, behaves like a logarithm:f(x)=clog⁡xf(x) = c\log x. Recognizing an equation as one of these named shapes can save serious time.

Worked example

Problem: Suppose f(x+y)=f(x)+f(y)−1f(x+y) = f(x) + f(y) - 1 for all reals, and f(1)=3f(1)=3. Find f(5)f(5).

Solution: Try x=y=1x=y=1: f(2)=f(1)+f(1)−1=3+3−1=5f(2) = f(1)+f(1)-1 = 3+3-1=5. Thenx=2,y=1x=2,y=1: f(3)=f(2)+f(1)−1=5+3−1=7f(3)=f(2)+f(1)-1=5+3-1=7. The pattern f(1)=3,f(2)=5,f(3)=7f(1)=3, f(2)=5, f(3)=7suggests f(n)=2n+1f(n) = 2n+1. Check: does this satisfy the original equation? f(x+y)=2(x+y)+1f(x+y) = 2(x+y)+1and f(x)+f(y)−1=(2x+1)+(2y+1)−1=2x+2y+1f(x)+f(y)-1 = (2x+1)+(2y+1)-1 = 2x+2y+1 — they match. So f(5)=2(5)+1=11f(5) = 2(5)+1 = 11.

Whenever you spot a pattern like this from small cases, always verify it satisfies the originalfunctional equation before trusting it — a pattern that fits three data points isn't automatically the unique answer.

Practice problems

1.If f(x+y)=f(x)f(y)f(x+y) = f(x)f(y) for all reals and f(1)=2f(1) = 2, find f(3)f(3).

2.If f(x)+2f(1−x)=x2f(x) + 2f(1-x) = x^2 for all xx, find f(2)f(2).

3.If f(xy)=f(x)+f(y)f(xy)=f(x)+f(y) for positive reals and f(2)=1f(2)=1, find f(8)f(8).

4.If f(x)=f(−x)f(x) = f(-x) for all xx, what property does this describe, and what does it tell you about the graph?

Sequences & RecursionsExponential & Log Equations