AMC 10 · 2006 · #18

Grade 9 algebra
Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination
functional-equationdomain-restrictionlogical-deduction extreme-principleguess-and-check ↑ Prerequisites: functional-equationdomain-restriction
📏 Long solution 💡 3 insights
Problem
A function's domain is closed under reciprocals, and an identity relates its value at a number to its value at the reciprocal. Find the largest possible domain.

Pick an answer.

(A)
$\{x|x\ne 0\}$
(B)
$\{x|x<0\}$
(C)
$\{x|x>0\}$
(D)
$\{x|x\ne -1\;\rm{and}\; x\ne 0\;\rm{and}\; x\ne 1\}$
(E)
$\{-1,1\}$
How to solve
Strategy Organize Information in More Ways

There is only one equation but it is quietly two, because the closure rule guarantees that the partner point 1/x is also a legal place to read the identity. Tool #15 (Organize Information in More Ways) is the whole trick: rewrite the same given rule at that partner point, line the two versions up, and the tension between them appears on its own. That comparison pins down which numbers are even allowed, which is Tool #14 (Extreme Principle) territory, since the word 'largest' asks for a boundary. But a boundary argument only half-answers the question. Ruling numbers out shows no domain can be bigger than {-1, 1}; it does not show that any f with that domain exists. So a short Tool #6 (Guess and Check) construction actually builds f on the surviving points, turning a ceiling into an answer. Tool #3 (Eliminate Possibilities) then serves as an independent audit of the five choices.

1STEP 1

Unpack the two domain rules

Closure guarantees the partner point is also in the domain.

x ∈ D ⟹ 1/x ∈ D, 0 ∉ D
2STEP 2

Read the identity at the partner point

Reading the identity there gives a second equation with the same left side.

f(1/x) + f (1/(1/x)) = 1/x ⟹ f(1/x) + f(x) = 1/x
3STEP 3

One left side, two right sides

One number cannot equal two, so every point equals its own reciprocal.

f(x) + f(1/x) = x and f(x) + f(1/x) = 1/x ⟹ x = 1/x
4STEP 4

Solve for the ceiling on the domain

That puts a ceiling of two points on the domain.

x = 1/x, x ≠ 0 ⟹ x² = 1 ⟹ x = ± 1 ⟹ D ⊆ {-1, 1}
5STEP 5

Build an f that reaches the ceiling

An explicit function reaches the ceiling, so the answer is {-1,1}, choice (E).

2f(1) = 1 → f(1) = 1/2, 2f(-1) = -1 → f(-1) = -1/2
Answer
{-1,1}
Test the constructed f head-on. At x = 1: f(1) + f(1/1) = 1/2 + 1/2 = 1 = x, correct. At x = -1: f(-1) + f(1/-1) = -1/2 + (-1/2) = -1 = x, correct. So (E) is not merely the last set standing, it is realized by an explicit function, which is what 'can be in the domain' demands. The distractors fail for a shared reason: any set containing a number x with x ≠ 1/x is dead, because the identity then asserts one sum equals both x and 1/x. Choice (A) admits every nonzero real, including 2; (B) admits -2; (C) admits 2; and (D) is the sharpest trap, since it deletes exactly -1, 0 and 1 and so throws away the only two numbers that could have survived, leaving a set where the identity fails everywhere. One more sanity note: {1} and {-1} are also legal domains, just smaller ones, which is consistent with {-1, 1} being the maximum rather than the only option.
💡Key takeaway

The sum f(x) + f(1/x) looks the same whether you start from x or from 1/x, but it is supposed to equal x, so the only numbers allowed in the domain are the ones equal to their own reciprocal: 1 and -1.

  • Unpack the two domain rules
  • Read the identity at the partner point
  • One left side, two right sides
  • Solve for the ceiling on the domain
  • Build an f that reaches the ceiling

More like this

Same archetype — closest grade level first.