AMC 10 · 2008 · #12

Grade 9 algebra
Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination
domain-restrictionfunction-compositioninterval-arithmetic identify-subproblemswork-backwards ↑ Prerequisites: function-composition
📏 Medium solution 💡 2 insights
Problem
A function has a known set of inputs and a known set of outputs. A new function shifts the input and subtracts the result from one. Find the new function's inputs and outputs.

Pick an answer.

(A)
[-1,1],[-1,0]
(B)
[-1,1],[0,1]
(C)
[0,2],[-1,0]
(D)
[1,3],[-1,0]
(E)
[1,3],[0,1]
How to solve
Strategy Identify Subproblems

The rule changes f in two places that do not interfere: the input is shifted by +1 before f sees it, and the output is sent through "1 minus" after f is done. The shift alone decides the domain, the outer step alone decides the range, so split the question in two and answer each separately. For each half, run the transformation backwards as well as forwards. Forwards only tells you where things land, which is a containment; backwards shows every candidate is actually reached, which is what makes the two intervals exact rather than merely big enough. Finish by matching the pair against the five options.

1STEP 1

Split into two separate questions

The inputs and outputs are two separate questions.

g(x) = 1 - f(x+1)_decides the domain and g(x) = 1 - t_decides the range, t = f(x+1)
2STEP 2

Reverse the shift to get the domain

Reversing the shift slides the input set the opposite way.

0 ≤ x+1 ≤ 2 ⇔ -1 ≤ x ≤ 1
3STEP 3

Check that the shift loses no values of f

The shift loses none of the original outputs.

u = x+1, x ∈ [-1,1] ⇔ u ∈ [0,2], {f(x+1) : x ∈ [-1,1]} = {f(u) : u ∈ [0,2]} = [0,1]
4STEP 4

Send those values through 1 minus, both ways

Subtracting from one maps that set onto itself, both ways.

0 ≤ t ≤ 1 ⇔ -1 ≤ -t ≤ 0 ⇔ 0 ≤ 1-t ≤ 1; y ∈ [0,1] → t = 1-y ∈ [0,1] → g(x) = y for some x
5STEP 5

Match the pair against the options

So the pair is the shifted inputs with the same outputs, choice (A).

domain [-1,1], range [0,1]
Answer
[-1,1],[0,1]
Nothing about f was used beyond its domain and range, so any concrete function with those two properties must give the same answer. Test f(x) = x/2 on [0,2], whose outputs sweep [0,1]. Then g(x) = 1 - (x+1)/2 needs -1 ≤ x ≤ 1, and it slides from g(-1) = 1 down to g(1) = 0, covering [0,1]. Test a non-monotone one too: f(x) = |x-1| on [0,2] also has range [0,1], and g(x) = 1 - |x| on [-1,1] again has range [0,1]. Both witnesses give the pair [-1,1] and [0,1], matching (B). It is worth seeing why the range came back unchanged, so the coincidence is not mistaken for a rule: the interval [0,1] is symmetric about 1/2, and t ↦ 1-t reflects across 1/2, so this particular interval maps onto itself. Had the range of f been [0,3], the same steps would have given [-2,1] instead.
💡Key takeaway

Changing what goes into a function moves its domain the opposite way, changing what comes out reshapes its range, and a range is only settled once you show every number in it really does get hit.

  • Split into two separate questions
  • Reverse the shift to get the domain
  • Check that the shift loses no values of f
  • Send those values through 1 minus, both ways
  • Match the pair against the options

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