Answer A: Anthropic Claude Opus 4.7
The student's claim contains a subtle but important error. Let's work through it carefully. **1. Simplifying g(x) for x ≠ 1.** For x ≠ 1 we may cancel the factor (x − 1) in the first fraction: (x² − 1)/(x − 1) = (x − 1)(x + 1)/(x − 1) = x + 1. Therefore, for x ≠ 1, g(x) = (x + 1)·|x − 1|. We can split this by cases, using the definition of |x − 1|: • If x > 1: g(x) = (x + 1)(x − 1) = x² − 1. • If x < 1: g(x) = (x + 1)(−(x − 1)) = −(x + 1)(x − 1) = 1 − x². • At x = 1: g(x) is not defined by the original formula, because (x² − 1)/(x − 1) is 0/0 there. **2. Can g be extended to a polynomial on all o...
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The student's claim contains a subtle but important error. Let's work through it carefully. **1. Simplifying g(x) for x ≠ 1.** For x ≠ 1 we may cancel the factor (x − 1) in the first fraction: (x² − 1)/(x − 1) = (x − 1)(x + 1)/(x − 1) = x + 1. Therefore, for x ≠ 1, g(x) = (x + 1)·|x − 1|. We can split this by cases, using the definition of |x − 1|: • If x > 1: g(x) = (x + 1)(x − 1) = x² − 1. • If x < 1: g(x) = (x + 1)(−(x − 1)) = −(x + 1)(x − 1) = 1 − x². • At x = 1: g(x) is not defined by the original formula, because (x² − 1)/(x − 1) is 0/0 there. **2. Can g be extended to a polynomial on all of ℝ?** No. Even if we fill in the removable hole by defining g(1) = (1 + 1)·|1 − 1| = 0, the extended function is g(x) = { x² − 1, if x ≥ 1; 1 − x², if x ≤ 1. } This is not a polynomial. Here is a clean argument. Suppose, for contradiction, that some polynomial p(x) satisfied p(x) = g(x) for all real x. Then on the interval (1, ∞) we would have p(x) = x² − 1, and on (−∞, 1) we would have p(x) = 1 − x². But two polynomials that agree on an interval (indeed, on any infinite set) must be identical. So p(x) would have to equal x² − 1 everywhere and also equal 1 − x² everywhere. These two polynomials are not the same (they differ at, e.g., x = 0), giving a contradiction. Hence no single polynomial can represent g on all of ℝ. Equivalently: every polynomial is infinitely differentiable, and in particular has a continuous second derivative. We will see in part 3 that g is not even differentiable at x = 1, so it cannot coincide with any polynomial there. **3. Differentiability at x = 1.** First extend g continuously by setting g(1) = 0 (both one-sided limits give 0, so g is continuous at 1). Now examine the one-sided derivatives using the definition g′(1) = lim_{h→0} [g(1 + h) − g(1)]/h = lim_{h→0} g(1 + h)/h. Right-hand derivative (h > 0, so 1 + h > 1, use g(x) = x² − 1): g(1 + h) = (1 + h)² − 1 = 2h + h². [g(1 + h)]/h = 2 + h → 2 as h → 0⁺. Left-hand derivative (h < 0, so 1 + h < 1, use g(x) = 1 − x²): g(1 + h) = 1 − (1 + h)² = −2h − h². [g(1 + h)]/h = −2 − h → −2 as h → 0⁻. Since the two one-sided limits are 2 and −2, they are unequal, so g′(1) does not exist. The graph of g has a corner at x = 1. Therefore g is not differentiable at x = 1. Because every polynomial is differentiable everywhere, this gives a second, independent proof that g cannot be a polynomial. **4. The conceptual mistake.** The student confused two different statements: (a) Two functions have the same values on a set S (here, S = ℝ \ {1}). (b) Two functions are equal as functions (i.e., have the same domain and the same values everywhere). It is true that (x² − 1)/(x − 1) and x + 1 agree on ℝ \ {1}, but they are not the same function: the first is undefined at x = 1 while the second is defined there. When you multiply by |x − 1|, the resulting function g(x) = ((x² − 1)/(x − 1))·|x − 1| is likewise undefined at x = 1, and even after filling in the hole, it equals (x + 1)|x − 1|, which contains an absolute value. Absolute value is not a polynomial operation: |x − 1| has a corner at x = 1, and multiplying it by the polynomial x + 1 (which is nonzero there) preserves that corner. A polynomial cannot have a corner, so g is not a polynomial. In short, "simplifies to" on a restricted domain is not the same as "equals" globally, and products involving |x − 1| generally break polynomiality even when the absolute value happens to be "hidden" inside a canceled factor.
Result
Winning Votes
3 / 3
Average Score
Total Score
Overall Comments
This is an outstanding answer. It is mathematically precise, complete, and exceptionally clear. Its primary strength lies in the quality of its reasoning, particularly the use of the identity theorem for polynomials to prove that the function cannot be a polynomial. This demonstrates a deeper level of mathematical understanding than simply pointing to the properties of the absolute value function. The structure is logical and easy to follow.
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Correctness
Weight 45%The answer is entirely correct across all four parts. The simplification, the piecewise definition, the one-sided derivatives (2 and -2), and the final conclusion are all accurate.
Reasoning Quality
Weight 20%The reasoning is exceptionally strong. The use of the identity theorem for polynomials in part 2 is a highly rigorous and insightful argument. The answer also provides a second, independent proof based on non-differentiability, demonstrating a deep understanding of the concepts.
Completeness
Weight 15%The answer is fully complete, addressing every component of the four-part prompt in detail. It provides the simplified form, the piecewise definition, the full derivative calculation, and a thorough conceptual explanation.
Clarity
Weight 10%The answer is exceptionally clear and well-organized. The use of bolded headers and bullet points makes the structure easy to follow. The explanation of the complex polynomial identity argument is particularly lucid.
Instruction Following
Weight 10%The answer perfectly follows all instructions. It addresses all four parts, provides justifications, and is written in a style appropriate for the target audience.
Total Score
Overall Comments
Answer A is mathematically rigorous, well-structured, and thorough. It correctly simplifies g(x), provides a clean contradiction-based proof that no polynomial can represent g, computes one-sided derivatives explicitly, and gives a nuanced conceptual explanation distinguishing domain restriction from global equality. The reasoning is layered (offering two independent proofs of non-polynomiality) and the language is accessible to a strong high-school student. Minor stylistic choices (e.g., using bullet points and bold headers) enhance readability without sacrificing rigor.
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Correctness
Weight 45%All mathematical claims are correct: the simplification, the piecewise form, the contradiction proof using polynomial identity on intervals, the one-sided derivative calculations (2 and -2), and the conclusion about non-differentiability. No errors detected.
Reasoning Quality
Weight 20%Exceptional reasoning. Uses the polynomial identity principle (two polynomials agreeing on an infinite set are identical) to give a clean contradiction proof. Also provides a second independent argument via non-differentiability. The logic is tight and multi-layered.
Completeness
Weight 15%All four parts are addressed fully and in depth. The extension discussion explicitly handles the undefined point at x=1 and the filled-in value. Both one-sided derivatives are computed with full algebraic detail.
Clarity
Weight 10%Well-organized with bold headers, clear case splits, and explicit limit calculations. The language is precise yet accessible. The two-proof structure for non-polynomiality is clearly signposted.
Instruction Following
Weight 10%Follows all four parts of the prompt in order, addresses the expected content (simplification, extension, differentiability, conceptual mistake), and maintains the appropriate level for a strong high-school student.
Total Score
Overall Comments
Answer A is mathematically strong, fully addresses all four requested parts, and gives two rigorous justifications that no polynomial extension is possible. It clearly distinguishes the original domain issue at x = 1 from the behavior of the extended function, computes the one-sided derivatives correctly, and explains the student's conceptual mistake precisely. Its only minor weakness is a small wording slip where one displayed piecewise line uses x ≤ 1 after saying x < 1, though the surrounding explanation makes the intended extension clear.
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Correctness
Weight 45%All core claims are correct: the simplification for x ≠ 1, the piecewise forms, the nonexistence of a polynomial extension, and the one-sided derivative calculation at x = 1. It also correctly distinguishes the original undefined point from the continuous extension. A tiny notation inconsistency appears in one displayed piecewise line using x ≤ 1 after earlier deriving x < 1, but the intended extension is still mathematically clear.
Reasoning Quality
Weight 20%Reasoning is rigorous and well structured. It gives a contradiction argument using the identity principle for polynomials on intervals and also notes non-differentiability as an independent obstruction. The conceptual distinction between equality on a restricted domain and equality as functions is especially strong.
Completeness
Weight 15%It fully answers every requested part, including simplification, extension analysis, differentiability, and the conceptual mistake. It also adds helpful supporting remarks about continuity and gives two different ways to see non-polynomiality.
Clarity
Weight 10%The explanation is clear, logically ordered, and pitched appropriately for a strong high-school student while remaining rigorous. The case split and derivative computation are especially readable.
Instruction Following
Weight 10%It follows the prompt very closely, answers all numbered parts explicitly, and maintains a rigorous but accessible essay style. It also directly evaluates the student's claim rather than merely restating computations.