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Analyze Why a Product Is Not a Polynomial

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Contents

Task Overview

Benchmark Genres

Education Q&A

Task Creator Model

Answering Models

Judge Models

Task Prompt

A student claims that because f(x) = (x^2 - 1)/(x - 1) simplifies to x + 1 for x ≠ 1, the function g(x) = ((x^2 - 1)/(x - 1)) · |x - 1| is a polynomial equal to (x + 1)|x - 1|. Evaluate this claim. Answer all parts: 1. Simplify g(x) as much as possible for x ≠ 1. 2. Determine whether g(x) can be extended to a polynomial on all real numbers. Justify your conclusion. 3. State whether g is differentiable at x = 1, and show the key calculation that supports your answer. 4. Briefly explain the conceptual mistake in the...

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A student claims that because f(x) = (x^2 - 1)/(x - 1) simplifies to x + 1 for x ≠ 1, the function g(x) = ((x^2 - 1)/(x - 1)) · |x - 1| is a polynomial equal to (x + 1)|x - 1|. Evaluate this claim. Answer all parts: 1. Simplify g(x) as much as possible for x ≠ 1. 2. Determine whether g(x) can be extended to a polynomial on all real numbers. Justify your conclusion. 3. State whether g is differentiable at x = 1, and show the key calculation that supports your answer. 4. Briefly explain the conceptual mistake in the student's reasoning. Your answer should be mathematically rigorous but understandable to a strong high-school student.

Judging Policy

A strong answer should correctly simplify the expression to (x + 1)|x - 1| for x ≠ 1, recognize that this equals x^2 - 1 for x ≥ 1 and 1 - x^2 for x < 1, and use this to analyze behavior at x = 1. It should clearly distinguish the originally defined function from any extension, and correctly conclude that no polynomial can agree with g on all x ≠ 1 because the two-sided formulas differ on intervals and a polynomial cannot match both unless they are the same polynomial. A good answer should identify the removable is...

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A strong answer should correctly simplify the expression to (x + 1)|x - 1| for x ≠ 1, recognize that this equals x^2 - 1 for x ≥ 1 and 1 - x^2 for x < 1, and use this to analyze behavior at x = 1. It should clearly distinguish the originally defined function from any extension, and correctly conclude that no polynomial can agree with g on all x ≠ 1 because the two-sided formulas differ on intervals and a polynomial cannot match both unless they are the same polynomial. A good answer should identify the removable issue in the rational factor separately from the non-polynomial absolute value behavior, and correctly assess differentiability at x = 1 by computing one-sided derivatives of the natural extension or by equivalent reasoning. The conceptual explanation should note that simplifying an expression on a restricted domain does not automatically make the resulting function a polynomial, especially when an absolute value and a missing point are involved.

Answers

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

#1 | Winner

Winning Votes

3 / 3

Average Score

94

Total Score

99

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%
100

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%
95

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%
100

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%
95

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%
100

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

90

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%
92

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%
90

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%
90

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%
85

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%
90

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.

Judge Models OpenAI GPT-5.4

Total Score

94

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%
94

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%
93

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%
96

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%
90

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%
97

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.

Let's analyze the function $g(x) = \frac{x^2 - 1}{x - 1} \cdot |x - 1|$. 1. **Simplifying g(x) for x ≠ 1:** For $x \neq 1$, we know that $x^2 - 1 = (x - 1)(x + 1)$. Therefore, $\frac{x^2 - 1}{x - 1} = x + 1$. Substituting this into the expression for $g(x)$, we get: $g(x) = (x + 1) \cdot |x - 1|$ for $x \neq 1$. 2. **Extending g(x) to a polynomial on all real numbers:** A polynomial is a function that can be expressed in the form $P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$, where $a_i$ are constants and $n$ is a non-negative integer. Polynomials are defined and continuous for all...

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Let's analyze the function $g(x) = \frac{x^2 - 1}{x - 1} \cdot |x - 1|$. 1. **Simplifying g(x) for x ≠ 1:** For $x \neq 1$, we know that $x^2 - 1 = (x - 1)(x + 1)$. Therefore, $\frac{x^2 - 1}{x - 1} = x + 1$. Substituting this into the expression for $g(x)$, we get: $g(x) = (x + 1) \cdot |x - 1|$ for $x \neq 1$. 2. **Extending g(x) to a polynomial on all real numbers:** A polynomial is a function that can be expressed in the form $P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0$, where $a_i$ are constants and $n$ is a non-negative integer. Polynomials are defined and continuous for all real numbers. The function $g(x) = (x + 1)|x - 1|$ involves the absolute value function $|x - 1|$. The absolute value function is not a polynomial because its graph has a sharp corner at $x = 1$. Specifically, for $x \geq 1$, $|x - 1| = x - 1$, so $g(x) = (x + 1)(x - 1) = x^2 - 1$. For $x < 1$, $|x - 1| = -(x - 1)$, so $g(x) = (x + 1)(-(x - 1)) = -(x^2 - 1) = -x^2 + 1$. Since the definition of $g(x)$ changes based on the sign of $(x - 1)$, and it involves an absolute value, it cannot be represented by a single polynomial expression for all real numbers. Therefore, $g(x)$ cannot be extended to a polynomial on all real numbers. 3. **Differentiability of g(x) at x = 1:** To determine if $g(x)$ is differentiable at $x = 1$, we need to examine the limit definition of the derivative: $g'(1) = \lim_{h \to 0} \frac{g(1 + h) - g(1)}{h}$. First, let's find $g(1)$. Since the original expression for $g(x)$ is undefined at $x=1$, we consider the limit as $x o 1$. As shown in part 1, for $x eq 1$, $g(x) = (x+1)|x-1|$. The limit as $x o 1$ of $g(x)$ is $(1+1)|1-1| = 2 imes 0 = 0$. So, if we were to define $g(1)=0$ to make the function continuous, we would proceed with the derivative calculation. Let's evaluate the limit of the difference quotient: For $h > 0$, $1 + h > 1$, so $|1 + h - 1| = |h| = h$. Thus, $g(1 + h) = (1 + h + 1)|1 + h - 1| = (2 + h)h = 2h + h^2$. For $h < 0$, $1 + h < 1$, so $|1 + h - 1| = |-h| = -h$. Thus, $g(1 + h) = (1 + h + 1)|1 + h - 1| = (2 + h)(-h) = -2h - h^2$. Now, we examine the left-hand and right-hand limits of the difference quotient: Right-hand derivative ($h o 0^+$): $\lim_{h \to 0^+} \frac{(2h + h^2) - 0}{h} = \lim_{h \to 0^+} \frac{h(2 + h)}{h} = \lim_{h \to 0^+} (2 + h) = 2$. Left-hand derivative ($h o 0^-$): $\lim_{h \to 0^-} \frac{(-2h - h^2) - 0}{h} = \lim_{h \to 0^-} \frac{h(-2 - h)}{h} = \lim_{h \to 0^-} (-2 - h) = -2$. Since the left-hand derivative ($-2$) and the right-hand derivative ($2$) are not equal, $g(x)$ is not differentiable at $x = 1$. 4. **Conceptual mistake in the student's reasoning:** The student's mistake lies in assuming that because a function can be simplified to a polynomial for *some* values of $x$, it is a polynomial everywhere. The simplification $\frac{x^2 - 1}{x - 1} = x + 1$ is only valid when $x \neq 1$. The function $g(x)$ includes the term $|x - 1|$, which introduces a non-polynomial behavior (a sharp corner) at $x = 1$. A function must be expressible by a single polynomial formula for *all* real numbers to be considered a polynomial. The presence of the absolute value function, which creates a piecewise definition with different algebraic forms on either side of $x=1$, prevents $g(x)$ from being a polynomial.

Result

#2

Winning Votes

0 / 3

Average Score

87

Total Score

96

Overall Comments

This is a very strong and correct answer. It successfully addresses all parts of the prompt with accurate calculations and clear explanations. The reasoning is sound and easy for the target audience to understand. Its only minor weakness, when compared to Answer A, is that its argument for why the function is not a polynomial is slightly less rigorous, relying more on the intuitive properties of the absolute value graph rather than a formal proof.

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Correctness

Weight 45%
100

The answer is entirely correct. All calculations, including the one-sided derivatives (2 and -2), are accurate, and the conclusions for each part of the prompt are correct.

Reasoning Quality

Weight 20%
85

The reasoning is correct and logical. However, the argument in part 2, while correct, is less mathematically rigorous than Answer A's. It relies on the intuitive idea of a 'sharp corner' rather than a more formal proof like the polynomial identity theorem.

Completeness

Weight 15%
100

The answer is fully complete. It systematically addresses all four questions posed in the prompt, providing detailed calculations and explanations for each.

Clarity

Weight 10%
90

The answer is very clear and well-written. The steps for each calculation are laid out logically, making it easy to follow. The use of LaTeX for mathematical notation enhances readability.

Instruction Following

Weight 10%
100

The answer perfectly adheres to the prompt's instructions. It answers all four parts, provides the required calculations and justifications, and maintains a tone suitable for a strong high-school student.

Total Score

82

Overall Comments

Answer B is correct and complete in its core mathematical content. It simplifies g(x) properly, identifies the piecewise nature, computes one-sided derivatives correctly, and explains the conceptual error. However, the justification for non-polynomiality in part 2 is weaker — it relies mainly on the informal observation that absolute value "has a sharp corner" and "changes definition," without the clean polynomial identity argument (two polynomials agreeing on an interval must be identical) that makes the proof rigorous. The conceptual explanation in part 4 is also somewhat superficial compared to Answer A.

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Correctness

Weight 45%
85

All core mathematical claims are correct: simplification, piecewise form, one-sided derivatives, and conclusion. However, the non-polynomiality argument in part 2 is informal (relies on 'sharp corner' intuition rather than a rigorous algebraic proof), which slightly weakens the correctness of the justification even if the conclusion is right.

Reasoning Quality

Weight 20%
70

Reasoning is adequate but not deep. The non-polynomiality argument rests on the observation that absolute value creates a piecewise definition and a 'sharp corner,' which is intuitive but not a rigorous proof. No use of the polynomial identity principle or other formal argument.

Completeness

Weight 15%
85

All four parts are addressed. The extension discussion covers the piecewise form and mentions the undefined point. One-sided derivatives are computed correctly. Slightly less depth in the conceptual explanation but overall complete.

Clarity

Weight 10%
80

Clear and readable with good use of LaTeX formatting. The structure follows the four parts cleanly. Slightly less polished in transitions between ideas, but overall clear.

Instruction Following

Weight 10%
85

Follows all four parts of the prompt in order and addresses each required element. Slightly less explicit about the 'removable hole' distinction but otherwise well-aligned with the prompt requirements.

Judge Models OpenAI GPT-5.4

Total Score

84

Overall Comments

Answer B is correct on the main mathematics and covers all four parts, including the correct simplification, piecewise formulas, and one-sided derivative calculation showing non-differentiability at x = 1. However, its justification for why no polynomial extension exists is less rigorous than Answer A's, because it mainly appeals to the presence of absolute value and a corner rather than explicitly proving impossibility via polynomial identity on intervals. It is clear and serviceable but not as precise or conceptually complete.

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Correctness

Weight 45%
84

The main conclusions are correct: g(x) = (x + 1)|x - 1| for x ≠ 1, the piecewise forms x^2 - 1 and 1 - x^2 are correct, and the derivative test at x = 1 is computed correctly. The weaker point is the argument about polynomial extension, which is stated correctly but not proved as rigorously as it could be.

Reasoning Quality

Weight 20%
76

Reasoning is generally sound and easy to follow, but the argument against polynomial extension leans on the presence of absolute value and a corner rather than fully proving impossibility through polynomial uniqueness on intervals. This is acceptable but less robust for a mathematically rigorous benchmark response.

Completeness

Weight 15%
88

It addresses all four requested parts and includes the key derivative calculation. The only missing element relative to a stronger benchmark answer is a more complete proof that no single polynomial can match both interval formulas.

Clarity

Weight 10%
86

The answer is clear and readable, with a straightforward step-by-step structure. Some phrasing is a bit repetitive, and the conceptual distinction between the original function and an extension could be made more sharply.

Instruction Following

Weight 10%
91

It follows the requested structure and answers each numbered part in an understandable way. It is slightly less aligned with the prompt's emphasis on rigorous justification for the polynomial-extension question.

Comparison Summary

Final rank order is determined by judge-wise rank aggregation (average rank + Borda tie-break). Average score is shown for reference.

Judges: 3

Winning Votes

3 / 3

Average Score

94
View this answer

Winning Votes

0 / 3

Average Score

87
View this answer

Judging Results

Judge Models OpenAI GPT-5.4

Why This Side Won

Answer A wins because it performs better on the most important weighted criteria, especially correctness and reasoning quality. Both answers reach the correct conclusions, but Answer A gives a more rigorous proof that no polynomial can agree with g on all x ≠ 1, carefully distinguishes the original function from a possible extension, and provides a sharper conceptual diagnosis of the student's mistake. Answer B is good but relies more on heuristic statements about absolute value and corners, so its justification is less mathematically robust.

Why This Side Won

Answer A wins primarily on the correctness and reasoning quality criteria (combined weight 65%). While both answers reach correct conclusions, Answer A provides a substantially more rigorous justification in part 2 using the polynomial identity principle (two polynomials agreeing on an infinite set must be identical), which is the mathematically proper argument. Answer A also offers a second independent proof via non-differentiability, and its conceptual explanation in part 4 is more precise and insightful. These advantages on the two highest-weighted criteria (correctness 45%, reasoning quality 20%) decisively favor Answer A.

Why This Side Won

Answer A is the winner due to its superior reasoning quality. While both answers are correct and complete, Answer A provides a more mathematically rigorous and insightful argument in part 2 by invoking the identity theorem for polynomials. This demonstrates a deeper conceptual understanding. Answer A also does a slightly better job of connecting the different parts of the problem, for example, by noting that the non-differentiability found in part 3 serves as an independent proof for the conclusion in part 2.

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