Continuity and types of discontinuity
Problem 1.482 · hard
Find every point where \( \displaystyle f(x) = \frac{- 6 x - 6}{x^{2} - x - 2} \) is discontinuous, and classify each discontinuity as removable, a jump, or infinite.
- A rational function is continuous wherever its denominator is not zero, so only the zeros of the denominator can be discontinuities.
- \[ x^{2} - x - 2 = \left(x - 2\right) \left(x + 1\right) \]Factor the denominator.✓ Proved
- \[ \frac{- 6 x - 6}{x^{2} - x - 2} = - \frac{6}{x - 2} \]The factor (x + 1) cancels.✓ Proved
- \[ \lim_{x \to -1^+}\left(\frac{- 6 x - 6}{x^{2} - x - 2}\right) = 2 \]At x = -1 the limit exists, but f(-1) is undefined: a removable discontinuity.✓ Proved
- \[ \left. -6 \right|_{\substack{ x=2 }} = -6 \]At x = 2 the numerator is not 0 while the denominator is, so |f(x)| grows without bound: an infinite discontinuity.✓ Proved
Answer \( \text{removable at } x = -1;\ \text{infinite at } x = 2 \)
Lines: 4 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. A reviewer's objection to the explanation stands; see the receipt.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | Not checked | — | a sentence; read, not computed |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 5 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | zeros of the denominator found by solve, each probed at ±1e-9: {-1: 'removable', 2: 'infinite'} |
Reviewers
gpt-oss:20b: fail (error) — The numerator at x=2 is -18, not -6. The solution incorrectly evaluates the numerator, though the conclusion about an infinite discontinuity at x=2 is still correct.qwen3.6:27b-mlx: fail (misleading) — Step 5 incorrectly states that the numerator is -6 at x=2; the numerator is actually -18. While the conclusion of an infinite discontinuity is correct, the reasoning provided is factually wrong and would teach a student to miscalculate function values.
Every verdict on record (4)
gpt-oss:20b: fail (error) 2026-10-09 — The numerator at x=2 is -18, not -6. The solution incorrectly evaluates the numerator, though the conclusion about an infinite discontinuity at x=2 is still correct.qwen3.6:27b-mlx: fail (misleading) 2026-10-09 — Step 5 incorrectly states that the numerator is -6 at x=2; the numerator is actually -18. While the conclusion of an infinite discontinuity is correct, the reasoning provided is factually wrong and would teach a student to miscalculate function values.gpt-oss:20b: fail (misleading) 2026-10-09 — Step 5 incorrectly substitutes the numerator at x=2; it should be -18, not -6. The conclusion about an infinite discontinuity is correct, but the sentence misleads a student about the value of the numerator.qwen3.6:27b-mlx: fail (error) 2026-10-09 — Step 5 incorrectly substitutes x=2 into the numerator of the simplified function (-6) instead of the original numerator (-6x-6), which is -18. While the conclusion of an infinite discontinuity is correct, the reasoning is mathematically flawed.
Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate
method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence,
not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by
generator:structured/continuity_classify, checked 2026-10-09 with SymPy 1.14.0.