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Continuity and types of discontinuity

Problem 1.543 · hard

Find every point where \( \displaystyle f(x) = \frac{- 3 x^{2} - 9 x - 6}{x^{2} - x - 6} \) 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.
Reviewed
\[ x^{2} - x - 6 = \left(x - 3\right) \left(x + 2\right) \]
Factor the denominator.✓ Proved
\[ \frac{- 3 x^{2} - 9 x - 6}{x^{2} - x - 6} = \frac{- 3 x - 3}{x - 3} \]
The factor (x + 2) cancels.✓ Proved
\[ \lim_{x \to -2^+}\left(\frac{- 3 x^{2} - 9 x - 6}{x^{2} - x - 6}\right) = - \frac{3}{5} \]
At x = -2 the limit exists, but f(-2) is undefined: a removable discontinuity.✓ Proved
\[ \left. - 3 x - 3 \right|_{\substack{ x=3 }} = -12 \]
At x = 3 the numerator is not 0 while the denominator is, so |f(x)| grows without bound: an infinite discontinuity.✓ Proved

Answer \( \text{removable at } x = -2;\ \text{infinite at } x = 3 \)

✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation.

The full receipt

Line	Status	Checked by	Detail
1	Reviewed	claude-sonnet-5-5	claude-sonnet-5-5: This is a sentence, not an equation. A rational function is continuous wherever its denominator is nonzero, so only the denominator's zeros can be discontinuities.
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: {-2: 'removable', 3: 'infinite'}

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: pass

Senior review claude-sonnet-5-5, 2026-10-11: pass — The numerator factors as -3(x+1)(x+2) and the denominator as (x-3)(x+2). The removable discontinuity at -2 (limit -3/5) and the infinite one at 3 (numerator -12, denominator 0) are both correct and match the stated answer.

gpt-oss:20b: dismiss — The objection's arithmetic is wrong: -3(x+1)/(x-3) at x=-2 is -3(-1)/(-5) = 3/(-5) = -3/5, which matches line 4. The one-sided limit is also valid, since the two-sided limit exists and equals it.

Every verdict on record (5)

qwen3.6:27b-mlx: pass 2026-10-11
gpt-oss:20b: pass 2026-10-11
claude-sonnet-5-5: pass 2026-10-11 — The numerator factors as -3(x+1)(x+2) and the denominator as (x-3)(x+2). The removable discontinuity at -2 (limit -3/5) and the infinite one at 3 (numerator -12, denominator 0) are both correct and match the stated answer.
gpt-oss:20b: fail (error) 2026-10-11 — The limit at x=-2 is mis‑calculated: -3(x+1)/(x-3) evaluated at x=-2 gives 3/5, not -3/5. The conclusion that the discontinuity is removable is correct, but the numerical value of the limit is wrong. The one‑sided limit is unnecessary; the two‑sided limit exists.
qwen3.6:27b-mlx: pass 2026-10-11

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 2026-10-11 with SymPy 1.14.0.