Home›Calculus 1›Continuity and types of discontinuity›Problem 1.196

Continuity and types of discontinuity

Problem 1.196 · hard

Find every point where \( \displaystyle f(x) = \frac{- 12 x - 24}{x^{2} + 3 x + 2} \) 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.
Reviewed
\[ x^{2} + 3 x + 2 = \left(x + 1\right) \left(x + 2\right) \]
Factor the denominator.✓ Proved
\[ \frac{- 12 x - 24}{x^{2} + 3 x + 2} = - \frac{12}{x + 1} \]
The factor (x + 2) cancels.✓ Proved
\[ \lim_{x \to -2^+}\left(\frac{- 12 x - 24}{x^{2} + 3 x + 2}\right) = 12 \]
At x = -2 the limit exists, but f(-2) is undefined: a removable discontinuity.✓ Proved
\[ \left. -12 \right|_{\substack{ x=-1 }} = -12 \]
At x = -1 the numerator is not 0 while the denominator is, so |f(x)| grows without bound: an infinite discontinuity.✓ Proved

Answer \( \text{removable at } x = -2;\ \text{infinite at } x = -1 \)

✓ 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, which is a standard true fact.
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', -1: 'infinite'}

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: fail (style) — [domain objection, downgraded to style] Step 5 incorrectly substitutes x = -1 into the simplified expression -12/(x+1) to get -12, which is a finite number. This contradicts the conclusion of an infinite discontinuity; the correct reasoning is that the limit of the simplified function as x approaches -1 is undefined (approaches +/- infinity), not that the substitution yields a non-zero finite value.

Senior review claude-sonnet-5-5, 2026-10-04: pass — The factorization is correct: f = -12(x+2)/((x+1)(x+2)). The discontinuity at x=-2 is removable with limit 12, and the one at x=-1 is infinite because the numerator is -12 there. Line 5 is terse, but the value -12 is the original numerator evaluated at -1, so it is true.

qwen3.6:27b-mlx: dismiss — The original numerator at x=-1 is -12(-1)-24 = -12, which is nonzero, so the numerator does not vanish there. The denominator does, so |f| blows up and the discontinuity is infinite; the claim that the limit is finite is false.

Every verdict on record (5)

qwen3.6:27b-mlx: fail (style) 2026-10-04 — [domain objection, downgraded to style] Step 5 incorrectly substitutes x = -1 into the simplified expression -12/(x+1) to get -12, which is a finite number. This contradicts the conclusion of an infinite discontinuity; the correct reasoning is that the limit of the simplified function as x approaches -1 is undefined (approaches +/- infinity), not that the substitution yields a non-zero finite value.
gpt-oss:20b: pass 2026-10-04
claude-sonnet-5-5: pass 2026-10-04 — The factorization is correct: f = -12(x+2)/((x+1)(x+2)). The discontinuity at x=-2 is removable with limit 12, and the one at x=-1 is infinite because the numerator is -12 there. Line 5 is terse, but the value -12 is the original numerator evaluated at -1, so it is true.
qwen3.6:27b-mlx: fail (error) 2026-10-04 — Step 5 incorrectly substitutes x = -1 into the simplified numerator -12 to claim the numerator is non-zero. The original numerator is -12x - 24, which equals 0 at x = -1. Since both numerator and denominator are zero at x = -1, the limit is finite (-12), making it a removable discontinuity, not infinite.
gpt-oss:20b: pass 2026-10-04

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 2026-10-04 with SymPy 1.14.0.