Home›Calculus 1›Continuity and types of discontinuity›Problem 1.392

Continuity and types of discontinuity

Problem 1.392 · hard

Find every point where \( \displaystyle f(x) = \frac{12 - 3 x}{x^{2} - 2 x - 8} \) 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.
\[ x^{2} - 2 x - 8 = \left(x - 4\right) \left(x + 2\right) \]
Factor the denominator.✓ Proved
\[ \frac{12 - 3 x}{x^{2} - 2 x - 8} = - \frac{3}{x + 2} \]
The factor (x − 4) cancels.✓ Proved
\[ \lim_{x \to 4^+}\left(\frac{12 - 3 x}{x^{2} - 2 x - 8}\right) = - \frac{1}{2} \]
At x = 4 the limit exists, but f(4) is undefined: a removable discontinuity.✓ Proved
\[ \left. -3 \right|_{\substack{ x=-2 }} = -3 \]
At x = -2 the numerator is not 0 while the denominator is, so |f(x)| grows without bound: an infinite discontinuity.✓ Proved

Answer \( \text{removable at } x = 4;\ \text{infinite at } x = -2 \)

Lines: 4 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in 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: {-2: 'infinite', 4: 'removable'}

Reviewers

gpt-oss:20b: fail (error) — Sentence 5 incorrectly evaluates the function at x=-2; it should state that the numerator is -3 (non‑zero) while the denominator is 0, leading to an infinite discontinuity, not that Subs(-3, x, -2) = -3.
qwen3.6:27b-mlx: fail (error) — Step 5 incorrectly substitutes x = -2 into the simplified numerator -3 to claim it is non-zero, ignoring that the original numerator 12 - 3x is also 0 at x = -2. Since both numerator and denominator are zero at x = -2, the limit must be evaluated (it is infinite), but the reasoning provided is mathematically false.

Senior review claude-sonnet-5-5, 2026-10-07: inconclusive — senior reply was not an object

Every verdict on record (5)

gpt-oss:20b: fail (error) 2026-10-07 — Sentence 5 incorrectly evaluates the function at x=-2; it should state that the numerator is -3 (non‑zero) while the denominator is 0, leading to an infinite discontinuity, not that Subs(-3, x, -2) = -3.
qwen3.6:27b-mlx: fail (error) 2026-10-07 — Step 5 incorrectly substitutes x = -2 into the simplified numerator -3 to claim it is non-zero, ignoring that the original numerator 12 - 3x is also 0 at x = -2. Since both numerator and denominator are zero at x = -2, the limit must be evaluated (it is infinite), but the reasoning provided is mathematically false.
claude-sonnet-5-5: inconclusive 2026-10-07 — senior reply was not an object
gpt-oss:20b: fail (style) 2026-10-07 — [domain objection, downgraded to style] The solution incorrectly evaluates the function at x = –2, claiming Subs(-3, x, -2) = –3. In fact, after cancellation the function is –3/(x+2), which is undefined at x = –2, leading to an infinite discontinuity. This misstatement could mislead a student about the behavior at that point.
qwen3.6:27b-mlx: fail (error) 2026-10-07 — Step 5 incorrectly substitutes x = -2 into the simplified numerator -3, ignoring that the original numerator 12 - 3x evaluates to 21 at x = -2. While the classification of an infinite discontinuity is correct, the reasoning that the numerator is -3 is factually wrong and misleading.

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 2026-10-07 with SymPy 1.14.0.