Home›Calculus 1›Continuity and types of discontinuity›Problem 1.478

Continuity and types of discontinuity

Problem 1.478 · hard

Find every point where \( \displaystyle f(x) = \frac{- 10 x - 40}{x^{2} + x - 12} \) 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
\[ x^{2} + x - 12 = \left(x - 3\right) \left(x + 4\right) \]
Factor the denominator.✓ Proved
\[ \frac{- 10 x - 40}{x^{2} + x - 12} = - \frac{10}{x - 3} \]
The factor (x + 4) cancels.✓ Proved
\[ \lim_{x \to -4^+}\left(\frac{- 10 x - 40}{x^{2} + x - 12}\right) = \frac{10}{7} \]
At x = -4 the limit exists, but f(-4) is undefined: a removable discontinuity.✓ Proved
\[ \left. -10 \right|_{\substack{ x=3 }} = -10 \]
At x = 3 the numerator is not 0 while the denominator is, so |f(x)| grows without bound: an infinite discontinuity.✓ Proved

Answer \( \text{removable at } x = -4;\ \text{infinite at } x = 3 \)

✓ 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 standard true fact: a rational function is continuous wherever its denominator is nonzero, so only zeros of the denominator can be discontinuities. It makes no algebraic claim that a script could test.
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: {-4: 'removable', 3: 'infinite'}

Reviewers

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

Senior review claude-sonnet-5-5, 2026-10-09: pass — The factorization, cancellation, removable discontinuity at x=-4 (limit 10/7, f undefined there) and infinite discontinuity at x=3 are all correct and match the stated answer. Line 5 is awkwardly phrased because it substitutes into the reduced numerator, but nothing in it is false.

gpt-oss:20b: dismiss — Line 5 evaluates the reduced numerator -10 (after cancelling x+4) at x=3, which is correct. The original numerator is -70 there, but both are nonzero, so the infinite-discontinuity classification is right and no false statement is made.

Every verdict on record (5)

gpt-oss:20b: pass 2026-10-09
qwen3.6:27b-mlx: pass 2026-10-09
claude-sonnet-5-5: pass 2026-10-09 — The factorization, cancellation, removable discontinuity at x=-4 (limit 10/7, f undefined there) and infinite discontinuity at x=3 are all correct and match the stated answer. Line 5 is awkwardly phrased because it substitutes into the reduced numerator, but nothing in it is false.
gpt-oss:20b: fail (misleading) 2026-10-09 — The solution incorrectly states that the numerator evaluates to -10 at x=3; it actually equals -70. This misleads the reader about the calculation of the infinite discontinuity.
qwen3.6:27b-mlx: pass 2026-10-09

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.