Continuity and types of discontinuity
Problem 1.194 · hard
Find every point where \( \displaystyle f(x) = \frac{3 x^{2} - 3 x - 36}{x^{2} - 7 x + 12} \) 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
- \[ x^{2} - 7 x + 12 = \left(x - 4\right) \left(x - 3\right) \]Factor the denominator.✓ Proved
- \[ \frac{3 x^{2} - 3 x - 36}{x^{2} - 7 x + 12} = \frac{3 x + 9}{x - 3} \]The factor (x − 4) cancels.✓ Proved
- \[ \lim_{x \to 4^+}\left(\frac{3 x^{2} - 3 x - 36}{x^{2} - 7 x + 12}\right) = 21 \]At x = 4 the limit exists, but f(4) is undefined: a removable discontinuity.✓ Proved
- \[ \left. 3 x + 9 \right|_{\substack{ x=3 }} = 18 \]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 | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | 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: {3: 'infinite', 4: 'removable'} |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass — The solution correctly identifies the points of discontinuity and classifies them based on the behavior of the simplified function and limits. The logic is sound and the steps are clearly justified.
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-10-04 — The solution correctly identifies the points of discontinuity and classifies them based on the behavior of the simplified function and limits. The logic is sound and the steps are clearly justified.gpt-oss:20b: pass 2026-10-04qwen3.6:27b-mlx: pass 2026-10-04gpt-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.