Continuity and types of discontinuity
Problem 1.540 · hard
Find every point where \( \displaystyle f(x) = \frac{- 3 x^{2} + 15 x - 18}{x^{2} - 8 x + 15} \) 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
- \[ x^{2} - 8 x + 15 = \left(x - 5\right) \left(x - 3\right) \]Factor the denominator.✓ Proved
- \[ \frac{- 3 x^{2} + 15 x - 18}{x^{2} - 8 x + 15} = \frac{6 - 3 x}{x - 5} \]The factor (x − 3) cancels.✓ Proved
- \[ \lim_{x \to 3^+}\left(\frac{- 3 x^{2} + 15 x - 18}{x^{2} - 8 x + 15}\right) = \frac{3}{2} \]At x = 3 the limit exists, but f(3) is undefined: a removable discontinuity.✓ Proved
- \[ \left. 6 - 3 x \right|_{\substack{ x=5 }} = -9 \]At x = 5 the numerator is not 0 while the denominator is, so |f(x)| grows without bound: an infinite discontinuity.✓ Proved
Answer \( \text{removable at } x = 3;\ \text{infinite at } x = 5 \)
✓ 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: 'removable', 5: 'infinite'} |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-10-11gpt-oss:20b: pass 2026-10-11gpt-oss:20b: pass 2026-10-11qwen3.6:27b-mlx: pass 2026-10-11 — The solution correctly identifies the points of discontinuity by factoring the denominator and numerator. It properly classifies x=3 as a removable discontinuity because the limit exists, and x=5 as an infinite discontinuity because the simplified numerator is non-zero at that point while the denominator is zero.
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.