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Continuity and types of discontinuity

Problem 1.286 · hard

Find every point where \( \displaystyle f(x) = \frac{- 3 x - 3}{x^{2} - 3 x - 4} \) is discontinuous, and classify each discontinuity as removable, a jump, or infinite.
  1. A rational function is continuous wherever its denominator is not zero, so only the zeros of the denominator can be discontinuities.
    Reviewed
  2. \[ x^{2} - 3 x - 4 = \left(x - 4\right) \left(x + 1\right) \]
    Factor the denominator.✓ Proved
  3. \[ \frac{- 3 x - 3}{x^{2} - 3 x - 4} = - \frac{3}{x - 4} \]
    The factor (x + 1) cancels.✓ Proved
  4. \[ \lim_{x \to -1^+}\left(\frac{- 3 x - 3}{x^{2} - 3 x - 4}\right) = \frac{3}{5} \]
    At x = -1 the limit exists, but f(-1) is undefined: a removable discontinuity.✓ Proved
  5. \[ \left. -3 \right|_{\substack{ x=4 }} = -3 \]
    At x = 4 the numerator is not 0 while the denominator is, so |f(x)| grows without bound: an infinite discontinuity.✓ Proved
Answer \( \text{removable at } x = -1;\ \text{infinite at } x = 4 \)

✓ 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
LineStatusChecked byDetail
1Reviewedclaude-sonnet-5-5claude-sonnet-5-5: This is a standard fact: a rational function is continuous wherever its denominator is nonzero, so only the denominator's zeros can be discontinuities. It is a sentence, not an algebraic equivalence, so no script applies.
2✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
3✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
4✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
5✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
answer, a second way✓ Checked independentlysympy 1.14.0 + mpmath 1.3.0zeros of the denominator found by solve, each probed at ±1e-9: {-1: 'removable', 4: 'infinite'}

Reviewers

  • gpt-oss:20b: pass
  • qwen3.6:27b-mlx: fail (error) — Step 5 incorrectly evaluates the numerator of the simplified function -3/(x-4) at x=4 as -3, whereas the numerator is the constant -3, so the value is -3, but the reasoning 'numerator is not 0' refers to the simplified form. However, the original numerator at x=4 is -15. The step claims 'Subs(-3, x, 4) = -3', which is syntactically confusing (substituting into a constant?) and logically weak. More importantly, step 3 cancels (x+1) to get -3/(x-4). Step 5 says 'At x=4 the numerator is not 0'. In the simplified form -3/(x-4), the numerator is -3, which is not 0. This is correct. But the label 'Subs(-3, x, 4)' is nonsense syntax for a CAS check. The real issue is that step 5's justification is slightly garbled but the conclusion is correct. Wait, let's look closer. Step 5 says 'Subs(-3, x, 4) = -3'. This is likely checking the numerator of the simplified fraction. The numerator is -3. It is not zero. So the limit is infinite. This is correct. Is there an error? Step 4 calculates the limit at -1 as 3/5. Let's check: -3/(-1-4) = -3/-5 = 3/5. Correct. Step 5: limit at 4. Numerator -3, denominator 0. Infinite discontinuity. Correct. Why would it fail? Maybe the 'Subs' line is misleading? Or is there a subtle error? The problem asks to classify. The solution does. The steps are logically sound. Is 'Subs(-3, x, 4)' a valid equation? It's not an equation in x. It's a constant evaluation. The CAS proved it. The sentence says 'numerator is not 0'. This is true. The conclusion is correct. I will pass it unless I find a hidden error. Ah, look at step 3: `(-3*x - 3)/(x**2 - 3*x - 4) = -3/(x - 4)`. This equality holds for x != -1. The solution uses this to find limits. This is standard. Step 5: `Subs(-3, x, 4) = -3`. This is trivial. The text says 'numerator is not 0'. In the simplified form, the numerator is -3. In the original form, the numerator is -15. Neither is 0. So the classification is correct. I see no error.

Senior review claude-sonnet-5-5, 2026-10-05: pass — The factorization, the cancellation, the limit 3/5 at x=-1 and the infinite behavior at x=4 are all correct, and they match the stated answer. Line 5's Subs(-3, x, 4) is a slightly odd way to show the reduced numerator is nonzero at 4, but it is not wrong.

  • qwen3.6:27b-mlx: dismiss — The objection contradicts itself, calling -3/(x-4) wrong and then giving the same expression as correct. Since -3x-3 = -3(x+1), cancelling (x+1) really does give -3/(x-4).
Every verdict on record (5)
  • qwen3.6:27b-mlx: fail (error) 2026-10-05 — Step 5 incorrectly evaluates the numerator of the simplified function -3/(x-4) at x=4 as -3, whereas the numerator is the constant -3, so the value is -3, but the reasoning 'numerator is not 0' refers to the simplified form. However, the original numerator at x=4 is -15. The step claims 'Subs(-3, x, 4) = -3', which is syntactically confusing (substituting into a constant?) and logically weak. More importantly, step 3 cancels (x+1) to get -3/(x-4). Step 5 says 'At x=4 the numerator is not 0'. In the simplified form -3/(x-4), the numerator is -3, which is not 0. This is correct. But the label 'Subs(-3, x, 4)' is nonsense syntax for a CAS check. The real issue is that step 5's justification is slightly garbled but the conclusion is correct. Wait, let's look closer. Step 5 says 'Subs(-3, x, 4) = -3'. This is likely checking the numerator of the simplified fraction. The numerator is -3. It is not zero. So the limit is infinite. This is correct. Is there an error? Step 4 calculates the limit at -1 as 3/5. Let's check: -3/(-1-4) = -3/-5 = 3/5. Correct. Step 5: limit at 4. Numerator -3, denominator 0. Infinite discontinuity. Correct. Why would it fail? Maybe the 'Subs' line is misleading? Or is there a subtle error? The problem asks to classify. The solution does. The steps are logically sound. Is 'Subs(-3, x, 4)' a valid equation? It's not an equation in x. It's a constant evaluation. The CAS proved it. The sentence says 'numerator is not 0'. This is true. The conclusion is correct. I will pass it unless I find a hidden error. Ah, look at step 3: `(-3*x - 3)/(x**2 - 3*x - 4) = -3/(x - 4)`. This equality holds for x != -1. The solution uses this to find limits. This is standard. Step 5: `Subs(-3, x, 4) = -3`. This is trivial. The text says 'numerator is not 0'. In the simplified form, the numerator is -3. In the original form, the numerator is -15. Neither is 0. So the classification is correct. I see no error.
  • gpt-oss:20b: pass 2026-10-05
  • claude-sonnet-5-5: pass 2026-10-05 — The factorization, the cancellation, the limit 3/5 at x=-1 and the infinite behavior at x=4 are all correct, and they match the stated answer. Line 5's Subs(-3, x, 4) is a slightly odd way to show the reduced numerator is nonzero at 4, but it is not wrong.
  • qwen3.6:27b-mlx: fail (error) 2026-10-05 — Step 3 incorrectly simplifies the function by canceling (x+1) without accounting for the factor of -3 in the numerator, resulting in -3/(x-4) instead of the correct -3/(x-4). While the final classification is correct, the intermediate algebraic simplification is wrong, which would mislead a student about the simplified form of the function.
  • gpt-oss:20b: pass 2026-10-05

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 2026-10-05 with SymPy 1.14.0.

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