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Continuity and types of discontinuity

Problem 1.289 · hard

Find every point where \( \displaystyle f(x) = \frac{- x^{2} + 3 x + 4}{x^{2} - 16} \) 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.
  2. \[ x^{2} - 16 = \left(x - 4\right) \left(x + 4\right) \]
    Factor the denominator.✓ Proved
  3. \[ \frac{- x^{2} + 3 x + 4}{x^{2} - 16} = \frac{- x - 1}{x + 4} \]
    The factor (x − 4) cancels.✓ Proved
  4. \[ \lim_{x \to 4^+}\left(\frac{- x^{2} + 3 x + 4}{x^{2} - 16}\right) = - \frac{5}{8} \]
    At x = 4 the limit exists, but f(4) is undefined: a removable discontinuity.✓ Proved
  5. \[ \left. - x - 1 \right|_{\substack{ x=-4 }} = 3 \]
    At x = -4 the numerator is not 0 while the denominator is, so |f(x)| grows without bound: an infinite discontinuity.✓ Proved
Answer \( \text{removable at } x = 4;\ \text{infinite at } x = -4 \)

Lines: 4 proved, 1 not checked. 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
1Not checked—a sentence; read, not computed
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: {-4: 'infinite', 4: 'removable'}

Reviewers

  • gpt-oss:20b: pass
  • qwen3.6:27b-mlx: fail (style) — [domain objection, downgraded to style] The solution fails to identify x = -1 as a discontinuity. The numerator -x^2 + 3x + 4 factors to -(x+1)(x-4), so x = -1 is a root of the numerator but not the denominator, meaning f(-1) = 0 and the function is continuous there. However, the problem asks for points where f is discontinuous. The zeros of the denominator are x=4 and x=-4. The solution correctly identifies these. Wait, let me re-read carefully. The numerator is -x^2 + 3x + 4. Roots: x = [-3 +/- sqrt(9 - 4(-1)(4))]/-2 = [-3 +/- 5]/-2. x1 = 2/-2 = -1. x2 = -8/-2 = 4. So numerator is -(x+1)(x-4). Denominator is (x-4)(x+4). f(x) = [-(x+1)(x-4)] / [(x-4)(x+4)]. For x != 4, f(x) = -(x+1)/(x+4). The domain excludes x=4 and x=-4. At x=4, limit exists (-5/8), so removable. At x=-4, limit is infinite, so infinite discontinuity. The solution lists x=4 and x=-4. Are there any other discontinuities? No. Rational functions are continuous everywhere else. The solution seems correct in its conclusions. Let me check the 'proved' equations. Eq 3: (-x-1)/(x+4). If x=4, (-4-1)/(8) = -5/8. Correct. If x=-4, num is 3, den is 0. Correct. The text says 'only the zeros of the denominator can be discontinuities'. This is true for rational functions. The solution finds x=4 and x=-4. It classifies them correctly. Why did I think it failed? Ah, I misread my own thought process. The solution is actually correct. Let me double check if I missed anything. 'Find every point...'. The points are 4 and -4. Classification: removable at 4, infinite at -4. The solution states exactly this. The steps are logical. The algebra is marked proved. The sentences are correct. Verdict should be pass.
Every verdict on record (4)
  • qwen3.6:27b-mlx: fail (style) 2026-10-05 — [domain objection, downgraded to style] The solution fails to identify x = -1 as a discontinuity. The numerator -x^2 + 3x + 4 factors to -(x+1)(x-4), so x = -1 is a root of the numerator but not the denominator, meaning f(-1) = 0 and the function is continuous there. However, the problem asks for points where f is discontinuous. The zeros of the denominator are x=4 and x=-4. The solution correctly identifies these. Wait, let me re-read carefully. The numerator is -x^2 + 3x + 4. Roots: x = [-3 +/- sqrt(9 - 4(-1)(4))]/-2 = [-3 +/- 5]/-2. x1 = 2/-2 = -1. x2 = -8/-2 = 4. So numerator is -(x+1)(x-4). Denominator is (x-4)(x+4). f(x) = [-(x+1)(x-4)] / [(x-4)(x+4)]. For x != 4, f(x) = -(x+1)/(x+4). The domain excludes x=4 and x=-4. At x=4, limit exists (-5/8), so removable. At x=-4, limit is infinite, so infinite discontinuity. The solution lists x=4 and x=-4. Are there any other discontinuities? No. Rational functions are continuous everywhere else. The solution seems correct in its conclusions. Let me check the 'proved' equations. Eq 3: (-x-1)/(x+4). If x=4, (-4-1)/(8) = -5/8. Correct. If x=-4, num is 3, den is 0. Correct. The text says 'only the zeros of the denominator can be discontinuities'. This is true for rational functions. The solution finds x=4 and x=-4. It classifies them correctly. Why did I think it failed? Ah, I misread my own thought process. The solution is actually correct. Let me double check if I missed anything. 'Find every point...'. The points are 4 and -4. Classification: removable at 4, infinite at -4. The solution states exactly this. The steps are logical. The algebra is marked proved. The sentences are correct. Verdict should be pass.
  • gpt-oss:20b: pass 2026-10-05
  • qwen3.6:27b-mlx: fail (style) 2026-10-05 — [domain objection, downgraded to style] The solution fails to identify x = -1 as a discontinuity. The numerator -x^2 + 3x + 4 factors to -(x+1)(x-4), so x = -1 is a zero of the numerator but not the denominator, meaning the function is defined and continuous there. However, the problem asks for points where f is discontinuous. Wait, if f is defined at x=-1, it is continuous there. Let me re-read. f(x) is a rational function. It is continuous everywhere except where the denominator is zero. The denominator is zero at x=4 and x=-4. So the only discontinuities are at 4 and -4. The solution correctly identifies these two points. Is there an error? Let's check the classification. At x=4, limit exists (-5/8), so removable. Correct. At x=-4, limit is infinite, so infinite. Correct. Why did I think x=-1 was a discontinuity? It's not. The function is defined at x=-1. So the solution is actually correct. Let me double check the factorization. Numerator: -x^2 + 3x + 4. Roots: x = [-3 +/- sqrt(9 - 4(-1)(4))]/-2 = [-3 +/- sqrt(25)]/-2 = [-3 +/- 5]/-2. x1 = 2/-2 = -1. x2 = -8/-2 = 4. So numerator is -(x+1)(x-4). Denominator is (x-4)(x+4). f(x) = [-(x+1)(x-4)] / [(x-4)(x+4)]. For x != 4, f(x) = -(x+1)/(x+4). Discontinuities are at x=4 and x=-4. At x=4, limit is -(4+1)/(4+4) = -5/8. Removable. At x=-4, denominator is 0, numerator is -(-4+1) = 3 != 0. Infinite. The solution is correct. Wait, did I miss something? "Find every point where f is discontinuous". The solution finds x=4 and x=-4. It classifies them correctly. The steps are logical. Step 1 is correct. Step 2 is correct. Step 3 is correct simplification for x!=4. Step 4 calculates limit at 4. Step 5 checks behavior at -4. Everything seems fine. Why would it fail? Maybe the prompt implies I should find an error? Let me look closer. "Stated answer: removable at x = 4; infinite at x = -4". This matches the solution. Is there a subtle error? Step 3 says "The factor (x-4) cancels". This is valid for the limit calculation. Step 5 says "Subs(-x-1, x, -4) = 3". This evaluates the simplified numerator at -4. Since the simplified denominator is x+4, which is 0 at -4, and the numerator is 3, the limit is infinite. This is correct reasoning. I see no error. I will mark it as pass.
  • 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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