Continuity and types of discontinuity
Problem 1.397 · hard
Find every point where \( \displaystyle f(x) = \frac{8 x - 24}{x^{2} + x - 12} \) 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.
- \[ x^{2} + x - 12 = \left(x - 3\right) \left(x + 4\right) \]Factor the denominator.✓ Proved
- \[ \frac{8 x - 24}{x^{2} + x - 12} = \frac{8}{x + 4} \]The factor (x − 3) cancels.✓ Proved
- \[ \lim_{x \to 3^+}\left(\frac{8 x - 24}{x^{2} + x - 12}\right) = \frac{8}{7} \]At x = 3 the limit exists, but f(3) is undefined: a removable discontinuity.✓ Proved
- \[ \left. 8 \right|_{\substack{ x=-4 }} = 8 \]At x = -4 the numerator is not 0 while the denominator is, so |f(x)| grows without bound: an infinite discontinuity.✓ Proved
Answer \( \text{removable at } x = 3;\ \text{infinite at } x = -4 \)
Lines: 4 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. A reviewer's objection to the explanation stands; see the receipt.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | Not checked | — | 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: {-4: 'infinite', 3: 'removable'} |
Reviewers
gpt-oss:20b: fail (misleading) — Step 5 incorrectly substitutes 8 for the numerator; the numerator is 8x-24, which equals -56 at x=-4, but the conclusion about an infinite discontinuity is still correct. The sentence should state that the numerator is nonzero at x=-4, not that it equals 8.qwen3.6:27b-mlx: fail (error) — Step 5 incorrectly substitutes x = -4 into the simplified numerator 8, claiming it is not zero. The original numerator is 8x - 24, which equals -56 at x = -4. While the conclusion (infinite discontinuity) is correct because the non-zero numerator leads to a vertical asymptote, the reasoning 'numerator is not 0' is factually wrong for the original function and misleading regarding the simplified form (where the numerator is constant 8, which is indeed non-zero, but the step implies checking the original numerator's value incorrectly).
Every verdict on record (4)
gpt-oss:20b: fail (misleading) 2026-10-07 — Step 5 incorrectly substitutes 8 for the numerator; the numerator is 8x-24, which equals -56 at x=-4, but the conclusion about an infinite discontinuity is still correct. The sentence should state that the numerator is nonzero at x=-4, not that it equals 8.qwen3.6:27b-mlx: fail (error) 2026-10-07 — Step 5 incorrectly substitutes x = -4 into the simplified numerator 8, claiming it is not zero. The original numerator is 8x - 24, which equals -56 at x = -4. While the conclusion (infinite discontinuity) is correct because the non-zero numerator leads to a vertical asymptote, the reasoning 'numerator is not 0' is factually wrong for the original function and misleading regarding the simplified form (where the numerator is constant 8, which is indeed non-zero, but the step implies checking the original numerator's value incorrectly).qwen3.6:27b-mlx: fail (error) 2026-10-07 — The solution incorrectly identifies x = 3 as a root of the denominator x^2 + x - 12. The actual roots are x = 3 is not a root (3^2+3-12 = 0? 9+3-12=0, yes it is). Wait, let me re-check. 3^2 + 3 - 12 = 9 + 3 - 12 = 0. So x=3 IS a root. Let me check the numerator at x=3: 8(3)-24 = 0. So it is 0/0. Let me check x=-4: (-4)^2 + (-4) - 12 = 16 - 4 - 12 = 0. Numerator at -4: 8(-4)-24 = -56 != 0. So the classification is correct. Why did I think it was wrong? Let me re-read the factorization. (x-3)(x+4) = x^2 + 4x - 3x - 12 = x^2 + x - 12. This is correct. The limit at x=3 is 8/(3+4) = 8/7. This is correct. The limit at x=-4 is infinite. This is correct. The stated answer is correct. The solution steps are correct. Is there a subtle error? Step 5 says "Subs(8, x, -4) = 8". This refers to the simplified numerator 8. The text says "At x = -4 the numerator is not 0". The original numerator is 8x-24. At x=-4, it is -56. The simplified numerator is 8. At x=-4, it is 8. Both are non-zero. The logic holds. Is there a style issue? "Subs(8, x, -4)" is weird notation but the text explains it. Let's look closer. Step 3 cancels (x-3). This is valid for x != 3. Step 4 uses the limit. Step 5 uses the simplified form to check the numerator at -4. Since -4 != 3, the simplification is valid at -4. The logic is sound. Wait, did I miss something? "Find every point... classify...". The points are 3 and -4. The classifications are removable and infinite. Everything seems correct. Why would it fail? Maybe the prompt implies I should find an error? Let me check the factorization again. x^2+x-12. Roots are (-1 +/- sqrt(1+48))/2 = (-1 +/- 7)/2. 6/2=3, -8/2=-4. Correct. Numerator 8x-24 = 8(x-3). Correct. So f(x) = 8(x-3)/((x-3)(x+4)). For x!=3, f(x) = 8/(x+4). Limit x->3 is 8/7. Removable. Limit x->-4 is +/- infinity. Infinite. The solution is correct. I will mark it pass.gpt-oss:20b: fail (error) 2026-10-07 — Sentence 5 incorrectly substitutes the numerator at x=-4; it should be 8*(-4)-24 = -56, not 8. The conclusion about an infinite discontinuity is correct, but the justification is mathematically wrong.
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 2026-10-07 with SymPy 1.14.0.