Critical numbers
Problem 3.235 · easy
Find the critical numbers of \( \displaystyle f(x) = x^{3} + \frac{15 x^{2}}{2} + 12 x + 4 \).
- Critical numbers are where f'(x) = 0 or f' does not exist; this f' is a polynomial, so it exists everywhere.Reviewed
- \[ \frac{d}{d x} \left(x^{3} + \frac{15 x^{2}}{2} + 12 x + 4\right) = 3 x^{2} + 15 x + 12 \]Differentiate.✓ Proved
- \[ 3 x^{2} + 15 x + 12 = \left(x + 4\right) \left(3 x + 3\right) \]Factor f'(x).✓ Proved
- f'(x) = 0 at x = -4 and x = -1.Reviewed
Answer \( x = -4, -1 \)
✓ Nihil obstat Lines: 2 proved, 2 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 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | a sentence; read, not computed |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | f' changes sign exactly that many times on [-8, 8], sampled finely |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-09-27gpt-oss:20b: pass 2026-09-27qwen3.6:27b-mlx: pass 2026-09-27 — The solution correctly identifies the definition of critical numbers, computes the derivative accurately, factors it correctly, and solves for the roots. The logic is sound and complete.gpt-oss:20b: pass 2026-09-27
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/critical_numbers, checked 2026-09-27 with SymPy 1.14.0.