The Mean Value Theorem and Rolle's theorem
Problem 3.761 · medium
Verify that \( \displaystyle f(x) = - 2 x^{3} + 14 x + 12 \) satisfies the hypotheses of Rolle's theorem on \( \displaystyle [-1, 3] \), and find every number \( \displaystyle c \) the theorem guarantees.
- f is a polynomial, so it is continuous on [a, b] and differentiable on (a, b): the hypotheses hold.
- \[ \frac{\left(-1\right) \left. - 2 x^{3} + 14 x + 12 \right|_{\substack{ x=-1 }}}{4} + \frac{\left. - 2 x^{3} + 14 x + 12 \right|_{\substack{ x=3 }}}{4} = 0 \]The slope of the secant line.✓ Proved
- \[ \frac{d}{d x} \left(- 2 x^{3} + 14 x + 12\right) = 14 - 6 x^{2} \]Differentiate.✓ Proved
- \[ \left. 14 - 6 x^{2} \right|_{\substack{ x=\frac{\sqrt{21}}{3} }} = 0 \]c = sqrt(21)/3 solves f′(c) = 0 and lies in (-1, 3).✓ Proved
Answer \( c = \frac{\sqrt{21}}{3} \)
Lines: 3 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in the receipt. How we verify
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 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | each c checked by a difference quotient; a scan of (a, b) finds no others |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: fail (error) — The solution fails to find all numbers c guaranteed by the theorem. The equation f'(c) = 0 yields two solutions, c = ±sqrt(21)/3, both of which lie in (-1, 3). The solution only lists the positive root.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-11-01 — The solution fails to find all numbers c guaranteed by the theorem. The equation f'(c) = 0 yields two solutions, c = ±sqrt(21)/3, both of which lie in (-1, 3). The solution only lists the positive root.gpt-oss:20b: pass 2026-11-01qwen3.6:27b-mlx: fail (error) 2026-10-31 — The solution fails to verify that f(-1) = f(3), which is a required hypothesis of Rolle's Theorem. Additionally, it omits the second valid solution c = -sqrt(21)/3, which also lies in the interval (-1, 3).gpt-oss:20b: pass 2026-10-31qwen3.6:27b-mlx: fail (error) 2026-10-31 — The solution fails to verify the hypothesis f(-1) = f(3), which is required for Rolle's Theorem (f(-1)=14, f(3)=0). Additionally, it misses the second valid solution c = -sqrt(21)/3, which also lies in the interval (-1, 3).gpt-oss:20b: pass 2026-10-31
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/mean_value_theorem, checked 2026-11-01 with SymPy 1.14.0.