The Mean Value Theorem and Rolle's theorem
Problem 3.496 · medium
Verify that \( \displaystyle f(x) = - 2 x^{3} - 2 x^{2} - 2 x + 2 \) satisfies the hypotheses of the Mean Value Theorem on \( \displaystyle [-2, 0] \), 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.
- \[ - \frac{\left. - 2 x^{3} - 2 x^{2} - 2 x + 2 \right|_{\substack{ x=-2 }}}{2} + \frac{\left. - 2 x^{3} - 2 x^{2} - 2 x + 2 \right|_{\substack{ x=0 }}}{2} = -6 \]The slope of the secant line.✓ Proved
- \[ \frac{d}{d x} \left(- 2 x^{3} - 2 x^{2} - 2 x + 2\right) = - 6 x^{2} - 4 x - 2 \]Differentiate.✓ Proved
- \[ - 6 \left(- \frac{\sqrt{7}}{3} - \frac{1}{3}\right)^{2} - \frac{2}{3} + \frac{4 \sqrt{7}}{3} = -6 \]c = -sqrt(7)/3 - 1/3 solves f′(c) = -6 and lies in (-2, 0).✓ Proved
Answer \( c = - \frac{\sqrt{7}}{3} - \frac{1}{3} \)
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.
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 claims to find 'every' number c, but only provides one of the two solutions to f'(c) = -6. The quadratic equation 3c^2 + 2c + 1 = 0 has two roots, c = (-1 ± sqrt(7))/3. The root c = (-1 + sqrt(7))/3 is approximately 0.55, which lies outside the interval (-2, 0), so it is correctly excluded, but the solution fails to explicitly identify and reject this second root, leaving the claim of finding 'every' number incomplete and potentially misleading regarding the existence of other critical points.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-10-08 — The solution claims to find 'every' number c, but only provides one of the two solutions to f'(c) = -6. The quadratic equation 3c^2 + 2c + 1 = 0 has two roots, c = (-1 ± sqrt(7))/3. The root c = (-1 + sqrt(7))/3 is approximately 0.55, which lies outside the interval (-2, 0), so it is correctly excluded, but the solution fails to explicitly identify and reject this second root, leaving the claim of finding 'every' number incomplete and potentially misleading regarding the existence of other critical points.gpt-oss:20b: pass 2026-10-08gpt-oss:20b: pass 2026-10-08qwen3.6:27b-mlx: fail (error) 2026-10-08 — The solution fails to find all values of c guaranteed by the Mean Value Theorem. The equation f'(c) = -6 yields two solutions, c = (-1 ± sqrt(7))/3, but only one is listed. Both solutions lie within the interval (-2, 0), so omitting the second one is an error.
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 2026-10-08 with SymPy 1.14.0.