The Mean Value Theorem and Rolle's theorem
Problem 3.598 · medium
Verify that \( \displaystyle f(x) = - 2 x^{3} - 2 x^{2} - 1 \) 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.
- \[ - \frac{\left. - 2 x^{3} - 2 x^{2} - 1 \right|_{\substack{ x=-2 }}}{2} + \frac{\left. - 2 x^{3} - 2 x^{2} - 1 \right|_{\substack{ x=0 }}}{2} = -4 \]The slope of the secant line.✓ Proved
- \[ \frac{d}{d x} \left(- 2 x^{3} - 2 x^{2} - 1\right) = - 6 x^{2} - 4 x \]Differentiate.✓ Proved
- \[ - 6 \left(- \frac{\sqrt{7}}{3} - \frac{1}{3}\right)^{2} + \frac{4}{3} + \frac{4 \sqrt{7}}{3} = -4 \]c = -sqrt(7)/3 - 1/3 solves f′(c) = -4 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 fails to find all values of c guaranteed by the Mean Value Theorem. The equation f'(c) = -4 yields two solutions, c = (-1 ± sqrt(7))/3, but only one is presented and verified. Additionally, the verification in line 4 is algebraically incorrect as written (it does not equal -4), though the intended root is valid.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-12qwen3.6:27b-mlx: fail (error) 2026-10-12 — The solution fails to find all values of c guaranteed by the Mean Value Theorem. The equation f'(c) = -4 yields two solutions, c = (-1 ± sqrt(7))/3, but only one is presented and verified. Additionally, the verification in line 4 is algebraically incorrect as written (it does not equal -4), though the intended root is valid.gpt-oss:20b: pass 2026-10-12qwen3.6:27b-mlx: fail (error) 2026-10-12 — The solution fails to find all values of c guaranteed by the Mean Value Theorem. The quadratic equation f'(c) = -4 has two real roots, but only one is presented. The other root, c = (sqrt(7) - 1)/3, also lies in the interval (-2, 0) and must be included.
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 2026-10-12 with SymPy 1.14.0.