The Mean Value Theorem and Rolle's theorem
Problem 3.396 · medium
Verify that \( \displaystyle f(x) = x^{3} - x^{2} - 2 x + 2 \) satisfies the hypotheses of the Mean Value Theorem on \( \displaystyle [0, 3] \), 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.
- \[ - \frac{\left. x^{3} - x^{2} - 2 x + 2 \right|_{\substack{ x=0 }}}{3} + \frac{\left. x^{3} - x^{2} - 2 x + 2 \right|_{\substack{ x=3 }}}{3} = 4 \]The slope of the secant line.✓ Proved
- \[ \frac{d}{d x} \left(x^{3} - x^{2} - 2 x + 2\right) = 3 x^{2} - 2 x - 2 \]Differentiate.✓ Proved
- \[ - \frac{2 \sqrt{19}}{3} - \frac{8}{3} + 3 \left(\frac{1}{3} + \frac{\sqrt{19}}{3}\right)^{2} = 4 \]c = 1/3 + sqrt(19)/3 solves f′(c) = 4 and lies in (0, 3).✓ Proved
Answer \( c = \frac{1}{3} + \frac{\sqrt{19}}{3} \)
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 numbers c guaranteed by the Mean Value Theorem. The quadratic equation f'(c) = 4 has two roots, but only one is presented. The other root, c = 1/3 - sqrt(19)/3, is approximately -1.18, which lies outside the interval (0, 3), so it is correctly excluded, but the solution does not explicitly state that the other root was checked and rejected, nor does it acknowledge the existence of the second root in the derivation steps, making the claim 'find every number' incomplete in its justification.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-10-04 — The solution fails to find all numbers c guaranteed by the Mean Value Theorem. The quadratic equation f'(c) = 4 has two roots, but only one is presented. The other root, c = 1/3 - sqrt(19)/3, is approximately -1.18, which lies outside the interval (0, 3), so it is correctly excluded, but the solution does not explicitly state that the other root was checked and rejected, nor does it acknowledge the existence of the second root in the derivation steps, making the claim 'find every number' incomplete in its justification.gpt-oss:20b: pass 2026-10-04qwen3.6:27b-mlx: inconclusive 2026-10-04 — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The solution fails to find all numbers c guaranteed by the Mean Value Theorem. The equation f'(c) = 4 yields two solutions, c = (1 ± sqrt(19))/3. Thegpt-oss:20b: pass 2026-10-04
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 2026-10-04 with SymPy 1.14.0.