The Mean Value Theorem and Rolle's theorem
Problem 3.445 · hard
Verify that \( \displaystyle f(x) = - x^{3} - x^{2} + 2 x \) satisfies the hypotheses of Rolle's theorem on \( \displaystyle [-2, 1] \), 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.
- \[ - \frac{\left. - x^{3} - x^{2} + 2 x \right|_{\substack{ x=-2 }}}{3} + \frac{\left. - x^{3} - x^{2} + 2 x \right|_{\substack{ x=1 }}}{3} = 0 \]The slope of the secant line.✓ Proved
- \[ \frac{d}{d x} \left(- x^{3} - x^{2} + 2 x\right) = - 3 x^{2} - 2 x + 2 \]Differentiate.✓ Proved
- \[ - 3 \left(- \frac{\sqrt{7}}{3} - \frac{1}{3}\right)^{2} + \frac{2 \sqrt{7}}{3} + \frac{8}{3} = 0 \]c = -sqrt(7)/3 - 1/3 solves f′(c) = 0 and lies in (-2, 1).✓ Proved
- \[ - \frac{2 \sqrt{7}}{3} - 3 \left(- \frac{1}{3} + \frac{\sqrt{7}}{3}\right)^{2} + \frac{8}{3} = 0 \]c = -1/3 + sqrt(7)/3 solves f′(c) = 0 and lies in (-2, 1).✓ Proved
Answer \( c = - \frac{\sqrt{7}}{3} - \frac{1}{3},\ - \frac{1}{3} + \frac{\sqrt{7}}{3} \)
Lines: 4 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 |
| 5 | ✓ 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 explicitly verify that f(-2) = f(1), which is a required hypothesis of Rolle's Theorem. While the algebraic check in line 2 computes the slope of the secant line, it does not state that this slope is zero (i.e., f(a)=f(b)), nor does it explicitly confirm the values are equal, leaving a critical hypothesis unverified.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-06qwen3.6:27b-mlx: fail (error) 2026-10-06 — The solution fails to explicitly verify that f(-2) = f(1), which is a required hypothesis of Rolle's Theorem. While the algebraic check in line 2 computes the slope of the secant line, it does not state that this slope is zero (i.e., f(a)=f(b)), nor does it explicitly confirm the values are equal, leaving a critical hypothesis unverified.gpt-oss:20b: pass 2026-10-06qwen3.6:27b-mlx: pass 2026-10-06 — The solution correctly verifies the hypotheses of Rolle's Theorem (continuity, differentiability, and equal endpoint values) and correctly identifies the critical points within the interval.
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 2026-10-06 with SymPy 1.14.0.