The Mean Value Theorem and Rolle's theorem
Problem 3.401 · medium
Verify that \( \displaystyle f(x) = - 2 x^{3} + x^{2} + x + 1 \) satisfies the hypotheses of the Mean Value Theorem on \( \displaystyle [0, 4] \), 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.
- \[ - \frac{\left. - 2 x^{3} + x^{2} + x + 1 \right|_{\substack{ x=0 }}}{4} + \frac{\left. - 2 x^{3} + x^{2} + x + 1 \right|_{\substack{ x=4 }}}{4} = -27 \]The slope of the secant line.✓ Proved
- \[ \frac{d}{d x} \left(- 2 x^{3} + x^{2} + x + 1\right) = - 6 x^{2} + 2 x + 1 \]Differentiate.✓ Proved
- \[ -27 \]c = 7/3 solves f′(c) = -27 and lies in (0, 4).✓ Proved
Answer \( c = \frac{7}{3} \)
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 that c = 7/3 is the only number guaranteed by the theorem, but f'(x) = -27 is a quadratic equation with two real roots: 7/3 and -2. While -2 is outside the interval (0, 4), the solution fails to explicitly solve the quadratic or justify why 7/3 is the unique solution in the interval, instead just stating that 7/3 solves it. More critically, the phrasing 'find every number' implies a search process which is skipped; simply verifying one root is insufficient proof that no other roots exist in the interval without showing the other root is outside.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-10-04 — The solution claims that c = 7/3 is the only number guaranteed by the theorem, but f'(x) = -27 is a quadratic equation with two real roots: 7/3 and -2. While -2 is outside the interval (0, 4), the solution fails to explicitly solve the quadratic or justify why 7/3 is the unique solution in the interval, instead just stating that 7/3 solves it. More critically, the phrasing 'find every number' implies a search process which is skipped; simply verifying one root is insufficient proof that no other roots exist in the interval without showing the other root is outside.gpt-oss:20b: pass 2026-10-04qwen3.6:27b-mlx: fail (error) 2026-10-04 — The solution fails to actually solve for c. It states that c = 7/3 is a solution but does not show the algebraic steps solving -6c^2 + 2c + 1 = -27, nor does it verify that this is the only solution in the interval (0, 4).gpt-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.