The Mean Value Theorem and Rolle's theorem
Problem 3.550 · medium
Verify that \( \displaystyle f(x) = 2 x^{3} + 2 x^{2} - 4 x \) satisfies the hypotheses of Rolle's 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.
- \[ - \frac{\left. 2 x^{3} + 2 x^{2} - 4 x \right|_{\substack{ x=-2 }}}{2} + \frac{\left. 2 x^{3} + 2 x^{2} - 4 x \right|_{\substack{ x=0 }}}{2} = 0 \]The slope of the secant line.✓ Proved
- \[ \frac{d}{d x} \left(2 x^{3} + 2 x^{2} - 4 x\right) = 6 x^{2} + 4 x - 4 \]Differentiate.✓ Proved
- \[ - \frac{16}{3} - \frac{4 \sqrt{7}}{3} + 6 \left(- \frac{\sqrt{7}}{3} - \frac{1}{3}\right)^{2} = 0 \]c = -sqrt(7)/3 - 1/3 solves f′(c) = 0 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 verify that f(-2) = f(0), which is a required hypothesis of Rolle's Theorem. Additionally, it only provides one value for c, whereas the quadratic derivative yields two solutions, one of which (involving +sqrt(7)) also lies in the interval (-2, 0) and must be included.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-10-10 — The solution fails to verify that f(-2) = f(0), which is a required hypothesis of Rolle's Theorem. Additionally, it only provides one value for c, whereas the quadratic derivative yields two solutions, one of which (involving +sqrt(7)) also lies in the interval (-2, 0) and must be included.gpt-oss:20b: pass 2026-10-10gpt-oss:20b: fail (error) 2026-10-10 — The solution asserts that the hypotheses of Rolle’s theorem hold but never verifies that f(−2)=f(0). Without checking the equal endpoint values the application of the theorem is incomplete.qwen3.6:27b-mlx: fail (error) 2026-10-10 — The solution fails to verify that f(-2) = f(0), which is a required hypothesis of Rolle's Theorem. Additionally, it only provides one value for c, whereas the quadratic derivative yields two solutions, one of which lies within the interval (-2, 0) and the other outside, but the problem asks for 'every number c' guaranteed by the theorem (implying finding all valid c in the open 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 2026-10-10 with SymPy 1.14.0.