The Mean Value Theorem and Rolle's theorem
Problem 3.545 · medium
Verify that \( \displaystyle f(x) = - x^{3} - 4 x^{2} + x + 4 \) satisfies the hypotheses of Rolle's theorem on \( \displaystyle [-1, 1] \), 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.
- \[ - \frac{\left. - x^{3} - 4 x^{2} + x + 4 \right|_{\substack{ x=-1 }}}{2} + \frac{\left. - x^{3} - 4 x^{2} + x + 4 \right|_{\substack{ x=1 }}}{2} = 0 \]The slope of the secant line.✓ Proved
- \[ \frac{d}{d x} \left(- x^{3} - 4 x^{2} + x + 4\right) = - 3 x^{2} - 8 x + 1 \]Differentiate.✓ Proved
- \[ - \frac{8 \sqrt{19}}{3} - 3 \left(- \frac{4}{3} + \frac{\sqrt{19}}{3}\right)^{2} + \frac{35}{3} = 0 \]c = -4/3 + sqrt(19)/3 solves f′(c) = 0 and lies in (-1, 1).✓ Proved
Answer \( c = - \frac{4}{3} + \frac{\sqrt{19}}{3} \)
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 verify that f(-1) = f(1), which is a required hypothesis for Rolle's Theorem (f(-1)=4, f(1)=0). Additionally, it only identifies one of the two values of c guaranteed by the theorem.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-10-10 — The solution fails to verify that f(-1) = f(1), which is a required hypothesis for Rolle's Theorem (f(-1)=4, f(1)=0). Additionally, it only identifies one of the two values of c guaranteed by the theorem.gpt-oss:20b: pass 2026-10-10gpt-oss:20b: fail (error) 2026-10-10 — The solution incorrectly identifies the root of f′(x)=0. The correct root in (-1,1) is c=(4-√19)/3, not c=(√19-4)/3. This misidentification would mislead a student about the application of Rolle’s theorem.qwen3.6:27b-mlx: fail (error) 2026-10-10 — The solution fails to verify that f(-1) = f(1), which is a required hypothesis for Rolle's Theorem (f(-1)=4, f(1)=0). Additionally, it only provides one of the two values of c guaranteed by the theorem.
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.