The Mean Value Theorem and Rolle's theorem
Problem 3.395 · medium
Verify that \( \displaystyle f(x) = x^{3} - 5 x \) satisfies the hypotheses of the Mean Value Theorem on \( \displaystyle [-1, 3] \), 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.
- \[ - \frac{\left. x^{3} - 5 x \right|_{\substack{ x=-1 }}}{4} + \frac{\left. x^{3} - 5 x \right|_{\substack{ x=3 }}}{4} = 2 \]The slope of the secant line.✓ Proved
- \[ \frac{d}{d x} \left(x^{3} - 5 x\right) = 3 x^{2} - 5 \]Differentiate.✓ Proved
- \[ 2 \]c = sqrt(21)/3 solves f′(c) = 2 and lies in (-1, 3).✓ Proved
Answer \( c = \frac{\sqrt{21}}{3} \)
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 solve f'(c) = 2 for c, omitting the negative root c = -sqrt(21)/3 which also lies in (-1, 3). The problem asks for 'every number c', so the solution is incomplete.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-10-04 — The solution fails to solve f'(c) = 2 for c, omitting the negative root c = -sqrt(21)/3 which also lies in (-1, 3). The problem asks for 'every number c', so the solution is incomplete.gpt-oss:20b: pass 2026-10-04qwen3.6:27b-mlx: fail (error) 2026-10-04 — The solution fails to find all values of c guaranteed by the Mean Value Theorem. Solving 3c^2 - 5 = 2 yields c = ±sqrt(7)/3, but only the positive root is listed. Additionally, the secant slope calculation in step 2 is presented with confusing syntax and an incorrect intermediate value (the slope is 2, but the expression shown evaluates to 2 only if interpreted as the final result, not the calculation steps).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.