The Mean Value Theorem and Rolle's theorem
Problem 3.595 · medium
Verify that \( \displaystyle f(x) = x^{3} - x^{2} + x - 1 \) satisfies the hypotheses of the Mean Value 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.
- \[ - \frac{\left. x^{3} - x^{2} + x - 1 \right|_{\substack{ x=-2 }}}{3} + \frac{\left. x^{3} - x^{2} + x - 1 \right|_{\substack{ x=1 }}}{3} = 5 \]The slope of the secant line.✓ Proved
- \[ \frac{d}{d x} \left(x^{3} - x^{2} + x - 1\right) = 3 x^{2} - 2 x + 1 \]Differentiate.✓ Proved
- \[ \frac{1}{3} + 3 \left(\frac{1}{3} - \frac{\sqrt{13}}{3}\right)^{2} + \frac{2 \sqrt{13}}{3} = 5 \]c = 1/3 - sqrt(13)/3 solves f′(c) = 5 and lies in (-2, 1).✓ Proved
Answer \( c = \frac{1}{3} - \frac{\sqrt{13}}{3} \)
Lines: 3 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. A reviewer's objection to the explanation stands; see 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: fail (misleading) — The equation in step 4 misrepresents the derivative: it omits the –2c term (replacing it with +2√13/3) yet still claims to solve f′(c)=5. A student could be misled into thinking that expression is correct, even though it is algebraically incorrect. The correct verification would use f′(c)=3c²–2c+1=5. The rest of the solution is sound.qwen3.6:27b-mlx: fail (error) — The solution fails to find all numbers c guaranteed by the Mean Value Theorem. The equation f'(c) = 5 yields two solutions, c = (1 ± sqrt(13))/3. Only c = (1 - sqrt(13))/3 lies in the interval (-2, 1), but the solution presents this single value as if it were the only solution to the quadratic without explicitly discarding the other root or stating that the other root is outside the interval.
Every verdict on record (4)
gpt-oss:20b: fail (misleading) 2026-10-12 — The equation in step 4 misrepresents the derivative: it omits the –2c term (replacing it with +2√13/3) yet still claims to solve f′(c)=5. A student could be misled into thinking that expression is correct, even though it is algebraically incorrect. The correct verification would use f′(c)=3c²–2c+1=5. The rest of the solution is sound.qwen3.6:27b-mlx: fail (error) 2026-10-12 — The solution fails to find all numbers c guaranteed by the Mean Value Theorem. The equation f'(c) = 5 yields two solutions, c = (1 ± sqrt(13))/3. Only c = (1 - sqrt(13))/3 lies in the interval (-2, 1), but the solution presents this single value as if it were the only solution to the quadratic without explicitly discarding the other root or stating that the other root is outside the interval.gpt-oss:20b: fail (error) 2026-10-12 — Sentence 4 incorrectly expands f′(c)=5; the algebraic expression is wrong, though the final value of c is correct.qwen3.6:27b-mlx: fail (error) 2026-10-12 — The solution fails to find all numbers c guaranteed by the Mean Value Theorem. The equation f'(c) = 5 yields two solutions, c = (1 ± sqrt(13))/3, but the solution only lists one. Additionally, the verification equation in line 4 is algebraically incorrect (it does not equal 5).
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 2026-10-12 with SymPy 1.14.0.