∫Calc Practice

The Mean Value Theorem and Rolle's theorem

Problem 3.760 · medium

Verify that \( \displaystyle f(x) = x^{3} - x^{2} + x + 1 \) satisfies the hypotheses of the Mean Value Theorem on \( \displaystyle [0, 3] \), and find every number \( \displaystyle c \) the theorem guarantees.
  1. f is a polynomial, so it is continuous on [a, b] and differentiable on (a, b): the hypotheses hold.
  2. \[ \frac{\left(-1\right) \left. x^{3} - x^{2} + x + 1 \right|_{\substack{ x=0 }}}{3} + \frac{\left. x^{3} - x^{2} + x + 1 \right|_{\substack{ x=3 }}}{3} = 7 \]
    The slope of the secant line.✓ Proved
  3. \[ \frac{d}{d x} \left(x^{3} - x^{2} + x + 1\right) = 3 x^{2} - 2 x + 1 \]
    Differentiate.✓ Proved
  4. \[ \left. 3 x^{2} - 2 x + 1 \right|_{\substack{ x=1 \cdot \frac{1}{3} + \frac{\sqrt{19}}{3} }} = 7 \]
    c = 1/3 + sqrt(19)/3 solves f′(c) = 7 and lies in (0, 3).✓ Proved
Answer \( c = \frac{1}{3} + \frac{\sqrt{19}}{3} \)

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. How we verify

The full receipt
LineStatusChecked byDetail
1Not checked—a sentence; read, not computed
2✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
3✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
4✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
answer, a second way✓ Checked independentlysympy 1.14.0 + mpmath 1.3.0each c checked by a difference quotient; a scan of (a, b) finds no others

Reviewers

  • gpt-oss:20b: pass
  • qwen3.6:27b-mlx: fail (error) — The solution fails to find 'every' number c guaranteed by the theorem. The quadratic equation f'(c) = 7 has two real roots, but only one is presented. The other root, c = 1/3 - sqrt(19)/3, is negative and thus not in (0, 3), but the solution does not explicitly state that this second root was found and rejected, nor does it confirm that the presented root is the unique solution in the interval.
Every verdict on record (6)
  • qwen3.6:27b-mlx: fail (error) 2026-11-01 — The solution fails to find 'every' number c guaranteed by the theorem. The quadratic equation f'(c) = 7 has two real roots, but only one is presented. The other root, c = 1/3 - sqrt(19)/3, is negative and thus not in (0, 3), but the solution does not explicitly state that this second root was found and rejected, nor does it confirm that the presented root is the unique solution in the interval.
  • gpt-oss:20b: pass 2026-11-01
  • qwen3.6:27b-mlx: fail (error) 2026-10-31 — The problem asks to find *every* number c guaranteed by the Mean Value Theorem. The quadratic equation f'(c) = 7 has two real roots, but the solution only presents one and claims it is the solution without acknowledging or checking the second root.
  • gpt-oss:20b: pass 2026-10-31
  • qwen3.6:27b-mlx: inconclusive 2026-10-31 — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The solution fails to find all numbers c guaranteed by the theorem. The equation f'(c) = 7 yields two solutions, c = (1 ± sqrt(19))/3. The solution o
  • gpt-oss:20b: pass 2026-10-31

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 2026-11-01 with SymPy 1.14.0.