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The Mean Value Theorem and Rolle's theorem

Problem 3.763 · medium

Verify that \( \displaystyle f(x) = x^{3} - 9 x^{2} + 23 x - 15 \) satisfies the hypotheses of Rolle's 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.
\[ \frac{\left(-1\right) \left. x^{3} - 9 x^{2} + 23 x - 15 \right|_{\substack{ x=1 }}}{2} + \frac{\left. x^{3} - 9 x^{2} + 23 x - 15 \right|_{\substack{ x=3 }}}{2} = 0 \]
The slope of the secant line.✓ Proved
\[ \frac{d}{d x} \left(x^{3} - 9 x^{2} + 23 x - 15\right) = 3 x^{2} - 18 x + 23 \]
Differentiate.✓ Proved
\[ \left. 3 x^{2} - 18 x + 23 \right|_{\substack{ x=3 - \frac{2 \sqrt{3}}{3} }} = 0 \]
c = 3 - 2*sqrt(3)/3 solves f′(c) = 0 and lies in (1, 3).✓ Proved

Answer \( c = 3 - \frac{2 \sqrt{3}}{3} \)

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

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: pass
qwen3.6:27b-mlx: fail (error) — The solution fails to verify that f(1) = f(3), which is a required hypothesis of Rolle's Theorem. Additionally, it only identifies one value of c, whereas the quadratic derivative yields two solutions in the interval (1, 3), so it does not find 'every' number guaranteed by the theorem.

Every verdict on record (6)

qwen3.6:27b-mlx: fail (error) 2026-11-01 — The solution fails to verify that f(1) = f(3), which is a required hypothesis of Rolle's Theorem. Additionally, it only identifies one value of c, whereas the quadratic derivative yields two solutions in the interval (1, 3), so it does not find 'every' number guaranteed by the theorem.
gpt-oss:20b: pass 2026-11-01
qwen3.6:27b-mlx: fail (error) 2026-10-31 — The solution fails to verify that f(1) = f(3), which is a required hypothesis of Rolle's Theorem. Additionally, it only identifies one value of c, whereas the quadratic derivative yields two solutions in the interval (1, 3), so it does not find 'every' number guaranteed by the theorem.
gpt-oss:20b: pass 2026-10-31
qwen3.6:27b-mlx: fail (error) 2026-10-31 — The solution fails to verify that f(1) = f(3), which is a required hypothesis for Rolle's Theorem. Additionally, it only finds one of the two values of c guaranteed by the theorem (the other is 3 + 2*sqrt(3)/3, which is outside the interval, but the quadratic 3x^2 - 18x + 23 = 0 has two roots; wait, let's check the roots. x = (18 +/- sqrt(324 - 276))/6 = (18 +/- sqrt(48))/6 = 3 +/- 2*sqrt(3)/3. 3 + 2*sqrt(3)/3 approx 3 + 1.15 = 4.15, which is outside (1,3). So there is only one c in (1,3). However, the primary error is the missing verification of f(1)=f(3). The problem asks to verify the hypotheses. The solution states hypotheses hold because it's a polynomial, but ignores the endpoint condition.
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.