The Mean Value Theorem and Rolle's theorem
Problem 3.603 · medium
Verify that \( \displaystyle f(x) = - x^{2} + 2 x + 2 \) satisfies the hypotheses of the Mean Value Theorem on \( \displaystyle [0, 3] \), 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.
- \[ - \frac{\left. - x^{2} + 2 x + 2 \right|_{\substack{ x=0 }}}{3} + \frac{\left. - x^{2} + 2 x + 2 \right|_{\substack{ x=3 }}}{3} = -1 \]The slope of the secant line.✓ Proved
- \[ \frac{d}{d x} \left(- x^{2} + 2 x + 2\right) = 2 - 2 x \]Differentiate.✓ Proved
- \[ -1 \]c = 3/2 solves f′(c) = -1 and lies in (0, 3).✓ Proved
Answer \( c = \frac{3}{2} \)
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 (misleading) — The solution fails to explicitly state the Mean Value Theorem equation f'(c) = (f(b)-f(a))/(b-a) and solve for c. It jumps from computing the derivative to asserting the solution, omitting the crucial algebraic step of setting 2-2c = -1. This hides the actual work required to find c.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-12qwen3.6:27b-mlx: fail (misleading) 2026-10-12 — The solution fails to explicitly state the Mean Value Theorem equation f'(c) = (f(b)-f(a))/(b-a) and solve for c. It jumps from computing the derivative to asserting the solution, omitting the crucial algebraic step of setting 2-2c = -1. This hides the actual work required to find c.gpt-oss:20b: pass 2026-10-12qwen3.6:27b-mlx: fail (error) 2026-10-12 — The calculation of the secant slope is incorrect; it computes the average of the function values rather than the difference quotient (f(b)-f(a))/(b-a). The correct slope is -2, not -1.
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.