Absolute extrema on a closed interval
Problem 3.142 · hard Video
Find the absolute maximum and minimum values of \( \displaystyle f(x) = x^{3} - \frac{15 x^{2}}{2} + 18 x - 2 \) on \( \displaystyle [-2, 3] \).
- A continuous function on a closed interval has its extreme values at critical points or endpoints.
- \[ \frac{d}{d x} \left(x^{3} - \frac{15 x^{2}}{2} + 18 x - 2\right) = \left(x - 2\right) \left(3 x - 9\right) \]Differentiate and factor.✓ Proved
- Critical numbers inside [-2, 3]: 2.
- \[ \left. x^{3} - \frac{15 x^{2}}{2} + 18 x - 2 \right|_{\substack{ x=-2 }} = -76 \]f(-2).✓ Proved
- \[ \left. x^{3} - \frac{15 x^{2}}{2} + 18 x - 2 \right|_{\substack{ x=2 }} = 12 \]f(2).✓ Proved
- \[ \left. x^{3} - \frac{15 x^{2}}{2} + 18 x - 2 \right|_{\substack{ x=3 }} = \frac{23}{2} \]f(3).✓ Proved
- The largest value is 12 and the smallest is -76.
Answer \( \text{max } 12 \text{ at } x=2;\ \text{min } -76 \text{ at } x=-2 \)
Lines: 4 proved, 3 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 | Not checked | — | a sentence; read, not computed |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 5 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 6 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 7 | Not checked | — | a sentence; read, not computed |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | f sampled at 40,001 points across the interval reaches the same max and min |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: fail (error) — The solution incorrectly identifies the critical numbers inside the interval, omitting x=3 which is a root of the derivative. Although f(3) is evaluated as an endpoint, the classification of critical points is factually wrong.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-09-28 — The solution incorrectly identifies the critical numbers inside the interval, omitting x=3 which is a root of the derivative. Although f(3) is evaluated as an endpoint, the classification of critical points is factually wrong.gpt-oss:20b: pass 2026-09-28qwen3.6:27b-mlx: fail (error) 2026-09-27 — The solution fails to identify x=3 as a critical point (since f'(3)=0) and omits evaluating f(3) in the list of candidates in step 3, although f(3) is calculated later. This creates a logical gap where the comparison in step 7 relies on a value not established as a candidate in step 3.gpt-oss:20b: fail (error) 2026-09-27 — The solution incorrectly states that the only critical number inside the interval is 2, omitting the critical point at x=3. This omission could mislead a student into thinking that all critical points have been considered.qwen3.6:27b-mlx: fail (error) 2026-09-26 — Step 3 incorrectly identifies the critical numbers inside the interval, omitting x=3. Since x=3 is a critical point (derivative is zero) and also an endpoint, it must be considered in the comparison of values to ensure the absolute extrema are correctly identified.gpt-oss:20b: inconclusive 2026-09-26 — reviewer returned a non-object
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/absolute_extrema, checked 2026-09-28 with SymPy 1.14.0.