Absolute extrema on a closed interval
Problem 3.136 · hard
Find the absolute maximum and minimum values of \( \displaystyle f(x) = x^{3} - 3 x^{2} + 3 \) on \( \displaystyle [-1, 2] \).
- A continuous function on a closed interval has its extreme values at critical points or endpoints.
- \[ \frac{d}{d x} \left(x^{3} - 3 x^{2} + 3\right) = 3 x \left(x - 2\right) \]Differentiate and factor.✓ Proved
- Critical numbers inside [-1, 2]: 0.
- \[ \left. x^{3} - 3 x^{2} + 3 \right|_{\substack{ x=-1 }} = -1 \]f(-1).✓ Proved
- \[ \left. x^{3} - 3 x^{2} + 3 \right|_{\substack{ x=0 }} = 3 \]f(0).✓ Proved
- \[ \left. x^{3} - 3 x^{2} + 3 \right|_{\substack{ x=2 }} = -1 \]f(2).✓ Proved
- The largest value is 3 and the smallest is -1.
Answer \( \text{max } 3 \text{ at } x=0;\ \text{min } -1 \text{ at } x=-1 \)
Lines: 4 proved, 3 not checked. The answer was also checked a second way, without looking at the solution. The explanation has not been reviewed yet.
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 |
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 2026-09-26 with SymPy 1.14.0.