Increasing, decreasing and concavity
Problem 3.154 · medium
For \( \displaystyle f(x) = 2 x^{3} - 9 x^{2} + 12 x \), find the intervals where f is increasing or decreasing, where it is concave up or down, and its inflection points.
- \[ \frac{d}{d x} \left(2 x^{3} - 9 x^{2} + 12 x\right) = \left(x - 1\right) \left(6 x - 12\right) \]f' factored.✓ Proved
- f' > 0 outside [1, 2] and f' < 0 between them (a positive quadratic with roots 1, 2).
- \[ \frac{d^{2}}{d x^{2}} \left(2 x^{3} - 9 x^{2} + 12 x\right) = 12 x - 18 \]f''.✓ Proved
- \[ \left. 12 x - 18 \right|_{\substack{ x=\frac{3}{2} }} = 0 \]f'' = 0 at x = \frac{3}{2}, where it changes sign.✓ Proved
- f'' < 0 to the left of \frac{3}{2} (concave down) and f'' > 0 to the right (concave up).
Answer \( \uparrow (-\infty,1) \cup (2,\infty);\ \downarrow (1,2);\ \text{concave down } (-\infty,\frac{3}{2}),\ \text{up } (\frac{3}{2},\infty);\ \text{inflection at } x=\frac{3}{2} \)
Lines: 3 proved, 2 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 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 2 | Not checked | — | a sentence; read, not computed |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 5 | Not checked | — | a sentence; read, not computed |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | the signs of f' and f'' were evaluated at a point inside each interval |
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/increasing_concavity, checked 2026-09-26 with SymPy 1.14.0.