Local extrema and saddle points
Problem 10.105 · easy
Find and classify the critical points of \( \displaystyle f(x, y) = x^{2} + 6 x - y^{2} + 4 y + 5 \).
- \[ \left[\begin{matrix}\frac{\partial}{\partial x} \left(x^{2} + 6 x - y^{2} + 4 y + 5\right)\\\frac{\partial}{\partial y} \left(x^{2} + 6 x - y^{2} + 4 y + 5\right)\end{matrix}\right] = \left[\begin{matrix}2 x + 6\\4 - 2 y\end{matrix}\right] \]The partial derivatives.✓ Proved
- \[ \left[\begin{matrix}0\\0\end{matrix}\right] \]Both vanish at (-3, 2), the only solution.✓ Proved
- \[ -4 \]D = f_xx f_yy − f_xy².✓ Proved
- D = -4 < 0: a saddle point.
Answer \( \text{saddle point at } (-3, 2) \)
Lines: 3 proved, 1 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 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 4 | Not checked | — | a sentence; read, not computed |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | f on a small circle around the point is compared with its centre value |
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/critical_points_2var, checked 2026-09-26 with SymPy 1.14.0.