Second Derivative Test

Critical Points and the Second Derivative Test

For a function $f(x,y)$, critical points occur where $\nabla f = \mathbf{0}$ (both partial derivatives are zero).

The Discriminant

At a critical point $(x_0, y_0)$, classify it using:

$D = f_{xx} \cdot f_{yy} - (f_{xy})^2$

Where:

• $f_{xx} = \partial^2 f / \partial x^2$
• $f_{yy} = \partial^2 f / \partial y^2$
• $f_{xy} = \partial^2 f / \partial x \partial y$ (mixed partial)

Classification Rules

Examples

$f(x,y) = x^2 + y^2$ at $(0,0)$:

• $f_{xx} = 2$, $f_{yy} = 2$, $f_{xy} = 0$
• $D = 4 - 0 = 4 > 0$, $f_{xx} > 0$ → local minimum ✓

$f(x,y) = x^2 - y^2$ at $(0,0)$:

• $f_{xx} = 2$, $f_{yy} = -2$, $f_{xy} = 0$
• $D = -4 < 0$ → saddle point ✓

$f(x,y) = -(x^2 + y^2)$ at $(0,0)$:

• $f_{xx} = -2$, $f_{yy} = -2$, $f_{xy} = 0$
• $D = 4 > 0$, $f_{xx} < 0$ → local maximum ✓

Computing $f_{xy}$ Numerically

$f_{xy} \approx \frac{f(x+h,y+h) - f(x+h,y-h) - f(x-h,y+h) + f(x-h,y-h)}{4h^2}$

Your Task

Implement double discriminant_2d(double (*f)(double, double), double x, double y, double h) that computes $D = f_{xx} \cdot f_{yy} - f_{xy}^2$.