Testing the Critical Points Analytically

In order to develop a second derivative test, let's assume that the function f(x,y) has a critical point at (x₀,y₀). A second order (quadratic) Taylor approximation to the function near the critical point is

If we translate the point (x,y) to (x+x₀,y+y₀) we can write this as

Now, we have three canonical examples of critical points.

1.

The origin of the bowl shaped function h₁ = x² + y² is a minimum.

2.

The origin of the upside down bowl, h₂ = -x² - y² is a maximum.

3.

The origin of the function h₃ = x² - y² is a saddle point.

If we can make our function g(x,y) look like one of these three, then we can classify the critical point. Now $g(x,y) = \frac{1}{2}x^2f_{xx}(x_0,y_0) + \frac{1}{2}y^2f_{yy}(x_0,y_0) + xyf_{xy}(x_0,y_0)$ since the gradient of f is the zero vector at a critical point.

If we now complete the square in x and let $A = \frac{1}{2}f_{xx}(x_0,y_0)$, B = f_xy(x₀,y₀), and $C = \frac{1}{2}f_{yy}(x_0,y_0)$, we get that

$$g(x,y) = A \left[ \left( x + \frac{B}{2A}y\right)^2 + \left( \frac{4AC - B^2}{4A^2}\right)y^2 \right].$$ (8)

The type of critical point is related entirely to the second term in parentheses. Comparing the sign of this quantity to the three canonical examples tells us which type of critical point we have. The quantity in question is called the discriminant, D. Note that

$$D = 4AC - B^2 = 4\left( \frac{1}{2}f_{xx}\right) \left( \frac{1}{2}f_{yy}\right) - f_{xy}^2 = f_{xx}f_{yy} - f_{xy}^2.$$ (9)

Comparison of this with the three canonical examples gives the following

1.

If D > 0, f_xx(x₀,y₀) > 0 then the point (x₀,y₀) is a local minimum.

2.

If D > 0, f_xx(x₀,y₀) < 0 then the point is a local maximum.

3.

If D < 0 the point is a saddle.

4.

If D = 0 then no information can be obtained at all. You will need to try graphing the function or looking at higher than second order approximations to the function.