Critical Points and Local Extrema: Learn It 2

Giselle Miller

Critical Points and Local Extrema: Learn It 2

Second Derivative Test

Consider the function [latex]f(x)=x^{3}[/latex]. This function has a critical point at [latex]x=0[/latex] since [latex]f'(0)=3(0)^{2}=0[/latex]. However, [latex]f[/latex] does not have an extreme value at [latex]x=0[/latex]. The existence of a critical point at [latex]x=x_0[/latex] does not guarantee a local extremum there.

The same is true for functions of two or more variables. One way this can happen is at a saddle point. Here’s an example:

In this graph, the origin is a saddle point. The first partial derivatives of [latex]f(x, y)=x^{2}-y^{2}[/latex] are both zero at this point, but it’s neither a maximum nor a minimum for the function. The vertical trace corresponding to [latex]y=0[/latex] is [latex]z=x^{2}[/latex] (a parabola opening upward), but the vertical trace corresponding to [latex]x=0[/latex] is [latex]z=-y^{2}[/latex] (a parabola opening downward). Therefore, it’s both a global maximum for one trace and a global minimum for another.

saddle point

Given the function [latex]z=f(x, y)[/latex], the point [latex](x_0, y_0, f(x_0, y_0))[/latex] is a saddle point if both [latex]f_x(x_0, y_0)=0[/latex] and [latex]f_y(x_0, y_0)=0[/latex], but [latex]f[/latex] does not have a local extremum at [latex](x_0, y_0)[/latex].

The second derivative test for single-variable functions provides a method for determining whether an extremum occurs at a critical point. Let’s recall how it works before extending it to two variables.

Second Derivative Test (Single Variable Version)Suppose that [latex]f(x)[/latex] is twice-differentiable at [latex]x = x_0[/latex] and [latex]f'(x_0) = 0[/latex].

If [latex]f''(x_0) < 0[/latex], then [latex]f[/latex] has a local maximum at [latex]x_0[/latex].
If [latex]f''(x_0) > 0[/latex], then [latex]f[/latex] has a local minimum at [latex]x_0[/latex].
If [latex]f''(x_0) = 0[/latex], then the test is inconclusive.

When extending this to functions of two variables, an issue arises: there are four different second-order partial derivatives (though equality of mixed partials reduces this to three). The second derivative test for a function of two variables, stated in the following theorem, uses a discriminant [latex]D[/latex] that replaces [latex]f''(x_0)[/latex] in the second derivative test for a function of one variable.

second derivative test

Let [latex]z=f(x, y)[/latex] be a function of two variables for which the first- and second-order partial derivatives are continuous on some disk containing the point [latex](x_0, y_0)[/latex]. Suppose [latex]f_x(x_0, y_0)=0[/latex] and [latex]f_y(x_0, y_0)=0[/latex]. Define the quantity

[latex]\large{D=f_{xx}(x_0,y_0)f_{yy}(x_0,y_0)-(f_{xy}(x_0,y_0))^2}.[/latex]

If [latex]D>0[/latex] and [latex]f_{xx}(x_0,y_0)>0[/latex], then [latex]f[/latex] has a local minimum at [latex](x_0, y_0)[/latex].
If [latex]D>0[/latex] and [latex]f_{xx}(x_0,y_0)<0[/latex], then [latex]f[/latex] has a local maximum at [latex](x_0,y_0)[/latex].
If [latex]D<0[/latex], then [latex]f[/latex] has a saddle point at [latex](x_0, y_0)[/latex].
If [latex]D=0[/latex], then the test is inconclusive.

See Figure 2.

This figure consists of three figures labeled a, b, and c. Figure a has two bulbous mounds pointing down, and the two extrema are listed as the local minima. Figure b has two bulbous mounds pointed up, and the two extrema are listed as the local maxima. Figure c is shaped like a saddle, and in the middle of the saddle, a point is marked as the saddle point. — Figure 2. The second derivative test can often determine whether a function of two variables has a local minima (a), a local maxima (b), or a saddle point (c).

To apply the second derivative test, we first need to find the critical points of the function. Here’s a systematic approach.

Using the Second Derivative Test for Functions of Two Variables

Let [latex]z=f(x,y)[/latex] be a function of two variables for which the first- and second-order partial derivatives are continuous on some disk containing the point [latex](x_0, y_0)[/latex]. To apply the second derivative test to find local extrema:

Find critical points [latex](x_0, y_0)[/latex] where [latex]f_x(x_0, y_0)=f_y(x_0, y_0)=0[/latex]. Discard any points where at least one of the partial derivatives does not exist.
Calculate the discriminant [latex]D=f_{xx}(x_0,y_0)f_{yy}(x_0,y_0)-(f_{xy}(x_0,y_0))^2[/latex] for each critical point.
Apply the second derivative test to determine whether each critical point is a local maximum, local minimum, saddle point, or whether the test is inconclusive.

Find the critical points for each of the following functions, and use the second derivative test to find the local extrema:

[latex]f(x, y)=4x^{2}+9y^{2}+8x-36y+24[/latex]
[latex]g(x, y)=\frac{1}{3}x^{3}+y^{2}+2xy-6x-3y+4[/latex]

a. Step 1 of the problem-solving strategy involves finding the critical points of [latex]f[/latex]. To do this, we first calculate [latex]f_x(x, y)[/latex] and [latex]f_y(x, y)[/latex], then set each of them equal to zero:

[latex]\hspace{8cm}\begin{align} f_x(x,y)&=8x+8 \\ f_y(x,y)&=18y-36. \end{align}[/latex]

Setting them equal to zero yields the system of equations

[latex]\hspace{8cm}\begin{align} 8x+8&=0 \\ 18y-36&=0. \end{align}[/latex]

The solution to this system is [latex]x=-1[/latex] and [latex]y=2[/latex]. Therefore [latex](-1, 2)[/latex] is a critical point of [latex]f[/latex].
Step 2 of the problem-solving strategy involves calculating [latex]D[/latex]. To do this, we first calculate the second partial derivatives of [latex]f[/latex]:

[latex]\hspace{8cm}\begin{align} f_{xx}(x,y)&=8 \\ f_{xy}(x,y)&=0 \\ f_{yy}(x,y)&=18. \end{align}[/latex]

Therefore, [latex]D=f_{xx}(-1,2)f_{yy}(-1,2)-(f_{xy}(-1,2))^2=(8)(18)-(0)^2=144[/latex].

Step 3 states to check the Second Derivative Test for Functions of Two Variables.. Since [latex]D>0[/latex] and [latex]f_{xx}(-1,2)>0[/latex], this corresponds to case 1. Therefore, [latex]f[/latex] has a local minimum at [latex](-1, 2)[/latex] as shown in the following figure.

The function f(x, y) = 4x2 + 9y2 + 8x – 36y + 24 is shown with local minimum at (–1, 2, –16). The shape is a plane curving up on both ends parallel to the y axis. — Figure 3. The function [latex]\small{f(x,y)}[/latex] has a local minimum at [latex]\small{(-1,2,-16)}[/latex].

b. For step 1, we first calculate [latex]g_x(x,y)[/latex] and [latex]g_y(x,y)[/latex], then set each of them equal to zero:[latex]\hspace{8cm}\begin{align}g_x(x,y)&=x^2+2y-6 \\g_y(x,y)&=2y+2x-3.\end{align}[/latex]
Setting them equal to zero yields the system of equations
[latex]\hspace{8cm}\begin{align}x^2+2y-6&=0 \\2y+2x-3&=0.\end{align}[/latex]
To solve this system, first solve the second equation for [latex]y[/latex]. This gives [latex]y=\frac{3-2x}{2}[/latex]. Substituting this into the first equation gives[latex]\hspace{7.5cm}\begin{align}x^2+3-2x-6&=0 \\x^2-2x-3&=0 \\(x-3)(x+1)&=0.\end{align}[/latex]
Therefore, [latex]x=-1[/latex] or [latex]x=3[/latex]. Substituting these values into the equation [latex]y=\frac{3-2x}{2}[/latex] yields the critical points [latex](-1,\frac52)[/latex] and [latex](3,-\frac32)[/latex].

Step 2 involves calculating the second partial derivatives of [latex]g[/latex]:

[latex]\hspace{8cm}\begin{align} g_{xx}(x,y)&=2x \\ g_{xy}(x,y)&=2 \\ g_{yy}(x,y)&=2. \end{align}[/latex]

Then, we find a general formula for [latex]D[/latex]:

[latex]\hspace{7cm}\begin{align} D&=g_{xx}(x_0,y_0)g_{yy}(x_0,y_0)-(g_{xy}(x_0,y_0))^2 \\ &=(2x_0)(2)-2^2 \\ &=4x_0-4. \end{align}[/latex]

Next, we substitute each critical point into this formula:

[latex]\hspace{7cm}\begin{align} D(-1,\frac52)&=(2(-1))(2)-(2)^2=-4-4=-8 \\ D(3,-\frac32)&=(2(3))(2)-(2)^2=12-4=8. \end{align}[/latex]

In step 3, we note that, applying the Second Derivative Test for Functions of Two Variables to point [latex](-1,\frac{5}{2})[/latex] leads to case 3, which means that [latex](-1,\frac{5}{2})[/latex] is a saddle point. Applying the theorem to point [latex](3,-\frac{3}{2})[/latex] leads to case 1, which means that [latex](3,-\frac{3}{2})[/latex] corresponds to a local minimum as shown in the following figure.

The function f(x, y) = (1/3)x3 + y2 + + 2xy – 6x – 3y + 4 is shown with local minimum at (3, –3/2, –29/4) and saddle point at (−1, 5/2, 41/12). The shape is a plane curving up on the corners near (4, 3) and (−2, −2). — Figure 4. The function [latex]\small{g(x,y)}[/latex] has a local minimum and a saddle point.