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The second derivative test for a function of one variable provides a method for determining whether an extremum occurs at a critical point of a function. When extending this result to a function of two variables, an issue arises related to the fact that there are, in fact, four different second-order partial derivatives, although 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     D that replaces f ( x 0 ) in the second derivative test for a function of one variable.

Second derivative test

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

D = f x x ( x 0 , y 0 ) f y y ( x 0 , y 0 ) ( f x y ( x 0 , y 0 ) ) 2 .
  1. If D > 0 and f x x ( x 0 , y 0 ) > 0 , then f has a local minimum at ( x 0 , y 0 ) .
  2. If D > 0 and f x x ( x 0 , y 0 ) < 0 , then f has a local maximum at ( x 0 , y 0 ) .
  3. If D < 0 , , then f has a saddle point at ( x 0 , y 0 ) .
  4. If D = 0 , then the test is inconclusive.

See [link] .

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.
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, it is necessary that we first find the critical points of the function. There are several steps involved in the entire procedure, which are outlined in a problem-solving strategy.

Problem-solving strategy: using the second derivative test for functions of two variables

Let z = f ( x , y ) be a function of two variables for which the first- and second-order partial derivatives are continuous on some disk containing the point ( x 0 , y 0 ) . To apply the second derivative test to find local extrema, use the following steps:

  1. Determine the critical points ( x 0 , y 0 ) of the function f where f x ( x 0 , y 0 ) = f y ( x 0 , y 0 ) = 0 . Discard any points where at least one of the partial derivatives does not exist.
  2. Calculate the discriminant D = f x x ( x 0 , y 0 ) f y y ( x 0 , y 0 ) ( f x y ( x 0 , y 0 ) ) 2 for each critical point of f .
  3. Apply [link] to determine whether each critical point is a local maximum, local minimum, or saddle point, or whether the theorem is inconclusive.

Using the second derivative test

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

  1. f ( x , y ) = 4 x 2 + 9 y 2 + 8 x 36 y + 24
  2. g ( x , y ) = 1 3 x 3 + y 2 + 2 x y 6 x 3 y + 4
  1. Step 1 of the problem-solving strategy involves finding the critical points of f . To do this, we first calculate f x ( x , y ) and f y ( x , y ) , then set each of them equal to zero:
    f x ( x , y ) = 8 x + 8 f y ( x , y ) = 18 y 36 .

    Setting them equal to zero yields the system of equations
    8 x + 8 = 0 18 y 36 = 0 .

    The solution to this system is x = −1 and y = 2 . Therefore ( −1 , 2 ) is a critical point of f .
    Step 2 of the problem-solving strategy involves calculating D . To do this, we first calculate the second partial derivatives of f :
    f x x ( x , y ) = 8 f x y ( x , y ) = 0 f y y ( x , y ) = 18 .

    Therefore, D = f x x ( −1 , 2 ) f y y ( −1 , 2 ) ( f x y ( −1 , 2 ) ) 2 = ( 8 ) ( 18 ) ( 0 ) 2 = 144 .
    Step 3 states to check [link] . Since D > 0 and f x x ( −1 , 2 ) > 0 , this corresponds to case 1. Therefore, f has a local minimum at ( −1 , 2 ) 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.
    The function f ( x , y ) has a local minimum at ( −1 , 2 , −16 ) .
  2. For step 1, we first calculate g x ( x , y ) and g y ( x , y ) , then set each of them equal to zero:
    g x ( x , y ) = x 2 + 2 y 6 g y ( x , y ) = 2 y + 2 x 3 .

    Setting them equal to zero yields the system of equations
    x 2 + 2 y 6 = 0 2 y + 2 x 3 = 0 .

    To solve this system, first solve the second equation for y. This gives y = 3 2 x 2 . Substituting this into the first equation gives
    x 2 + 3 2 x 6 = 0 x 2 2 x 3 = 0 ( x 3 ) ( x + 1 ) = 0 .

    Therefore, x = −1 or x = 3 . Substituting these values into the equation y = 3 2 x 2 yields the critical points ( −1 , 5 2 ) and ( 3 , 3 2 ) .

    Step 2 involves calculating the second partial derivatives of g :

    g x x ( x , y ) = 2 x g x y ( x , y ) = 2 g y y ( x , y ) = 2 .

    Then, we find a general formula for D :
    D = g x x ( x 0 , y 0 ) g y y ( x 0 , y 0 ) ( g x y ( x 0 , y 0 ) ) 2 = ( 2 x 0 ) ( 2 ) 2 2 = 4 x 0 4 .

    Next, we substitute each critical point into this formula:
    D ( −1 , 5 2 ) = ( 2 ( −1 ) ) ( 2 ) ( 2 ) 2 = −4 4 = −8 D ( 3 , 3 2 ) = ( 2 ( 3 ) ) ( 2 ) ( 2 ) 2 = 12 4 = 8 .

    In step 3, we note that, applying [link] to point ( −1 , 5 2 ) leads to case 3 , which means that ( −1 , 5 2 ) is a saddle point. Applying the theorem to point ( 3 , 3 2 ) leads to case 1, which means that ( 3 , 3 2 ) 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).
    The function g ( x , y ) has a local minimum and a saddle point.
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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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