In mathematics, the second partial derivative test is a method in multivariable calculus used to determine if a critical point of a function is a local minimum, maximum or saddle point.
Contents
Functions of two variables
Suppose that f(x, y) is a differentiable real function of two variables whose second partial derivatives exist. The Hessian matrix H of f is the 2 × 2 matrix of partial derivatives of f:
Define D(x, y) to be the determinant
of H. Finally, suppose that (a, b) is a critical point of f (that is, fx(a, b) = fy(a, b) = 0). Then the second partial derivative test asserts the following:
- If
D ( a , b ) > 0 andf x x ( a , b ) > 0 then( a , b ) is a local minimum of f. - If
D ( a , b ) > 0 andf x x ( a , b ) < 0 then( a , b ) is a local maximum of f. - If
D ( a , b ) < 0 then( a , b ) is a saddle point of f. - If
D ( a , b ) = 0 then the second derivative test is inconclusive, and the point (a, b) could be any of a minimum, maximum or saddle point.
Note that other equivalent versions of the test are possible. For example, some texts may use the trace fxx + fyy in place of the value fxx in the first two cases above. Such variations in the procedure applied do not alter the outcome of the test.
Functions of many variables
For a function f of more than two variables, there is a generalization of the rule above. In this context, instead of examining the determinant of the Hessian matrix, one must look at the eigenvalues of the Hessian matrix at the critical point. The following test can be applied at any critical point (a, b, ...) for which the Hessian matrix is invertible:
- If the Hessian is positive definite (equivalently, has all eigenvalues positive) at (a, b, ...), then f attains a local minimum at (a, b, ...).
- If the Hessian is negative definite (equivalently, has all eigenvalues negative) at (a, b, ...), then f attains a local maximum at (a, b, ...).
- If the Hessian has both positive and negative eigenvalues then (a, b, ...) is a saddle point for f (and in fact this is true even if (a, b, ...) is degenerate).
In those cases not listed above, the test is inconclusive.
Note that for functions of three or more variables, the determinant of the Hessian does not provide enough information to classify the critical point, because the number of jointly sufficient second-order conditions is equal to the number of variables, and the sign condition on the determinant of the Hessian is only one of the conditions. Note also that this statement of the second derivative test for many variables also applies in the two-variable and one-variable case. In the latter case, we recover the usual second derivative test.
In the two variable case,
Examples
To find and classify the critical points of the function
we first set the partial derivatives
equal to zero and solve the resulting equations simultaneously to find the four critical points
In order to classify the critical points, we examine the value of the determinant D(x, y) of the Hessian of f at each of the four critical points. We have
Now we plug in all the different critical values we found to label them; we have
Thus, the second partial derivative test indicates that f(x, y) has saddle points at (0, −1) and (1, −1) and has a local maximum at