There is a version of the second derivative test in the case that the function and its first and second partial
derivatives are all continuous.
Definition 13.3.1The matrix H
whose ijthentry at the point x is
is called theHessian matrix. The eigenvalues of H
are the solutions λ to the equation det
(λI − H (x))
The following theorem says that if all the eigenvalues of the Hessian matrix at a critical point are
positive, then the critical point is a local minimum. If all the eigenvalues of the Hessian matrix at a critical
point are negative, then the critical point is a local maximum. Finally, if some of the eigenvalues of the
Hessian matrix at the critical point are positive and some are negative then the critical point is a saddle
point. The following picture illustrates the situation.
Theorem 13.3.2Let f : U → ℝ for U an open set in ℝnand let f be a C2function and supposethat at some x ∈ U, ∇f
= 0. Also let μ and λ be respectively, the largest and smallest eigenvaluesof the matrix H
. If λ > 0 then f has a local minimum at x. If μ < 0 then f has a local maximumatx.If either λ or μ equals zero, the test fails. If λ < 0 and μ > 0 there exists a direction in whichwhen f is evaluated on the line through the critical point having this direction, the resulting functionof one variable has a local minimum and there exists a direction in which when f is evaluated onthe line through the critical point having this direction, the resulting function of one variable has alocal maximum. This last case is called a saddle point.
Here is an example.
Example 13.3.3Let f
= 10xy + y2. Find the critical points and determine whether they arelocal minima, local maxima or saddle points.
(10xy + y2)
and so there is one critical point at the point
. What is it?
The Hessian matrix is
and the eigenvalues are of different signs. Therefore, the critical point
is a saddle point. Here is a
graph drawn by Maple.
Here is another example.
Example 13.3.4Let f
= 2x4− 4x3 + 14x2 + 12yx2− 12yx − 12x + 2y2 + 4y + 2. Find thecritical points and determine whether they are local minima, local maxima, or saddle points.
= 8x3− 12x2 + 28x + 24yx− 12y − 12 and fy
= 12x2− 12x + 4y + 4. The points at which
both fx and fy equal zero are
The Hessian matrix is
( 2 )
24x + 28 +24y − 24x 24x − 12
24x− 12 4
and the thing to determine is the sign of its eigenvalues evaluated at the critical points.
First consider the point
. The Hessian matrix is
( 16 0 )
and its eigenvalues are 16,4
showing that this is a local minimum.
at this point the Hessian matrix is
4 − 12
− 12 4
and the eigenvalues are 16,−8. Therefore, this point is a saddle point. To determine this, find the eigenvalues.
and the eigenvalues are 16,−8 so this point is also a saddle point.
Below is a graph of this function which illustrates the behavior near saddle points.
Or course sometimes the second derivative test is inadequate to determine what is going on. This should
be no surprise since this was the case even for a function of one variable. For a function of two variables, a
nice example is the Monkey saddle.
Example 13.3.5Suppose f
= 6xy2−2x3−3y4. Show that
is a critical point for whichthe second derivative test gives no information.
Before doing anything it might be interesting to look at the graph of this function of two variables
plotted using a computer algebra system.
This picture should indicate why this is called a monkey saddle. It is because the monkey can sit in the
saddle and have a place for his tail. Now to see
is a critical point, note that fx
= 6y2− 6x2, fy
= 12xy − 12y3 and so
is a critical point. So are
. Now fxx
= 0 and so are fxy
. Therefore, the Hessian matrix is the zero
matrix and clearly has only the zero eigenvalue. Therefore, the second derivative test is totally useless at
However, suppose you took x = t and y = t and evaluated this function on this line. This reduces to
= 4t3− 3t4), which is strictly increasing near t = 0. This shows the critical point
of f is
neither a local max. nor a local min. Next let x = 0 and y = t. Then p
= −3t4. Therefore,
along the line,
, f has a local maximum at
Example 13.3.6Find the critical points of the following function of three variables and classifythem as local minimums, local maximums or saddle points.
Note that in this simple example, the Hessian matrix is constant and so all that is left is to consider the
eigenvalues. Writing the characteristic equation and solving yields the eigenvalues are 2,−2,4. Thus the
given point is a saddle point.