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 22.3.1The matrix H
whose ijthentry at the point x is
is called the Hessian matrix. The eigenvalues of H
are the solutionsλ 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 22.3.2Let f : U → ℝ for U an open set in ℝnand let f bea C2function and suppose that at some x ∈ U, ∇f
= 0. Also let μ and λ berespectively, the largest and smallest eigenvalues of the matrix H
. If λ > 0 thenf has a local minimum at x. If μ < 0 then f has a local maximum atx.If either λor 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, theresulting function of one variable has a local minimum and there exists a direction inwhich when f is evaluated on the line through the critical point having this direction,the resulting function of one variable has a local maximum. This last case is calleda saddle point.
Here is an example.
Example 22.3.3Let f
= 10xy + y2. Find the critical points and determinewhether they are local 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 22.3.4Let f
= 2x4− 4x3 + 14x2 + 12yx2− 12yx − 12x + 2y2 +
4y + 2. Find the critical points and determine whether they are local minima, localmaxima, 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
24x2 + 28 + 24y − 24x 24x− 12
24x− 12 4
and the thing to determine is the sign of its eigenvalues evaluated at the critical
First consider the point
. The Hessian matrix is
( 16 0 )
eigenvalues are 16,4 showing that this is a local minimum.
at this point the Hessian matrix is
( 4 − 12 )
− 12 4
eigenvalues are 16,−8. Therefore, this point is a saddle point. To determine this, find 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
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
Example 22.3.5Suppose f
= 6xy2−2x3−3y4. Show that
is a criticalpoint for which the 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
= 0 because fx
= 6y2− 6x2, fy
= 12xy − 12y3
is a critical point. So are
. Now fxx
= 0 and so are
. Therefore, the Hessian matrix is the zero matrix and clearly has
only the zero eigenvalue. Therefore, the second derivative test is totally useless at this
However, suppose you took x = t and y = t and evaluated this function on this line.
This reduces to h
= 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
Example 22.3.6Find the critical points of the following function of threevariables and classify them as local minimums, local maximums or saddle points.
Next you need to set the gradient equal to zero and solve the equations. This yields
y = 5,x = 3,z = −2. Now to use the second derivative test, you assemble the Hessian
matrix which is
53 − 73 − 43
( − 73 53 − 43 ) .
− 43 − 43 23
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