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
(x)
whose ij^{th}entry at the point x is
-∂2f-
∂xi∂xj
(x)
is called theHessian matrix. The eigenvalues of H
(x )
are the solutions λ to the equation det
(λI − H (x))
= 0.
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.
PICT
Theorem 13.3.2Let f : U → ℝ for U an open set in ℝ^{n}and let f be a C^{2}function and supposethat at some x ∈ U, ∇f
(x)
= 0. Also let μ and λ be respectively, the largest and smallest eigenvaluesof the matrix H
(x)
. 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
(x,y)
= 10xy + y^{2}. Find the critical points and determine whether they arelocal minima, local maxima or saddle points.
First ∇
(10xy + y2)
=
(10y,10x+ 2y)
and so there is one critical point at the point
(0,0)
. What is it?
The Hessian matrix is
( )
0 10
10 2
and the eigenvalues are of different signs. Therefore, the critical point
(0,0)
is a saddle point. Here is a
graph drawn by Maple.
PICT
Here is another example.
Example 13.3.4Let f
(x,y)
= 2x^{4}− 4x^{3} + 14x^{2} + 12yx^{2}− 12yx − 12x + 2y^{2} + 4y + 2. Find thecritical points and determine whether they are local minima, local maxima, or saddle points.
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.
PICT
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
(x,y)
= 6xy^{2}−2x^{3}−3y^{4}. Show that
(0,0)
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.
PICT
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
(0,0)
is a critical point, note that f_{x}
(0,0)
= f_{y}
(0,0)
= 0
because f_{x}
(x,y)
= 6y^{2}− 6x^{2}, f_{y}
(x,y)
= 12xy − 12y^{3} and so
(0,0)
is a critical point. So are
(1,1)
and
(1,− 1)
. Now f_{xx}
(0,0)
= 0 and so are f_{xy}
(0,0)
and f_{yy}
(0,0)
. 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 point.
However, suppose you took x = t and y = t and evaluated this function on this line. This reduces to
h
(t)
= f
(t,t)
= 4t^{3}− 3t^{4}), which is strictly increasing near t = 0. This shows the critical point
(0,0)
of f is
neither a local max. nor a local min. Next let x = 0 and y = t. Then p
(t)
≡ f
(0,t)
= −3t^{4}. Therefore,
along the line,
(0,t)
, f has a local maximum at
(0,0)
.
Example 13.3.6Find the critical points of the following function of three variables and classifythem 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
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.