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
that replaces
in the second derivative test for a function of one variable.
Second derivative test
Let
be a function of two variables for which the first- and second-order partial derivatives are continuous on some disk containing the point
Suppose
and
Define the quantity
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
be a function of two variables for which the first- and second-order partial derivatives are continuous on some disk containing the point
To apply the second derivative test to find local extrema, use the following steps:
Determine the critical points
of the function
where
Discard any points where at least one of the partial derivatives does not exist.
Calculate the discriminant
for each critical point of
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:
Step
of the problem-solving strategy involves finding the critical points of
To do this, we first calculate
and
then set each of them equal to zero:
Setting them equal to zero yields the system of equations
The solution to this system is
and
Therefore
is a critical point of
Step 2 of the problem-solving strategy involves calculating
To do this, we first calculate the second partial derivatives of
Therefore,
Step 3 states to check
[link] . Since
and
this corresponds to case 1. Therefore,
has a local minimum at
as shown in the following figure.
The function
has a local minimum at
For step 1, we first calculate
and
then set each of them equal to zero:
Setting them equal to zero yields the system of equations
To solve this system, first solve the second equation for
y. This gives
Substituting this into the first equation gives
Therefore,
or
Substituting these values into the equation
yields the critical points
and
Step 2 involves calculating the second partial derivatives of
Then, we find a general formula for
Next, we substitute each critical point into this formula:
In step 3, we note that, applying
[link] to point
leads to case
which means that
is a saddle point. Applying the theorem to point
leads to case 1, which means that
corresponds to a local minimum as shown in the following figure.
The function
has a local minimum and a saddle point.
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