# Second derivative test

**Last time:** Max/Min problems. Least squares approximation.

**Today: **Second derivative test. Global max/min: Boundaries and infinity.
Level curves and level surfaces.

**Reading Material:** From Simmos 19.1, 19.7. From the Lecture Notes SD.

## 2 Second derivatives

Given a function z = f(x, y) we denoted the first partial derivatives of f as

Then it is simple to define the second partial derivatives:

**Remark.** There is a very surprising theorem: if all second derivatives
are continuous then

and this in fact is going to be the case for all functions we will be dealing with in this class->

## 3 2^{nd} derivative test

Last time we learned that for a function z = f(x, y) local max/min occur on
critical pints, that is

on points (x_{0}, y_{0}) such that

We also learned that not all critical points are either a local max or min,
take for example the point

(0, 0) critical point for the function z = x^2 − y^2 which is a saddle point.

**Question:** Is there a way to recognize if a critical point is a max a
min or a saddle?

The answer to this question is a “partial” **YES**. We first have to
define the **Discriminant D:**

**Definition 1.** Given a function z = f(x, y) we define

We have the following result:

**Theorem 1.** Assume that (x_{0}, y_{0}) is a critical point for
the function z = f(x, y). Let D be the

discriminant above evaluated at (x_{0}, y_{0}). Then

1. if D < 0 -> the critical point is a saddle

2. if D > 0 (note f_{xx} = 0 impossible)

(a) and f_{xx} > 0 -> the critical point is a min

3. if D > 0

(a) and f_{xx} < 0 -> the critical point is a max

4. if D = 0 no info

Exercise 1. Consider first the function f(x, y) = x^2 − y^2. We already know
that (0, 0) is a critical

point. We have: f_{x} = 2x, f_{y} = −2y and hence

which implies D = −4 < 0 and (0, 0) is a saddle point:

On the other hand if f(x, y) = x^2 + y^2

we still get the same critical point (0, 0), but in this case

so max or min but f_{xx} = 2 > 0 so min.

We finally consider the function f(x, y) = x^4 + y^4

It is easy to check that again (0, 0) is a critical point, but now f_{xx} =
12x^2, f_{yy}= 12y^2 and f_{xy} = 0

and at the point (0, 0)

D = 0,

so there is no conclusion that we can make. We need to make a sort of
“manual” inspection: since

for any point near zero we have

it follows that (0, 0) is a local min point.

**Remark.** A proof of the **Second Derivative Test** is rather
involved. But there are two fundamental

steps in it : the quadratic approximation of a function near a point (otherwise
called the Taylor

polynomial of degree 2) and the Second Derivative Test for quadratic functions.
These two facts are

stated in the following lemmas:

**Lemma 2. [ Quadratic Approximation of a Function of two variables ]**
Given a function

z = f(x, y) and a point P_{0} = (x_{0}, y_{0}) we have
the following second order approximation for f near

P_{0}:

**Lemma 3. [Second Derivative Test in a Special Case]** Consider the
quadratic function

Clearly O = (0, 0) is a critical point. Then

is a minimum point | |

is a minimum point | |

is a saddle point. |

For the proof of this lemma see SD in Lecture Notes.

**Remark.** Notice that in Lemma 3

and hence D = AC −B^2. To prove the Second Derivative Test for arbitrary
functions one first uses

Lemma 2 to approximate the function f(x, y) near the critical point by a
quadratic function . Then

one uses Lemma 3.

## 4 Global max/min: Boundaries and Infinity

In this section we pass from considering local max/min to global ones. Let’s
start with the following:

**Exercise 2.** Consider the function f(x, y) = x^2 + y^2 defined on the
domain

Compute global max and min of f on this domain.

** Solution :** We first observe that if the global max and min occur inside W then
it must be also a

local max and min. We know from the previous example that there is no local max
for f, but there

is a local min at (0, 0). This is also the global minimum since for any other
point P = (x_{0}, y_{0}) one

has

But what about the global max? For this we have to look at what happens to f
at the boundary

(perimeter) of W, also denoted with @W. We have to evaluate f on this boundary
and decide if f

has a max on it. Since the function and the domain W are symmetric with respect
to the x−axis

we will only check the values of f on the upper boundary of W, let’s denote it
:

It is easy to parametrize by

The value of f on is given by

Now h(x) is a function of one variable and it is not difficult to show that
on its domain its max

occurs at x = 0, which translate into the fact that f has two max on the
boundary of W, more

precisely at P_{+} = (0, 1) and P_{-} = (0,−1). We also have that

and this is the global max of f on W.

**Question:** What is the global max of f when we allow
it to be defined on the whole plane?

Since there is no boundary we can evaluate f on points P =
(x, y) “going to infinity”, that is such

that |x| or |y|
. Clearly on these points the value of f = x^2
+ y^2 will tend to infinity as

well, so the function does not admit a global max.

## 5 Level curves and level surfaces

You already have seen an example of level curves: any **
topographic map** is made by drawing several

**level curves** relative to the 3D landscape that it represents. In fact if
h(x, y) is the function that

represents the altitude at point P = (x, y), then the graph of this function is
the landscape and the

topographic map is made by representing the (level ) curves that are obtained by
considering all the

points at a certain altitude.

We have the following definition:

**Definition 2**. Given a two variable function z = f(x,
y), for any scalar C we call the level curve

γC of f the set

These level curves are represented on the xy−plane and
their collection is called** contour map:**

**Exercise 3.** Consider the function f(x, y) = y −
x^2. Draw the contour map of this function.

We have a similar definition for functions of three variables :

**Definition 3.** Given a three variable function w =
f(x, y, z), for any scalar C we call the level

surface SC of f the set

These level surfaces are represented on the xyz−space and
their collection is also called **contour
map.**

**Exercise 4.** Consider the function f(x, y, z) = 4x^2
+ y^2 + 9z^2. Draw the contour map of this

function.

**Study Guide 1.** Think about the following questions:

• Why in Theorem 1 I said that if D > 0 then it must be
f_{xx} ≠ 0?

• Draw several contour maps and try to see how the picture
looks like at max, min and saddle

point.

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