MAT 1221Survey of Calculus

Section 3.2Extrema and the First

Derivative Test

http://myhome.spu.edu/lauw

Expectations Check your algebra. Check your calculator works. Make sure you have all the important

details for each point you test.

0, 0

1 Minute… You can learn all the important concepts

in 1 minute.

1 Minute… High/low points – most of them are at points

with horizontal tangent

1 Minute… High/low points – most of them are at points

with horizontal tangent.

Highest/lowest points – at points with horizontal tangent or endpoints

1 Minute… You can learn all the important concepts

in 1 minute. We are going to develop the theory

carefully so that it works for all the functions that we are interested in.

There are a few definitions…

Preview Define

• Local (relative) max./min.• Absolute max./min.

First Derivative Test Closed Interval method

Absolute Max has an absolute maximum at on if for all in ( =Domain of )

c

D

Absolute Min has an absolute maximum at on if for all in ( =Domain of )

cD

Extreme Values The absolute maximum and minimum

values of are called the extreme values of .

x

Example 0y

Absolute max.

Absolute min.

Local (Relative) Max/Min has an local maximum at if for all in some open interval containing

has an local minimum at if for all in some open interval containing

x

Example 0y

Local max.

Local min.

Note An end point is not consider as a local

max/min.

The First Derivative TestSuppose that is a critical number of a continuous function (a) If changes from positive to negative at

, then has a local maximum at .(b) If changes from negative to positive at , then has a local

minimum at .

(c) If does not changes sign at , then has no local max. or min. at .

The First Derivative Test

x

y

a b

)(xfy

c

>0

=0

<0

Local max.x

y

a b

)(xfy

c

>0

=0

<0

Local min.

Example 1 (Continue from 3.1)

The local min. value of is f(3)=1

3

)3,( ),3(

2( ) 6 103

f x x xf

Expectations The first part of the problem is to find the

interval of increasing/decreasing. The second part is to find the local

max./min. from the results of the first part You are expected to answer the problem

formally with a statement“The local min. value(s) of is .”

How to find Absolute Max./Min.? It is easy if the domain is a closed

interval Fact: It is difficult to do that if the domain

is NOT a closed interval.

The Extreme Value Theorem If is continuous on a closed interval ,

then attains an absolute max value and an absolute min value at some numbers and d in .

No guarantee of absolute max/min if one of the 2 conditions are missing.

x

How to find Absolute Max./Min.?y

Local max.

Local min.

Absolute min.

Absolute max.

The Closed Interval Method Idea: the absolute max/min values of a continuous function on a closed interval only occurs at1. the local max/min (the critical numbers) 2. end points of the interval

The Closed Interval Method To find the absolute max/min values of a

continuous function f on a closed interval :1. Find the values of at the critical numbers of in .2. Find the values of f at the end points.3. The largest of the values from steps 1 and 2 is the

absolute maximum value; the smallest of the those values from is the absolute minimum value.

The Closed Interval Method To find the absolute max/min values of a

continuous function f on a closed interval :1. Find the values of at the critical numbers of in .2. Find the values of at the end points.3. The largest of the values from steps 1 and 2 is the

absolute maximum value; the smallest of the those values from is the absolute minimum value.

The Closed Interval Method To find the absolute max/min values of a

continuous function on a closed interval :1. Find the values of at the critical numbers of in .2. Find the values of at the end points.3. The largest of the values from steps 1 and 2 is the

absolute maximum value; the smallest of the those values from is the absolute minimum value.

Example 1Find the absolute max/min values of

]5,3[on 112)( 3 xxxf

2 17, 2 15

3 10, 5 66

f f

f f

ExpectationsFormally answer the problem in this form: