maxima and minimaottummath.com/2413/files/handouts/4.1.pdf · 1 maxima and minima 1 extrema 2 there...

1

Maxima and minima

1

Extrema

2

There are two kinds of extrema

A global (or absolute) maximum (or minimum) has the largest (or smallest) possible y-value on the domain

A local (or relative) maximum (or minimum) has the largest (or smallest) possible y-value on an interval (neighborhood)

Remarks

3

We will learn how to

Find absolute exterma on closed intervals

Find intervals where a curve is increasing /decreasing

Find intervals where a curve is concave up / down

Find relative exterma on open intervals

First, we must learn the language of exterma

2

Absolute Maximum

Absolute Minimum

Absolute Maximum or Minimum

Relative Maximum

Relative Minimum

4

Closed Intervals

5

Extrema can occur at end points of closed intervals

At a Maximum

0 xf 0 xf 0 cf

6

The derivative changes from positive to negative

The curve changes from increasing to decreasing

The curve is concave down

Tangent lines are above the curve

3

At a Minimum

0 xf 0 xf 0 cf

7

The derivative changes from negative to positive

The curve changes from decreasing to increasing

The curve is concave up

Tangent lines are below the curve

8

Difference – Concave and Convex

Concave Surface Convex Surface

Concave Lens Convex LensRemember

Concave

as

Caved In

Fermat’s Theorem

9

If has a local maximum or minimum at , and

if exists, then 0

f c

f c f c

Note: Extrema also occur where the derivative is not defined and at end points of closed intervals

1601-1665

French lawyer and amateur mathematician

4

Critical Points

Let be defined at . If 0 or

is undefined at , then is a of

f x c f c f c

c c f

critical number

and , is a c f c critical point

10

Extrema occur where the slope is:

zero or undefined

These points may be maximum or minimum points, but they also be neither

Critical Points

11

The slope is zero at this point, but it neither a maximum or a minimum

Example

12

3 22 15 36f x x x x

Find critical numbers

26 30 36f x x x

6 3 2x x

3 0 3x x

We take the derivative, and factor

To find where the derivative is zero, we use the factors

2 0 2x x

2,3x Together these are the critical numbers

5

Example

13

4ln 1g x x x


4

11

g xx

5

1

x

x

5 0 5x x

We take the derivative, and rewrite as one term

To find where the derivative is zero, we use the numerator

1 0 1x x To find where the derivative is undefined, we use the denominator


5

lnf x xx


2

1 5f x

x x

2

5x

x

5 0 5x x

We take the derivative, and rewrite as one term

To find where the derivative is zero, we use the numerator

2 0 0x x To find where the derivative is undefined, we use the denominator


Practice

14

Extrema

15

Maximum and minimums occur at either critical points or end points of a closed interval

Extrema refers to maximum and minimum points on a curve

Critical points are not always a maximum or minimum

End points of a closed interval are not always a maximum or minimum

6

The Usual Suspects

16

The points of interest when exploring a function to find it’s extrema are

(1) points where the first derivative is zero

(2) points where the first derivative is undefined

(3) endpoints, if on a closed interval

Extreme Value Theorem

A function that is continuous on a closed interval

[a,b] will have both an absolute maximum and an

absolute minimum on the interval

f

If:• f is continuous• on closed interval

Then:• f has absolute max and• f has absolute min

The extreme value theorem tells us that all continuous function reach a top and a bottom on a closed interval17

Karl Weierstrass

18

Proved the Extreme Value Theorem in 1860

A German mathematician, known as the “father of modern analysis”

1815-1897

7

19

Remarks

We can refer to the Extreme Value Theorem as the EVT

This theorem is an existence theorem in the sense that it doesn’t tell us where the extrema occurs, but promises there are extrema

In some problems, all that we need to know is that they do exist

Every global extrema is a relative extrema or an endpoint extrema

Example Find the location of the indicated absolute extremum for the function

Minimum y = -3

20

Maximum y = 3

Closed Interval Method

To find the absolute maximum and minimum values of a continuous function f on a closed interval [a,b]

1. Find the values of f at the critical points of f in (a,b)

2. Find the values of f at the endpoints of the intervali.e., f(a) and f(b)

3. The largest of these values is the absolute maximum value; the smallest of these values is the absolute minimum value

21

8

f(x) = x3 - 9x2 + 24x - 13

on the interval [0,5]

3 2

2

9 24 13

3 18 24

3 2 4

Critical numbers

2, 4

y x x x

y x x

x x

x

Example

22

x f(x)

endpoint 0 -13 min

critical point 2 7 max

critical point 4 3relative

min

endpoint 5 7 max

f(x) = x3 - 9x2 + 24x - 13

on the interval [0,5]Example

23

f(x) = x3 - 9x2 + 24x - 13


Using TI to create table

24

9

f(x) = x3 - 9x2 + 24x - 13



25

f(x) = x3 - 9x2 + 24x - 13



26

f(x) = x3 - 9x2 + 24x - 13


TI Solution on closed interval

Note: These are the x-values, the absolute min is at f(0) and the absolute max is at f(2) or f(5) 27

Did not find x=5

10

28

f(x) = x3 - 9x2 + 24x - 13


Using TI graph

29

f(x) = x3 - 9x2 + 24x - 13


Using TI graph

30

y = f(x) = x 2

on the interval [2,5]

x y

2 4

2.5 6.25

3 9

3.5 12.25

4 16

4.5 20.25

5 25

The maximum value of y on the interval[2,5] is at x=5

Example

11

31

y = f(x) =x2

on the interval [2,5)

x y

2 4

2.5 6.25

3 9

3.5 12.25

4 16

4.5 20.25

4.9 24.01

4.999 24.99001

This function does not have a maximum value of y on the interval[2.5)

A maximum or minimum may not exist on an interval that is not closed

Example

Remarks

Let us be certain we understand why y = f(x) = x2

does not have a maximum on the interval [2,5)

The problem is that no matter how close we make x to 5 there are still points between x and 5 that are greater

The only way we can avoid this is to have a closed interval [2,5]

24.99999999y

25y

points

32

2 on 5,1xf x x e

Find the absolute maxima and minima of the function

2 2x xf x x e xe 2xe x x

2 0

0 0

x f x

x f x

2,0 are critical numbersx

Example

33

12

x f(x)

-5 0.168449 Relative Minimum

-2 0.541341 Relative Maximum

0 0 Absolute Minimum

1 2.71828 Absolute Maximum

2 on 5,1xf x x e

Find the absolute maxima and minima of the function

Example

34

Example

35

2ln 1 on 1,1x x

Find absolute extrema

2

2 1

1

x

x x

We take the derivative f’ = 0 when x = -1/2

We construct our table

x f(x) f(x)

-1 0 0

-1/2 ln(3/4) -0.287682

1 ln(3) 1.09861

Our min is ln(3/4) at x = -1/2 and max is ln(3) at x = 1

x

yExample

36

2ln 1 on 1,1x x


Our min is ln(3/4) at x = -1/2and max is ln(3) at x = 1

13

Example

37

2ln 1 on 1,1x x


TI Result

Example

38

on 1,5x x


1 xxy x x

1 lnln ln x x

y xx

We will use logs to take the derivative

2

11 lnx x

y x

y x

2

1 ln x

x

2

1 ln xy y

x

2

1 lnx x x

x

Example

39

on 1,5x x


We will find where f’ = 0 2

1 lnx x xy

x

1 ln 0x x e

We will build the table

x f(x) f(x)

1 1 1

e e 1/e 1.44467

5 5 1/5 1.37973

Min at (1,1) and Max at (e, e1/e)

2 0 0 1,5x x

14

x

Example

40

on 1,5x x


Min at (1,1) and Max a (e,e1/e)

1 0 1 2,6x x

4ln 1 on 2,6g x x x


4

11

g xx

5

1

x

x

5 0 5x x

We will build the table

x f(x) f(x)

2 2 2

5 5-8ln2 -0.545177

6 6-4ln5 -0.437752

Min at (5,5-8ln2) and Max at (2,2)

Practice

41

maxima and minimaottummath.com/2413/files/handouts/4.1.pdf · 1 maxima and minima 1 extrema 2 there...

Documents