Local Extrema and Extrema on Open Intervals在线视频

Today we study the Calculus (I)

The topic is

Local Extrema and Extrema on Open Intervals

We have studied

global maximum values on the given domain

What is our global maximum

Means the highest marks in your school

What means local

means local maximum value

or relative maximum value

How to get it

So now look at

for example

what is the local maximum value in the following figure

Look at the figure

Yes

y equals f(x)

changeable

Who can tell me

what are our local maximum value

or minimum value

Get it x1 minimum

corresponding x2 maximum

x4

x5

these two points are our

corresponding maximum or minimum values

We can find

in the special points

the corresponding tangent line equals what

It is a horizontal line

means the corresponding slopes are 0's

How to get it

So now look at the special point

may be here

Look at the definition of local extrema

Let S

the domain of function f

contain the point c

Case 1

f(c) is a local maximum value of f

if there is an interval from a to b

It is our open interval

Containing c such that

f(c) is the maximum value of f

on the intersection interval from a to b

intersection the given domain S

means local maximum value

Similarly meaning

f(c) is our local minimum value of f

if there is an interval a, b

containing c such that

f(c) is the minimum value on the interval a, b

intersection domain S

local minimum value

f(c) is local extreme value of f

if it is either a local maximum

or a local minimum

means our local extreme value

Try to understand

means global maximum value means

the highest marks in your school

Local means the highest marks market in your class

So now look at Theorem of Local Extrema

The First Derivative Test

look at the theorem

Name First Derivative

means using the first derivative signs

Let f be continuous on an open interval (a, b)

that contains a critical point c

If f prime x larger than 0

for all x in (a, c)

corresponding f prime x less than 0

for all x in (c, b)

and then

f(c) is a local maximum value of f

After that we can see Figure 1

Now look at case 2

f prime x less than 0 for all x in (a, c)

and f prime x larger than 0 for all x in (c, b)

and then f(c) is a local minimum value

This means f prime less than 0

and then larger than 0

At this time

it is our local minimum value

f prime x has the same signs

means both are positive

or both are negative

on both sides of c

and then

f(c) is not a local extreme value of f

Pay attention to the key words

larger than 0

less than 0

or less than 0

larger than 0

At this time

it is our local extrema

Otherwise have the same signs

may be positive

may be negative

At this time

it is not our local extreme value

So now look at the figure

From a, c

positive

from c, b

negative

At this time

it is our local maximum value

Corresponding look at Figure 2

from a, c

negative sign

from c, b

positive sign

At this time

it is our local minimum

a local extreme value

Otherwise

have the same signs

positive sign

it is not a local extreme value

Negative sign

negative sign

not a local extreme value

means not a local extreme value

Why

Because they have the same signs

may be positive sign

may be negative sign

At this time

not a local extreme value

Why These two

positive means increasing

decreasing

maximum

Decreasing

increasing

minimum value

So now look at Theorem of Local Extrema

The Second Derivative Test

Look at the name

second derivative

means using double prime

So our condition means

let f prime and double prime exist

means the first derivative and the second derivative exist

at every point

in an open interval (a, b) containing point c

and suppose that

f prime c equals 0

Pay attention to the given condition

f prime c equals 0

means c is our stationary point

At this time

we will consider the double prime

How to determine the local extrema

If Case 1

f double prime c less than 0

and then f(c) is a local maximum value

Pay attention to

less than 0

and then

maximum value

Case 2

if f double prime larger than 0

at this time

f(c) is a local minimum value

larger than 0 minimum value

Try to understand the theorem

Pay attention to the given condition

f prime c equals 0

And the condition c is our stationary point

and then

according to the signs of double prime

determine

which value is our local maximum

or minimum values

o how to remember

It is a theorem

need us to prove

So now we prove the case 1

Case 1

the condition means

f double prime c less than 0

What means double prime

The second derivative

how to get

From the first

f double prime c equals

the limit f prime x minus f prime c

over x minus c

It is our the definition of limit

Write it f prime c

Why

The given condition

our stationary point

so minus 0 over x minus c

Given condition less than 0

Less than 0

if x doesn't equal c means

corresponding this expression less than 0

Why

According to the meaning of limit

we can get

f prime x time x minus c

The product of the two terms less than 0

We can get

if x less than c

less than c means

x minus c less than 0

At this time

f prime larger than 0

Corresponding

x larger than c

f prime less than 0

So

from a to c positive

c to b negative

This time it is our f(c)

is a local maximum value

In the similar way

try to do

prove the case 2

So work by yourself

Try to prove case 2

Using our theorems

to finish our examples

Now look at Example 1

If f(x) equals

x plus 1 cube time x minus 1 2 over three

Use the first Derivative Test

to identify the local extrema

What is our the First Derivative Test

Yes

f the first prime

Get the corresponding signs

What are signs

The signs between a and c

and then c and b

According to the signs

determine it is our local

or not local extrema

So the first

we should do the f prime

f prime x equals 3 time

bracket x plus 1 square

time something plus another prime

It is our the first derivative

How to do

Yes

get the corresponding common factor

It is our common factor

taken out

Together we can get another expression

Pay attention to

this expression

How to get

the corresponding f prime equals 0

or does not exist

Look at the stationary point

means

this term equals 0

This term equals 0

We can get

the point x equals minus 1

or x equals 7 over 11

And then

this term

pay attention to this term equals 0

We can get our singular poin

Singular point means x equals 1

Together

1 2 3

three points

So we can get

three points

Get the corresponding table

Remember stationary points

and singular points

And from f prime

we can get

the corresponding table

1 2 3

three points

The original domain is divided by 4 intervals

So we can get x

ok

minus infinity to minus 1

minus 1 is our special point

means stationary point

Minus 1 to 7 over 11

7 over 11 is another stationary point

And then

7 over 11 to 1

1 is our singular point

from 1 to positive infinity

So 1, 2, 3, three points

Our original domain is divided by

1 2 3 4

four intervals

So now look at

the corresponding f prime signs

According to this expression

x between this interval

f prime

corresponding positive

This point 0

Corresponding positive 0

Minus means negative

Not exist

Larger than 0 at this time positive

How to determine

Yes

consider the signs of the stationary points

or singular points

This is increasing

This is increasing

So the point is not local extrema

Increasing

increasing means the same signs

Together

increasing

decreasing

means the point is our maximum

means local maximum value

Positive means increasing

decreasing function

increasing function

so local minimum

We can get our conclusion

f at the point 7 over 11

close to 2 point 2

is a local maximum value

Corresponding

f(1) equals 0 is a local minimum value

Try to understand the example

Four steps

the first step get f prime

The second step

try to get the special point

Step 3

making up the table

consider the special points

Step 4

we will determine

the point is our local maximum

or minimum points

So now look at Example 2

If f(x) is given

1 over 3 x cube minus x square minus 3 x plus 4

Use the Second Derivative Test

to identify the local extrema

Do you remember the Second Derivative Test

Yes

according to the sign of the double prime

But do you remember the given condition

And the condition

f prime c equals 0

means c is our stationary point

So now look at f prime x

f is given

We can get

f prime x equals x square minus double x minus 3

We can get this formula

x plus 1 time x minus 3

f prime equals 0

We can get

two stationary points

One is minus 1

another is 3

f double prime

what to do

Yes

from here

The first time derivative

the second derivative

means derivative again

Prime again

double x minus 2

We should determine the signs of f double prime

At what points

At the stationary points

means minus 1 and 3

So determine the sign of

double prime at minus 1

equals minus 4

minus 4 less than 0

Do you remember

less than 0 means local maximum

So f minus 1

is a local maximum value

less than 0

Similar meaning

We will determine

the sign of double prime at the point 3

f double prime at 3 equals 4

4 is larger than 0

So f(3) is a local minimum value

Try to understand

according to our stationary point minus 1 and 3

and the signs of double prime

determine local maximum

or local minimum values

Now look at Example 3

Try to find the local extrema value

as to this expression

f(x) equals x square minus 6 x plus 5

on the given interval

the real number system

How to write it

Yes the first

we should get the first derivative

f prime x equals double x minus 6

2 is our common factor taken out

equals 2 time bracket x minus 3

f prime x equals 0

We can get x equals 3

prime 3 equals 0

3 is our stationary point

determine the sign of f double prime

Look at this

f double prime x from here equals 2

This means at every point

corresponding f the second derivative

always equal 2

means always is a positive number

So f double prime at every point

of course

at 3 point equals 2 larger than 0

Why we select the point 3

Not 30

Not 300

Because 3 is our stationary point

We should consider the sign of

double prime at point 3

Larger than 0 this means

f(3) is our local minimum value

So our conclusion is

f(3) is a local minimum value

Remember local minimum value

So we can get

the following summary

The First Derivative Test

f prime larger than 0

and then

f prime less than 0

between the different intervals

And then f(c) is maximum value

Corresponding

less than 0 and larger than 0

means local minimum value

Together

if f prime has the same signs

all positive or all negative

At this time

f(c) is not a local extreme value

Summary 2

try to remember

the Second Derivative Test

f the first prime

f double prime

exist at every point

in an open interval (a, b)

The (a, b) containing the point c

and suppose that means

and the condition f prime c equals 0

Try to understand

f prime c equals 0

means c is our stationary point

Case 1

f double prime less than 0

corresponding local maximum value

f double prime is larger than 0

it is local minimum value

So try to remember less than 0

maximum value

larger than 0

minimum value

According to our summary

try to finish the question

Try to find the local extrema value as to f(x)

It is a polynomial

equals x cube plus 3 x square minus 24 x minus 20

How to do

Yes

find the first derivative

Look at f the first prime

equals 3 x square plus 6 x minus 24

Get the points ok

what points

Write it this

equals 3 time x plus 4 time x minus 2

If we let f prime x equals 0

Try to get

the corresponding points means

obtain the stationary points

How many points

Two points

one is minus 4

the other is positive 2

We found it

the two stationary points

And then

determine the signs

corresponding stationary points

the double prime

f double prime from this

6 x plus 6

We can determine

f double prime at the first stationary point

minus 4 equals minus 18 less than 0

Less than 0 remember

Yes

so the corresponding local maximum value

is f minus 4

The value is 60

Corresponding

determine f double prime

at another stationary point 2

determine the sign

equals positive 18 larger than 0

At this time

the local minimum value f(2) equals

minus 48

Remember larger than 0

minimum

less than 0

maximum

So the class is over

See you next time