Maxima and Minima慕课视频播放-微积分1（Calculus I）-MOOC慕课视频教程-柠檬大学

Today we study the Calculus (I)

The topic is Maxima and Minima

Now look at the figure

can you find the maximum value and minimum value

Maybe some points are our end points

maybe some points are the points between a and b

And then you can ask yourself some questions

does f(x) have a maximum or minimum value

on the given closed interval S between a and b

If it does have a maximum or minimum value

where are they attained

If they exist

what are the maximum and minimum values

How to determine

So now we will tell you some idea

to show the questions

Now look at

let S is our domain of the function f

contain the point c

we will say that

f(c) is our maximum value of function f on the given domain S

Under the condition

if f(c) is larger than or equal to f(x)

for all x in the domain

Pay attention to

it is our maximum value

larger than or equal to

And then f(c) is our minimum value of function f on the given S

if f(c) is less than or equal to f(x)

for all x in the given domain

minimum value

Together

if f(c) is the extreme value of f on the given domain

If it is either the maximum value

or the minimum value

This means both a and c

maximum value and minimum value means extreme value

extreme value

The function we want to maximize or minimize

is our objective function

means objective function

So now we look at the theorem

how to get the Maxima and Minima

Theorem A

Maxima–Minima Existence Theorem

If f is a continuous function

on the given closed interval from a to b

we can get

f will attain both a maximum value

and a minimum value there

means

we can get maximum value and minimum value

So now look at the definition

If f prime c equals 0

we can get

give a new definition for the special point c

c is a stationary point

means stationary point

Look at Definition 2

if prime c fails to exist

this means does not exist

At this time

give c a new name

c is a singular point

means special case

So it is our singular point

The following three kinds of points

end points

stationary points and singular points

are our critical points

If you ask me what are our critical points

You should consider the three kinds of points

end points

stationary points and singular points

Remember the definitions

Now look at the following Theorem B

Critical Point Theorem

f(c) is an extreme value

at this time

c must be a critical point

Do you remember

what is our critical point

Yes there are three cases

means three kinds of points

So c may be an end point of the domain

f prime c equals 0

means maybe c is a stationary point

Maybe Case 3

f prime c does not exist

means singular point

So we have the theorem B

f(c) is an extreme value

at this time

c must be one kind of critical point

end point

stationary point or singular point

So now look at how to get the maxima and minima

Try to remember

Step 1

find f prime x equals 0

or f prime x does not exist

What

What is our goal

Try to get the corresponding critical points

from the given domain

We will know the end points

the end points

From f prime x equals 0

we can get the corresponding stationary points

f prime x does not exist

at this time

we can get the singular points

And then evaluate f at the critical points

evaluate the values

The last step

compare

compare the values to find the maxima and minima

Try to remember

So now look at Example 1

Try to find

the critical points of f(x) equals

minus 2 x cube plus 3 x square on the given domain

What are our critical point

Do you remember the definition

Yes there are three kinds of points

end points

so and then

stationary points and singular points

So now look at the solving idea

f prime x equals

minus 2 time 3 equals minus 6 x square

plus 6x equals 0

We can get

if f prime equals 0

get the two solutions

x equals 0 or x equals 1

It is our stationary point

No singular point

And then consider the end point

There are two

one is minus half

the other is 2

So together

the critical points are

minus half and 2

and 0 and 1

We have got it

the critical points

Now look at Example 2

Try to find

maximum and minimum values of the given function

f(x) equals x cube on the given domain

minus 2 to 2

How to get it

Do you remember the four steps

Yes I remember

Step 1

try to get the corresponding f prime x

What is our f prime x

Yes

f is given

x cube

So corresponding

the first derivative equals 3 x square equals 0

f prime x equals 0

There is only one solution

x equals 0

It is our stationary point

Step 2

count the number of critical points

What are our critical points

Yes consider end points

minus 2

2 and the stationary point

Together

there are three points

minus 2

0 and 2

Step 3

evaluate the values of the function

at the different critical points

f(0) together f(0) equals 0

f(2) equals 2 cube means 8

f minus 2 equals minus 2 cube

means minus 8

The last step

do you remember

Yes compare the values

We can find

8 is our maximum value

minus 8 is our minimum value

So together

f maximum value is 8

get it at the point 2

Minimum value is minus 8

we can get it

at the point minus 2

So now the question is ok

We will look at Example 3

Try to find

the maximum and minimum values

of the given expression

Now maybe it is complex

But the domain is the same

the closed domain from minus 2 to 2

the same idea

So now we will give the idea

If you know how to draw the figure

it is ok

But we don't know how to draw the figure

How to do

According to our computation steps

now look at

how to compute

Yes

get the f prime x

We can get f prime x

the first derivative for our function

x 2 over 3

equals 2 over 3 x

2 over 3 minus 1

equals minus 1 over 3

and then minus 1 by 3 bracket

x square minus 1 minus 2 over 3

time u prime

why

Chain Rule Yes

Chain Rule by the derivative

time u prime means double x

How to get f prime equals 0

or doesn't exist

Taking out the common factor

what is our common factor

Pay attention to

x power is 1

This

x power minus 1 over 3

which is smaller

Yes

minus 1 by 3

so taking out

Together this term

corresponding minus 2 over 3

known

this term means the power is 0

Together minus 2 over 3

it is our common factor

It is an equation

We can get another term

And now it is our expression

Write it in this form

Pay attention to this

and this term

become denominator

And then this is our numerator

What get it

Yes critical points

How to get it

From numerator equals 0

we can get our stationary points

Denominator equals 0

we can get some singular points

So we can get singular points

x equals 0

or x equals positive or minus 1

Stationary points

we can get x equals positive

or minus 1 over square root of 2

And then end points

it is very easy

We can find it

positive or minus 2

Together

1 2 3 4 5 6 7

how many critical points

Yes seven points

Do you remember Step 3

Yes evaluate the values

Seven points may be seven values

But you should pay attention to the fact

f(x) is an even function

So we only consider four values

it is ok

Because it is an even function

so f(0) equals 1

f positive or minus

they have the same values

So we only need to compute one value

it is ok

And then f(1) equals 1

f(2) we can get it

So now the last step

compare the values

at different critical points

compare each other

This is our maximum value

This is our minimum value

So we have our final result

f corresponding maximum value

and minimum are got

So now we look at the figure

Compare

yes

corresponding maximum value

get it at the 2 points ok

positve 1 over square root of 2

or minus 1 over square root of 2

And then

minimum values are got

What points

Yes

at the end points

may be positive 2

may be minus 2

Why

It is even function

symmetrical

So at the positive 2 or minus 2

we get the minimum value

So try to understand the question

using the property of even function

and symmetrical property

So now we look at the summary

How to get the maximum and minimum value

Step 1

get f prime x

find it equals 0 or does not exist

the corresponding solutions

Step 2

together

consider the critical points

including end points

stationary points

and singular points

Step 3

evaluate the corresponding values

at each of these critical points

Last step

compare the values

and then find the maximum and minimum

Try to use the four steps

So now

using the four steps

We will finish our questions

Remember the smallest of these values

is the minimum value

corresponding

the largest of these values

is our maximum value

Using the four steps

we will solve some questions

Now look at Question 1

Try to find

the maximum and minimum values

of the function f(x)

f(x) equals x 2 over 3

on the given closed interval from minus 1 to 2

How to do

Yes

according to our summary

the first step

try to find f prime x

This question is very easy

f prime is our x-λ form

So the corresponding first derivative

equals 2 over 3 time x

the power is minus 1 over 3

So f prime x equals 0

We can find

f prime 0 does not exist

no stationary point

So f prime 0 does not exist

This means 0 is our critical point

And then

count the number of critical points

How many

0 is our

the corresponding critical point

Consider the end points

so together

0 minus 1 and 2

1 2 3

there are three points

And then

evaluate function values

f at the given point minus 1 equals 1

f(0) equals 0

f(2) equals the third root of 4

Compare

corresponding

which one is the biggest

which one is the smallest

We found 0 is our minimum

and this number is our maximum

In the similar way

we can do the question 2

Try to find

the maximum and minimum values of f(x)

equals minus 2 x cube plus 3 x square

on the given closed interval

Yes using the four steps

Step 1

try to find f prime x

What is our f prime

Yes

minus 2 time 3 x square equals

minus 6 x square plus 6 x

equals 0

f prime equals 0

we can find two points

means there are two solutions

One is x equals 0

the other is x equals 1

The corresponding stationary points

2 points

Count the number of critical points

critical points 0, 1

and then

end points minus half and 2

1, 2 3 4

together four points

Evaluate the corresponding function values

f at the given point minus half

the value is 1

The 0 point the value is 0

f value at the given point 1 is 1

Value at the point 2 is minus 4

Compare the values

We can find

f maximum value is 1

minimum value is minus 4

So try to remember the four steps

The class is over

See you next time

Maxima and Minima在线视频

Maxima and Minima课程教案、知识点、字幕

微积分1（Calculus I）课程列表：

Course Introduction

Chapter 1 Limits

Homework 1

Chapter 2 The Derivative

Homework 2

Chapter 3 Applications of the Derivative

Test 1

Chapter 4 The Definite Integral

Homework 4

Chapter 5 Applications of the Integral

Homework 5

Chapter 6 Transcendental and Functions

Chapter 7 Techniques of Integration

Homework 7

Chapter 8 Indeterminate Forms and Improper Integrals

Test 2

Maxima and Minima笔记与讨论

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