The Mean Value Theorem for Derivatives在线视频

Today we study Calculus (I)

The topic is

the Mean Value Theorem for Derivatives

So now look at the problem

We have the following geometrical fact

Look at the figure

it is a curve

In the pink curve

we can find that

there is a red line

How to write it

If the pink curve is smooth enough

we can find

the red line

which is parallel to the connection line

of A and B

How to find it

It is our question

So now look at this fact

In the picture

the pink curve satisfies

y equals f(x)

The two points are little a and little b

There is a connection line

through a and b

If f(a) equals f(b)

we can find

there are two points ξ1 and ξ2

whose corresponding horizontal tangent lines

1 and 2

are parallel to x-axis

How to find it

Yes I know something from the picture

the right horizontal tangent line

the corresponding

f prime ξ equals 0

means the slope is 0

Now look at this fact

This figure

the pink curve

The equation is y equals f(x)

The two end points

are little a and little b

There are still two points ξ1 and ξ2

If f(a) does not equal f(b)

if the curve is smooth enough

we can find

the two red straight lines are still parallel

to connection line of capital A and capital B

You can ask yourself

if f prime ξ still equals 0

No

why

The lines are only parallel to

the connection line of A and B

not the x-axis

Of course

you ask yourself

how about the prime

It means

how about the slope of the red tangent lines

So now look at

the Mean Value Theorem for Derivatives

There are two conditions

Look at Condition 1

if f is continuous on

a closed interval from a to b

sometimes we can write it as

f belongs to continuous functions

on the closed interval of a and b

Look at Condition 2

and f is differentiable on

its interior little a and little b

means open interval

Sometimes condition 2

we will say

f prime x exists on its interior

between little a and little b

At this time

and then

there is at least one number

“there is at least”

means there exists one number ξ in a and b

satisfying f prime ξ

equals f(b) minus f(a) over b minus a

Sometimes

the equation is written in this form

f(b) minus f(a)

equals f prime ξ time b minus a

It is our Mean Value Theorem

How to prove

From here

we can find

the theorem is true from the figure

Look at the figure

it is a curve smooth enough

with the two end points little a and little b

There are two points ξ1 and ξ2

We can find that

if f(a) does not equal f(b)

there exist two tangent lines parallel to

the cutting lines passing A and B

How to write it

Yes

f prime ξ is parallel to

the connection of A and B

the black line

The corresponding black line slope is two-point formula

equals f(b) minus f(a) over b minus a

From the geometrical meaning

it is obvious

How to prove it

in mathematical language

Now look at our concrete proof

Proof

let a new function

g(x) minus f(a) equals

it is our k

means what

Slope

x minus a

according to the basic form

the point-slope form

What is the point

Yes

it is the given point (a,f(a))

of course

you can write (b,f(b))

In this question

we write (a,f(a))

means the given point

How about the slope

Yes

the slope of two points

capital A and capital B

write it as k

k equals f(b) minus f(a) over b minus a

This is our new function

If let little s(x) function

equals f(x) minus g(x)

means the difference of the two functions

f(x) minus g(x)

taking in the expression for g(x)

g(x) equals f(a) plus k

bracket x minus a

So now pay attention to

two values at the two end points

One is x equals a

taken in

s(a) equals f(a) minus g(a)

f(x)

x equals a

taken in x equals a

this term equals 0

f(a) minus f(a) minus 0 is still 0

So s(a) equals 0

In the similar meaning

we can get

s(b) is still 0

It means

at the two end points

s(a) equals s(b) equals 0

At this time

we can get

there exists

means there is at least one x

belongs to (a,b)

satisfying s prime x equals

what is our f prime x

Pay attention to

this equation

Do the same computation

s prime equals f prime minus g prime

so equals this form

and then

f prime

f(a) prime equals 0

it is our kx

kx prime equals itself k

so f(b) minus f(a) over b minus a

is our s prime x for x

between (a,b)

as an interior point

Now select a special point ξ

means there is at least one number s prime ξ

equals f prime ξ minus the constant

equals 0

Why Write it as the ξ

in the same variables

s prime ξ

equals f prime ξ minus g prime ξ

yes

taking the same values

So we can get

s prime ξ equals 0

Now we can write it

as f prime ξ minus k equals 0

so f prime ξ equals k

k is our constant

f(b) minus f(a) over b minus a

Pay attention to the formula

you can think about something

about the slope

means it is the slope of the connection line of

passing capital A and capital B

It is our Mean Value Theorem

So now look at

Example 1

How to use

the Mean Value Theorem

to prove some question

Now look at the question

Prove that

arctan(x2) minus arctan(x1)

less than or equal to x2 minus x1

x1 and x2 satisfying this equation

How to apply

Do you remember

the basic formula

f(b) minus f(a)

equals f prime ξ time bracket b minus a

so you have some idea

between a and b

x1 and x2

Think it in your mind

Yes

f(x)

means arctan(b) minus arctan(a)

less than or equal to some special value b minus a

according to this idea

We can let f(x)

equals arctan x on [a,b]

means the closed interval

the idea is important

means the step is very important

From the given equation

or the given question

we can find

how to let f(x)

In the question

f(x) equals arctan x

so now

by the Mean Value Theorem for Derivatives

we will have

f(b) minus f(a)

means arctan(b) minus arctan(a)

What is our b

Yes

x2 is our b

What is our a

a is x1

So correspondingly

arctan(x2) minus arctan(x1) equals

equals what

Bracket b minus a

b means x2

a means x1

so x2 minus x1

So now look at this term

f prime ξ

what is our f prime

This function

f prime equals 1 over 1 plus x square

so now means f prime ξ

Corresponding variable

taking in the value ξ

so f prime ξ

means 1 over 1 plus ξ square

Pay attention to the step

How to write

the f prime ξ

From here

we can get

some new results

Pay attention to ξ

ξ between a and b

means x1 and x2

Because 1 over 1 plus ξ square

less than or equal to 1

You ask me why

Because ξ square is a positive number

1 plus nonnegative number

larger than or equal to 1

over

together

1 over 1 plus ξ square less than or equal to1

And then we can get

arctan(x2) minus arctan(x1)

our left side less than

or equal to 1 times x2 minus x1

Together

it is our final result

arctan(x2) minus arctan(x1)

less than or equal to x2 minus x1

How to use the idea

Pay attention to f(x)

Try to find

the function f(x)

its expression

Use the idea

to do Example 2

Prove that

ln bracket 1 plus x between the two values

if x is a positive number

Maybe the question is

more difficult than Example 1

How to understand

In Example 1

we write

arctan(b) minus arctan(a)

how to think about this question

arctan

f prime

yes

Let f(x) equals ln bracket 1 plus x

on the given closed interval

What is the given

This condition

x is larger than 0

gives you some idea

little a is 0

little b is x

So f(x) equals ln bracket 1 plus x

on the closed interval between a and b

means 0 and x

x is a positive number

So the idea is very important

If you know this idea

you will do this step by yourself

So now we together

by the Mean Value Theorem for derivatives

we have f(b) minus f(a)

What is our b

b is x

so f(b) means f(x)

minus f(a)

f(a) means f(0)

Equals f prime ξ time b minus a

means x minus 0

ξ is between a and b

means ξ between 0 and x

According to the basic Mean Value Theorem

we can write it out

And then

the next

how to write it

Look at f(0)

what is our f(0)

f(0) means the value function

at given point 0

f(0) equals ln 1 plus 0 equals 0

ln1 equals 0

So f(0) equals 0

And then f(x)

f(x) itself equals

ln bracket 1 plus x

How about f prime ξ

We should know

f prime x

So f prime x equals 1 over 1 plus x

of course

we know f prime 2 equals 1 over 1 plus 2

f prime 6 equals 1 over 1 plus 6

so f prime ξ equals 1 over 1 plus ξ

So we have

ln 1 plus x minus 0

the left side equals ln bracket 1 plus x

equals this term for x

this term

1 over 1 plus ξ

together

x over 1 plus ξ

So now

we know the concrete expression

for ln 1 plus x

We will do some work for 1 plus ξ

according to what

According to the domain

So now

ξ is between 0 and x

ξ is between 0 and x in this question

x is a positive number

plus 1

0 plus 1 less than ξ plus 1 less than x plus 1

together

get this

Positive number

positive number

positive number

we can get this formula

1 over 1 plus x less than

1 over 1 plus ξ less than 1

And then

pay attention

times x

times x

times x

Together they are equal

So we can get this expression

every term times x

equals this term

Pay attention to

the middle term

middle term means x over 1 plus ξ

this term

it is our ln 1 plus x

so together

get this result

So in this question

you should pay attention to

how to get our f(x)

If we know f(x)

how to make about our 0 and x

means little a and little b

If you know the idea

the question's proof is very easy

So now

we have our summary

about the Mean Value Theorem

The basic theorem tells us this expression

f prime ξ is some slope

for the connection line

of passing capital A and capital B

the two points

f(b) minus f(a) over b minus a

if ξ is between a and b

So now

look at Case 1

we can write it in this form

f(b) minus f(a) means

the difference of the two function values

equals some prime at some given point

times the difference of the two variables

means times b minus a

If a is larger than b

the formula still holds

Case 2

we can write it in another form

special case

If b equals x plus △x

a equals x

at this time

we can write

f(b) minus f(a) equals f prime ξ

copy this term

how about b minus a

In this question

b

x plus △x

a

x

the difference of the two values equals △x

so equals f prime ξ time △x

Pay attention to

ξ is between a and b

it is our interior point

Look at Case 3

△y means the increment of the function value

△y equals f prime ξ time △x

how about △y

We can write it

as x plus θ time △x

θ between 0 and 1

△y means the increment of the function value

means the accurate expression of increment △y

How to write it in this term

There are some relationship

between ξ and this expression

pay attention

ξ is between x and △x plus x

so ξ is an interior point

between the end points

so ξ equals x plus △x times some constant

x plus △x times some proportion

some ratio

How about the ratio

Yes

the ratio is between 0 and 1

Pay attention

θ equals 0

θ equals 0 means the left end point

θ equals 1 means x plus 1 time △x

the right end point

Try to understand

the three different forms

So now

use the Mean Value Theorem

to finish our questions

Look at the question

proof question

Prove arcsin(x) plus arccos(x) equals half π

x is between minus 1 and 1

the closed interval

How to prove

Now look at

we can write it

Let the function f(x)

equals arcsin(x) plus arccos(x)

the corresponding domain of x is minus 1 and 1

the closed interval

How to write it

f(x) always equals half π

What means always equals half π

Correspondingly

do you remember

C prime equals 0

So we try to get f prime

Corresponding f prime x

arcsin(x) prime equals

1 over square root of 1 minus x square

do you remember

arccos(x) prime

Correspondingly

equals minus 1 over square root of 1 minus x square

They are the same expressions

the difference is positive

minus

together

and summation equals 0

f prime x equals 0

how about f(x)

What kind of function's prime equals 0

Yes

f(x) equals C forever

for every x

always equals C

f(x) equals C for every real number x

in the given domain

How to understand this question

This means

x equals 0

f(x) equals C

x equals minus 1

f(x) equals C

x equals positive 1

f(x) still equals C

So we try to find the C

So now

take special point 0

f(0) equals arcsin(0) plus arccos(0)

we can get

f(0) equals this

Do you know this

Yes

arcsin(0) equals 0

how about arccos(0)

Equals half π

together

equals 0 plus half π

means half π

From this expression

we know f(0) equals half π

This means

capital C is half π

It is a constant

f(x) always equals C for every x

So together

arcsin(x) plus arccos(x) equals half π

Try to remember this formula

Prove by yourself in the similar way

arctan(x) plus arccot(x) equals half π

if x is any real number

means the domain is our real number system

So now look at this question

The question is try to find c

c means some concrete number for f(x)

equals double square root of x on 1 to 4

the closed interval

How to find the c

satisfying the Mean Value Theorem

We can find it

f prime x equals 1 over square root of x

Ok

we know

the basic formula f prime

and then

f(b) minus f(a) over b minus a

In this special case

a equals 1

b equals 4

So f(4) minus f(1)

f(4) means double square root of 4

means 2 times 2

equals 4

minus f(1)

f(1) means double square root of 1

so minus 2

Over b minus a

4 minus 1

so equals 3

Together

the value is 2 over 3

Try to find

the c satisfying f prime c equals 2 over 3

So we must have

f prime c equals 2 over 3

what is our f prime c

Look at this

f prime x means for every x

in the given domain

always equals 1 over square root of

the corresponding value

so f prime c

square root of c

f prime 2

square root of 2

In this question

f prime c equals 1 over square root of c

equals 2 over 3

c is an interior point

between a and b

Solving this equation

we can get

the single solution

means the single point is c equals 9 over 4

The question means

how to find the concrete point ξ

In this question

means c

satisfying our Mean Value Theorem

Try to understand the theorem

So now class is over

see you next time