Tuan Nguyen – Community College of Philadelphia 12/9/2012

MEAN VALUE THEOREM AND ITS APPLICATION

MEAN VALUE THEOREM

1. ROLLE’S THEOREM

Formal Statement: If ( )f x is a continuous function on [ ; ]a b , is differentiable on ( ; )a b and ( ) ( )f a f b

so there is a number ( ; )c a b so that '( ) 0f c .

Proof:

Because ( )f x is continuous on [a; b], according to Weierstrass’s theorem, we have ( )f x has its maximum

M and its minimum m on [a; b].

- If M = m we have ( )f x is a constant on [a; b], so for every ( ; )c a b we have '( ) 0f c .

- If M > m, because ( ) ( )f a f b so there is a number c (a; b) so that ( )f c m or ( )f c M , according to

Fermat’s theorem we have '( ) 0f c .

Consequence 1: If ( )f x is differentiable on (a; b) and ( )f x has n real roots (n is bigger than 1) on (a; b) so

that '( )f x has at least n - 1 real roots on (a; b).

Consequence 2: If ( )f x is differentiable on (a; b) and '( )f x has no real root (a; b) so that ( )f x has

maximum one real root on (a; b).

Consequence 3: If ( )f x is differentiable on (a; b) and '( )f x has maximum n real roots (n is a positive

integer) on (a; b) so that ( )f x has maximum n + 1 real roots on (a; b).

These consequence is proved by Rolle’s theorem and they are still true when some roots have the same value

(when ( )f x is a polynomial).

2. MEAN VALUE THEOREM

Formal Statement: If ( )f x is a continuous function on ( ; )a b so there is a

number ( ; )c a b so that ( ) ( )

'( )f b f a

f cb a

.

Proof:

Let’s consider a function:

( ) ( )( ) ( )

f b f aF x f x x

b a

.

We have F(x) is a continuous function on [ ; ]a b , is differentiable on ( ; )a b

and ( ) ( )F a F b .

According to Rolle’s theorem, we have there is a number ( ; )c a b so that

'( ) 0F c .

Joseph Louis Lagrange (1736 - 1813)

And ( ) ( )

'( ) '( )f b f a

F x f xb a

, so

( ) ( )'( )

f b f af c

b a

.

Geometric Interpretation:

Mean Value theorem allows us estimate ( ) ( )f b f a

b a

and then gives us an ideal about the theorems of how

a function changes and leads us to some application of derivatives.

Theorem: Let ( )f x is a differentiable function on ( ; )a b .

- If '( ) 0, ( ; ) f x x a b then ( )f x is increasing on ( ; )a b .

- If '( ) 0, ( ; ) f x x a b then ( )f x is decreasing on ( ; )a b .

- If '( ) 0, ( ; ) f x x a b then ( )f x is constant on ( ; )a b .

Proof:

Suppose '( ) 0, ( ; ) f x x a b and for every pair of numbers 1 2 1 2, ( ; ),x x a b x x , according to the Mean

Value theorem, there is a number 1 2c (x ; x ) so that 2 1

2 1

( ) ( )'( )

f x f xf c

x x

.

And 1 2'( ) 0 ( ) ( ) ( )f c f x f x f x is increasing on (a; b).

If in the statement of the Mean Value theorem, we have the condition '( )f x is increasing or decreasing on

[a; b] then we can compare ( ) ( )f b f a

b a

with '( ), '( )f a f b .

For specific: '( )f x is increasing on [a;b] ( ) ( )

'( ) '( )f b f a

f a f bb a

'( )f x is decreasing on [a;b] ( ) ( )

'( ) '( )f b f a

f a f bb a

This gives us an ideal about using the Mean Value theorem to prove inequality and evaluate finite sum.

Similarly, if in the statement of the Mean Value theorem, we have the condition '( )f x is increasing or

decreasing on [a; b] then we can compare ( ) ( )f c f a

c a

with

( ) ( )f b f c

b c

when [ ; ]c a b , this gives us an

ideal to prove a lot of inequalities, for example, the Jensen inequality…

In addition, the Mean Value theorem is also stated as a form of integral:

A C

B

b a c O

y

x

Let is a function as the Mean Value theorem

stated, has the graph (C), and two points A(a;f(a)),

B(b;f(b)).

On (C) there is a point C(c;f(c)), so that the

tangent line of (C) at C is parallel to AB.

Theorem: If ( )f x is a continuous function on [a; b] then there is a number ( ; )c a b so that:

( ) ( )( )

b

a

f x dx f c b a

The Mean Value theorem is used to solve many problems of integral and limit of a function.

APPLICATION OF THE MEAN VALUE THEOREM

1. ROOT OF AN EQUATION

Problem 1. Prove that the equation acosx + bcos2x + ccos3x has at least one real root for every real numbers

a, b, c.

Proof:

Consider bsin2x sin3x

( ) asinx+ '( ) osx+bcos2x+ccos3x, x .2 3

cf x f x ac R

We have 0 0(0) ( ) 0 (0; ), '( ) 0f f x f x , Q.E.D.

Note: This problem can be generalized as:

Give a continuous function f(x) on [a; b], prove that the equation f(x) = 0 has at least on real root on (a; b).

Proof:

Consider F(x) in such a way that F(x) is continuous on [a; b], F’(x) = f(x).g(x) for every x in (a; b), g(x) has

no real root on (a;b) and F(a) = F(b). According to the Rolle’s theorem, the proof is done.

Problem 2. Given a positive integer m and real numbers a, b, c so that:

02 1

a b c

m m m

.

Prove that the equation ax2 + bx + c = 0 has real root on (0; 1).

Hint: Consider 2 1. . .

( )2 1

m m ma x b x c xf x

m m m

.

Similarly, we have a generalized problem.

Problem 3. Given a positive integer m, positive integer n and real numbers 0 1, ,..., na a a in such a way that:

1 0... 01

n na a a

m n m n m

.

Prove that 1

1 1 0... 0n n

n na x a x a x a

has real root on (0; 1).

Hint: Consider 11 0( ) ...1

m n m n mn na a af x x x x

m n m n m

Problem 4.(Cauchy’s theorem)

If ( ), ( )f x g x are continuous functions on [ ; ]a b , differentiable on ( ; )a b and '( )g x is not zero on ( ; )a b ,

then there is a number ( ; )c a b so that ( ) ( )

'( )( ) ( )

f b f af c

g b g a

.

Proof: According to the Mean Value theorem, there is a number 0 ( ; )x a b so that 0

( ) ( )'( )

g b g ag x

b a

( ) ( )g a g b .

Consider ( ) ( )

( ) ( ) ( )( ) ( )

f b f aF x f x g x

g b g a

, we have: F(x) is continuous on [ ; ]a b , differentiable on ( ; )a b

and ( ) ( ) ( ) ( )

( ) ( )( ) ( )

f a g b f b g aF a F b

g b g a

.

According to the Rolle’s theorem, there is a number ( ; )c a b so that '( ) 0F c .

We have ( ) ( )

'( ) '( )( ) ( )

f b f aF x f x

g b g a

, so

( ) ( )'( )

( ) ( )

f b f af c

g b g a

.

Note: The Mean Value Theorem is a consequence of the Cauchy’s theorem (in the case of ( )g x x )

Problem 5: Given a + b – c = 0. Prove that: asinx+9bsin3x+25csin5x = 0 has at least 4 real roots on 0; .

Note: This problema is similar with other problems. In order to prove ( )f x has at least n real roots we got to

prove F(x) has at least n + 1 real roots F(x) is a primitive function of ( )f x on (a;b)

Proof: Consider ( ) 3 5f x asinx bsin x csin x , we have:

'( ) os 3 os3 5 os5f x ac x bc x cc x , ''( ) 9 3 25 5f x asinx bsin x csin x .

We have 1 2 3

3 3 3(0) ( ) ( ) ( ) 0 (0; ), ( ; ), ( ; )

4 4 4 4 4 4f f f f x x x

so that

1 2 3 4 1 2 5 2 3 4 5(0) '( ) ' ( ) '( ) 0 ( ; ), ( ; ) | ''( ) ''( ) 0f f x f x f x x x x x x x f x f x

but ''(0) ''( ) 0f f Q.E.D.

Problem 6. Given two polynomials P(x) and Q(x) = aP(x) + bP’(x) in which a, b are real numbers, a 0.

Prove that if Q(x) has no roots then P(x) also has no roots.

Proof: We have degP(x) = degQ(x)

Because Q(x) has no roots, so degQ(x) is even. Suppose P(x) has at least one real root, because degP(x) is

even so P(x) has at least two real roots.

- When P(x) has two roots which have the same value x = x0 , we have x0 is also a root of P’(x) then Q(x) has

a real root ( impossible ).

- When P(x) has two distinct real roots x1 < x2.

If b = 0 then Q(x) has a root ( impossible ).

If b 0 : Consider ( ) ( )a

xbf x e P x we have: ( )f x has at least two distinct real roots x1 < x2

a a a a

b b b b1 1

'( ) ( ) '( ) ( ( ) '( )) ( )x x x xa

f x e P x e P x e aP x bP x e Q xb b b

Because ( )f x has two real roots, we have '( )f x has at least one real root and then Q(x) has at least one real

root ( impossible ).

2. SOLVING AN EQUATION

Problem 7: Solve the equation: 3 5 2.4x x x (1)

Proof:

Note: 0; 1x x are roots of that equation (1).

Let x0 be a root of equation (1). We have:

0 0 0 0 0 0 03 5 2.4 5 4 4 3 (1a)x x x x x x x

Consider 0 0( ) ( 1)x x

f t t t , we have (1a) (4) (3)f f

Because f(t) is continuous on [3; 4] and differentiable on (3; 4), according to the Rolle’s theorem, there is a

number c (3; 4) so that: 0 001 1

0

0

0'( ) 0 [( 1) ]=0

1

x xx

f c x c cx

So the equation (1) has two roots x = 0 and x = 1.

Problem 8: Solve the equation: x5 3 2x (2)x

Proof:

Note: 0; 1x x are roots of equation (2).

Let x0 be a root of the equation (2), then we have: 0 0x

0 05 5 3 3x (2a)x

x

Consider 0

0( )x

f t t tx , then: (2a) (5) (3)f f

Because ( )f t is continuous on [3; 5] and differentiable on (3; 5), then according to the Mean Value theorem,

there is a number c (3; 5) so that: 001

0

0

0'( ) 0 ( 1)=0

1

xx

f c x cx

So the equation (2) has two roots x = 0 and x = 1.

Problem 9. Solve the equation: 3194.23 xxx (3).

Proof:

(5) 03194.23 xxx .

Consider 3194.23)( xxfy xx then: 194ln4.23ln3)(' xxxf

Rxxf xx ,0)4(ln4.2)3(ln3)('' 22 so ''( )f x has no roots, then '( )f x has maximum one root, so

( )f x has maximum 2 roots.

We have 0)2()0( ff so the equation (3) has exactly two roots 2,0 xx .

Problem 10. Solve the equation: cos(1 cos )(2 4 ) 3.4cos (4)xx x

Proof:

Let cos , ( [-1;1]) t x t

( ) ( )( ) ( )( )

Consider ( ) (1 )(2 4 ) 3.4 t tf t t

2'( ) 2 4 ( - 2)4 ln 4, ''( ) 2.4 ln 4 ( - 2)4 ln 4 t t t tf t t f t t

We have: 2

''( ) 0 2ln 4

f t t ''( )f t has only one root

'( )f t has maximum 2 roots ( )f t has maximum 3 roots

We can easily see that1

(0) ( ) (1) 02

f f f , so ( )f t has 3 roots 1

0, ,12

t .

And then, the roots of equation (4) are:

2 , 2 , 2 , 2 3

x k x k x k k Z

3. INEQUALITY

Problem 11. Given two positive real numbers a, b and a < b. Prove that:

lnb a b b a

b a a

Proof:

Consider 1

( ) ln '( ) , (0; ).f x x f x xx

According to the Mean Valua Theorem, there is a number c (a; b) so that ( ) ( )

'( )f b f a

f cb a

then

1 ln lnln

b a a b b

c b a c a

, we also have

1 1 10 a b c

b c a ln

b a b b a

b a a

.

Problem 12. Prove that: 11 1(1 ) (1 ) , (0; ).

1

x x xx x

Proof:

We have: 11 1(1 ) (1 ) [ln( 1) - ln ] ( 1)[ln( 2) - ln( 1)]

1

x x x x x x x xx x

Consider ( ) [ln( 1) - ln ]f x x x x

We have 1

'( ) ln( 1) ln 1 ln( 1) ln1 1

xf x x x x x

x x

Apply the Mean Value Theorem to y = lnt on [x; x+1], there is a number c (x; x+1) so that:

1'( ) ln( 1) ln ln( 1) ln .f c x x x x

c

1 1 1

0 11

x c xx c x

1 1 1ln( 1) ln ln( 1) ln 0

1 1x x x x

x x x

'( ) 0, (0;+ )f x x ( )f x is increasing on (0;+ ).

So ( 1) ( ), (0; )f x f x x Q.E.D.

Note:

We can prove the above problem by another way.

Consider ( ) ln(1 )F x x

For every real numbers x, y so that 0 < x < y, according to the Mean Value theorem, there is a number

0 0(0; ), ( ; )x x y x y so that:

0 0

( ) (0) ( ) ( )'( ) , '( )

0

f x f f y f xf x f y

x y x

then 0 0

1 ln(1 ) 1 ln(1 ) ln(1 );

1 1

x y x

x x y y x

.

We have 0 0

1 1

1 1x y

ln(1 ) ln(1 ) ln(1 )ln(1 ) ln(1 ).

x y xy x x y

x y x

So for every real numbers x, y in such a way that 0 < x < y, we have ln(1 ) ln(1 ),y x x y substitute x by

1

y and y by

1

x we have:

1 1 1 1 1 1ln(1 ) ln(1 ) (1 ) (1 )y x

x y y x y x

Problem 13. (Jensen inequality)

Given a function ( )f x is two-level differentiable on (a; b) and ''( ) 0, ( ; )f x x a b .

Prove that: 1 2 1 21 2

( ) ( )( ), , ( ; )

2 2

f x f x x xf x x a b

Proof:

An equality is equivalent to 1 2x x .

When 1 2x x , according to the Mean Value Theorem, there is a number 1 2 1 21 2( ; ), ( ; )

2 2

x x x xc x d x

so

that

1 2 1 21 2

2 1 2 1

( ) ( ) ( ) ( )2 2'( ) , '( )

2 2

x x x xf f x f x f

f c f dx x x x

.

''( ) 0, ( ; ) '( )f x x a b f x is increasing on (a; b)

1 2 1 2 1 2 1 21 2

( ) ( )'( ) ( ) ( ) ( ) ( ) ( ) ( ).

2 2 2 2

x x x x f x f x x xf c f d f f x f x f f

Problem 14. (Bernoulli inequality)

For every real number x so that x > -1, prove that (1 ) 1 .nx nx

Proof:

- When x > 0: consider ( ) (1 )nf t t , according to the Mean Value Theorem, there is a number (0; )a x so

that 1( ) (0) '( ) (1 ) 1 (1 ) (1 ) 1n n nf x f xf a x nx a nx x nx

- When -1< x < 0: consider ( ) (1 )nf t t , according to the Mean Value Theorem, there is a number

( ;0)a x so that 1( ) (0) '( ) (1 ) 1 (1 ) (1 ) 1n n nf x f xf a x nx a nx x nx

So (1 ) 1 , (-1; )nx nx x . The equality is equivalent to x = 0.

Problem 15. Given a function ( )f x has its second derivative on R, ''( ) 0,f x x R ( ''( ) 0f x has finite

real roots). Prove that:

*

1

( ) (0) '( ) ( 1) (1),n

i

f n f f i f n f n N

.

Proof:

Because ''( ) 0,f x x R ( ''( ) 0f x has finite roots) '( )f x is increasing on R.

According to the Mean Value theorem, there is a number ( ; 1)ix i i so that:

'( ) ( 1) ( ),if x f i f i i R .

Because '( )f x is increasing on R '( ) '( ) '( 1)if i f x f i

'( ) ( 1) ( ) '( 1),f i f i f i f i i R .

*

1 1

'( ) [ ( 1) ( )] ( 1) (1),n n

i i

f i f i f i f n f n N

and *

1 1

'( ) [ ( ) ( 1)] ( ) (0),n n

i i

f i f i f i f n f n N

Note: If ''( ) 0,f x x R so the inequality is reverse.

Problem 16. Prove that: *

1

11 ln ln( 1),

n

i

n n n Ni

.

Proof:

Consider1

( ) ln '( )f x x f xx

and '( )f x is decreasing on (0 : )

Similarly to the above problem, we have: *

1

( ) (1) '(1) '( ) ( 1) (1),n

i

f n f f f i f n f n N

*

1

11 ln ln( 1),

n

i

n n n Ni

Problem 17: Given a positive integer k, find

52

5 41

1 1

5

k

i i

( [x] is a biggest number and not over x).

Proof:

Consider 5( ) 5f x x , we have: 5 4

1'( ) '( )f x f x

x is decreasing on (0; ) .

So

525 5 55 5 5

5 41 1

1 1( ) (0) '( ) ( 1) (1) 2 2 1 1 2 1

5

knk k k

i i

f n f f i f n fi

510

5 41

1 12 .

5

k

k

i i

Note: From the above problems, in order to evaluate *

1

( ),n

i

f i n N

( ( )f x is increasing or decreasing on

(0 : ) ), we have to evaluate ( )F x is a primitive function of ( )f x on (0 : ) .

From evaluating *

1

( ),n

i

f i n N

we can think about finding 1

lim ( ) ( ),n

i

g n f i

let’s consider this problem.

Problem 18. Find 1

0

1 ilim os

2n

n

i

cn

.

Proof:

Consider 2

( ) sin '( ) os '( ) 0, [ ; ]2 2

n x xf x f x c f x x n n

n n

So f(x) is increasing on [ ; ]n n . So 1

( ) (0) '( ) ( 1) (1)n

i

f n f f i f n f

1 1

0 0

2 i 2 2 1 i 2 1os os sin os os sin

2n 2n 2n 2n 2n 2n

n n

i i

n nc c c c

n n

And 2 2 1 2

lim( os ) lim( sin )2n 2n

cn

1

0

1 i 2lim os

2n

n

i

cn

.

Problem 19. Given an equation: 1

1.

n

i

ni nx

Prove that: for every positive integer n, the equation has only one positive real root xn, find limxn.

Proof:

Consider1

1( )

n

i

f x ni nx

We have3

1

1'( ) 0, (0; ) ( )

2 ( )

n

i

f x x f xi nx

is continuous and decreasing on [0; ) .

1

1(0) 0, lim ( ) 0 ( ) 0

n

xi

nf n n f x n f x

i n

has exactly one positive real root.

Consider ( ) 2n nF x x nx , we have 1

'( ) '( )n n

n

F x F xx nx

is decreasing on (0; ) .

1

( ) (0) '( ) ( 1) (1)n

n n n n n

i

F n F F i F n F

1

12( ) 2( 1 1 )

n

n n n n

i n

n nx nx n nx nxi nx

2( ) 2( 1 1 )n n n nn nx nx n n nx nx

1 1 11 1

2n n n nx x x x

n n

1 1 21 2 1 1n n n n n nx x x x x x

n n n

22 1 2 lim( 1 ) 2n n n nx x x x

n

n

1 3 9lim( 1 ) lim lim x

2 4 16n n nx x x .

Problem 20: Given positive real numbers a, b, c, d so that

2 2 2 2 2

4 4 4 4 4

a b c d e

a b c d e

Prove that 3 3 3 3 3.a b c d e

Note: In this problem, from the statement

2 2 2 2 2

4 4 4 4 4

a b c d e

a b c d e

, we can see the Rolle’s theorem applied for

( ) ( (2) (4) 0)x x x x xf x a b c d e f f , then we need to prove (3) 0f . Because ( )f x is continuous

and (3) 0f , then there is an interval ( ; ) 3m n so that ( ) 0, ( ; )f x x m n , then problem is finding how

( )f x changes.

Proof:

Without generality, assume that 1,a b c d e

If 2 2 2 21 1 ( 0)d d x x b a e x

4 4 4 4 4 4 2 2 2 2 4( ) 1 (1 )a b c d e a e x a x e

2 2 2

2 2 2 2 2 2 2 2 2

2

0

( 1) ( ) 0 1 1

x e a b

e a x a a e e a e a

e a e a

( impossible ) 1d

Similarly, we have 1a b d e

Consider ( ) 1 (2) (4) 0x x x xf x a b d e f f

Suppose ( )f x has a root 0 2;4.x According to the Rolle’s theorem, there are 1 2x x so that:

1 2'( ) '( ) 0f x f x then 1 1 1 1ln ln ln ln ,x x x x

a a b b d d e e

2 2 2 2ln ln ln lnx x x x

a a b b d d e e

2 2 2 2

1 1 1 1

ln ln ln ln

ln ln ln ln

x x x x

x x x x

a a b b d d e e

a a b b d a e b

And 2 2 1 2 1 21 0 ln ln ln lnx x x x x x

a b d e a a b b a b a b b

2 2

2 1

1 1

ln ln

ln ln

x xx x

x x

a a b bb

a a b b

and 2 2

2 2 2 2 1 1 2 1

1 1

ln lnln ln ln ln 0

ln ln

x xx x x x x x x x

x x

d d e ed d e e d d d e b d

d d e e

2 2 2 2

1 1 1 1

ln ln ln ln

ln ln ln ln

x x x x

x x x x

a a b b d d e e

a a b b d a e b

(Impossible).

So ( )f x has exactly two roots x = 2, x = 4 and '( )f x has exactly one root, and it is in (2; 4). Because ( )f x is

continuous then ( )f x has the same sign on ( ;2),(2;4),(4; ). And we have

(0) 1 0 ( ) 0, ( ;0) ( ) 0, (2;4)f f x x f x x (if ( ) 0, (2;4)f x x then x = 2 is a root of

'( )f x ) (3) 0f (Q.E.D).

The Mean Value Theorem is also used to prove symmetric inequality, in order to decrease the number of

variables. If we need to prove a symmetric inequality with n variables 1 2, ,..., na a a then we consider the

polynomial 1 2( ) ( )( )...( )nf x x a x a x a , then ( )f x has n roots, then '( )f x has n – 1 roots 1 2 1, ,..., nb b b ,

and based on the Viet’s theorem, we prove an inequality with n – 1 variables 1 2 1, ,..., nb b b .

Problem 21. Given a < b < c, prove that:

2 2 2 2 2 23 3 3a a b c a b c ab bc ca b a b c a b c ab bc ca c

Proof:

Consider: ( ) ( )( )( ) ( ) ( ) ( ) 0f x x a x b x c f a f b f c

According to the Mean Value Theorem, there are two numbers and , 1 2a x b x c so that:

1( ) ( ) ( ) '( )f a f b a b f x , 1 1 2( ) ( ) ( ) '( ) '( ) '( ) 0f c f b c b f x f x f x

2'( ) 3 2( )f x x a b c x ab bc ca 2 2 2

13

a b c a b c ab bc cax

2 2 2

23

a b c a b c ab bc cax

Then, from 1 2a x b x c . We have:

2 2 23 3a a b c a b c ab bc ca b 2 2 2 3a b c a b c ab bc ca c

Problem 22. Given non negative real numbers a, b, c, d. Prove that:

3

4 6

abc bcd cda dab ab bc cd da ac db

Proof:

Consider ( ) ( )( )( )( )f x x a x b x c x d .

Let , , ,p a b c d q ab bc cd da ac bd r abc bcd cda dab s abcd

4 3 2 3 2( ) '( ) 4 3 2f x x px qx rx s f x x px qx r

We have ( ) ( ) ( ) ( ) 0f a f b f c f d , according to the Rolle’s theorem '( ) 0f x has three roots (if a =

b then a is a root of f’(x)).

So there are three numbers , ,w 0u v so that '( ) 4( )( )( w)f x x u x v x

3 24 4( ) 4( ) 4x u v w x uv vw wu x uvw

3

4

1

2

1

4

u v w p

uv vw wu q

uvw r

.Mà 23 3 3( )3 3 6 4

uv vw wu uv vw wu q ruvw uvw

3

4 6

abc bcd cda dab ab bc cd da ac db

An equality wu v a b c d .

4. LIMIT OF SEQUENCE

The Mean Value Theorem is used to find limit of a sequence, for sequences defined by ( )f x and sequences

defined by roots of an equation ( ) 0nf x , generally, ( ),f x ( )nf x

are differentiable and monotonic

functions, their derivatives can be estimated by some inequalities. Then if we can guess the limit is a, we can

compare ( ) ( ),nf x f a

( ) ( )n n nf x f a with nx a and then find the limit of xn.

Problem 23. Given a real sequence (xn) defined by:

1

*

12

2007

3 ,1

nn

n

x

xx n N

x

Find the limit of this sequence.

Proof: We have *3, .nx n N

Consider f(x) = 1

32

x

x, we have:

2 3

1'( )

( 1)f x

x

1'( ) , ( 3; )

2 2f x x .

If (xn) has limit then this limit is bigger than 3 and is a root of the equation ( )f x x . We have:

2( ) 3

1

xf x x x

x

22

2( 3)

1

xx

x

2 2 2( 3 ) 2( 3 ) 3 0x x x x

2

2

3 1

3 3

x x

x x

3 15

.2

x

Let 3 15

2a

, according to the Mean Value theorem, there is a number ( ; )n nc x a

or ( ; )na x so that:

( ) ( ) '( ) .n n nf x f a f c x a

1 1

1 1( ) ( ) '( ) ... ( )

2 2 2 2

n

n n n n nx a f x f a f c x a x a x a

And 1

1lim( ) 0

2 2

n x a , then limxn = a = 2

153 .

Note:

In the problem above, we don’t need to solve the equation ( )f x x , we just need to guess or find one root.

There is a generalized version of the above problem:

Given a real sequence (xn) defined by:1

*

1 ( ),n n

x a

x f x n N

. Prove that:

a) If ( )f x is differentiable on D containing a and '( ) 1,f x b x D then (xn) has limit.

b) If ( )f x is differentiable on D containing a, ( ) 0f a and '( ) 1,f x b x D then |xn| goes to positive

infinity.

Problem 24. (Vietnam Mathematical Olympiad 2008)

Given a real number a and a real sequence (xn) defined by:

x1 = a and xn+1 = ln(3+cosxn + sinxn) – 2008, *.n N

Prove that (xn) has limit.

Proof:

Let f(x) = ln(3+sinx+cosx) – 2008, then:cos sin

'( ) , R3 sin cos

x xf x x

x x

.

And 2|cossin|,2|sincos| xxxx , so: .123

2|)('|

qxf

According to the Mean Value theorem : for every numbers x, y (x < y), there is a number ( ; )z x y so that:

f(x) – f(y) = f’(z)(x-y).

then |f(x) – f(y)| q|x – y| for every x, y in R.

Apply the above property to m > n N, we have :

|xm – xn| = |f(xm-1) – f(xn-1)| q|xm-1- xn-1| … qn-1

|xm-n+1 – x1| qN-1

|xm-n+1 – x1|.

In addition, (xn) is restricted and q < 1 so for every > 0 there is a number N big enough so that:

qN-1

|xm-n+1 – x1| < .

So the sequence (xn) has a limit according to the Cauchy’s theorem.

Problem 25. (Vietnam Mathematical Olympiad 2007)

Given a real number a > 2 and 10 10 1( ) ... 1n n n

nf x a x x x x .

a) Prove that for every positive integer n, the equation ( )nf x a has exactly one positive root xn.

b) Prove that the sequence (xn) has limit 1a

a

.

Proof:

Let ( ) ( )n nF x f x a , then ( )nF x is continuous and increasing on [0; ) and

10(0) 1 0, (1) 1 0.n nF a F a n a So the equation ( )nf x a always has exactly one positive real

root xn

Let 9 *1( ) ( 1)[( 1) 1] ( ) ,n

n n n

ab f b b a a a f b a b x n N

a

According to the Mean Value theorem, there is a number ( ; )n nc x b so that:

( ) ( ) '( )( )n n n n nf b f x f c b x .

And '( ) 1nf c then 9

n( ) ( ) ( 1)[( 1) 1] limxn

n n n nb x f b f x b a a b

9

n( 1)[( 1) 1] limxn

nb b a a x b b (vì (0;1)b ).

Note:

How can we know what is the limit of this sequence if the problem doesn’t give us first.

We can solve this problem by this way:

First, the limit of the sequence has to be in (0; 1), suppose the limit is b, then we have:

10 10 1 1 1( ) ( 1) lim ( )

1 1 1

n

n nf b b a b f bb b b

( (0;1)b ).

And 1 1

( )1

n n

af x a a b

b a

.

Problem 26: (Vietnam Mathematical Olympiad 2002)

Given an equation 2

1

1 1

1 2

n

i i x

, with n is a positive integer.

a) Prove that for every number n, the above equation has exactly one root bigger than 1; let it be xn.

b) Prove that lim 4nx .

Proof:

a) Consider 2

1

1 1( )

1 2

n

n

i

f xi x

, we have: ( )nf x is continuous and decreasing on (1; ).

And 1

1lim ( ) , lim ( )

2n n

xxf x f x

( ) 0nf x has exactly one root bigger than 1.

b) For every number n we have:1

(4) 0 (4) ( ) 42(2 1)

n n n n nf f f x xn

According to the Mean Value theorem, there is a number ( ;4)n nc x so that:

(4) ( ) '( )(4 )n n n n nf f x f c x .

And 1 9

'( ) 4 9( (4) ( )) 49 2(2 1)

n n n n n nf c x f f x xn

94 4 lim 4.

2(2 1)n nx x

n

3. OTHER PROBLEMS

1. Solve these equations.

a) xxx 3)12(log)13(log 32

b) 2008 2010 2.2009x x x

c) (4 2)(2 ) 6x x

2. Prove that if function ( )f x has second derivative on [a; b] and '( ) '( )f a f b then the inequation

4

4''( ) ( ) '( )

( )f x f a f b

a b

has at least one root.

3. Find 1

1 1

21 sin

2

n

i

Limin

n

4. Given a real sequence (xn) defined by: 1

2

1 ln 1 2010, 1n n

x a

x x n

.

Prove that xn has limit.

5. Given an equation: 1

11.

n

i i nx

Prove that for every positive integer n , the above equation has exactly one positive real root xn, find limxn.

6. Prove that 1b aa b for every , 0a b .

7. Give two polynomials P(x) and Q(x) = aP(x) + bP’(x) + cP”(x) in which a, b, c are real numbers so that a

0 and b2 – 4ac > 0. Prove that if Q(x) has no real roots then also does P(x).

8. Given a real number a not zero, a polynomial ( ), deg ( ) 1P x P x n and a polynomial

2 ( )( ) ( ) '( ) "( ) ... ( )n nQ x P x aP x a P x a P x . Prove that if ( )P x has no real roots then also does ( )Q x .