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 () fx is a continuous function on [ ; ] ab , is differentiable on ( ; ) ab and () () fa fb so there is a number ( ; ) c ab so that '( ) 0 f c . Proof: Because () fx is continuous on [a; b], according to Weierstrass’s theorem, we have () fx has its maximum M and its minimum m on [a; b]. - If M = m we have () fx is a constant on [a; b], so for every ( ; ) c ab we have '( ) 0 f c . - If M > m, because () () fa fb so there is a number c (a; b) so that () fc m or () fc M , according to Fermat’s theorem we have '( ) 0 f c . Consequence 1: If () fx is differentiable on (a; b) and () fx 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 () fx is differentiable on (a; b) and '( ) f x has no real root (a; b) so that () fx has maximum one real root on (a; b). Consequence 3: If () fx is differentiable on (a; b) and '( ) f x has maximum n real roots (n is a positive integer) on (a; b) so that () fx 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 () fx is a polynomial). 2. MEAN VALUE THEOREM Formal Statement: If () fx is a continuous function on ( ; ) ab so there is a number ( ; ) c ab so that () () '( ) fb fa f c b a . Proof: Let’s consider a function: () () () () fb fa Fx fx x b a . We have F(x) is a continuous function on [ ; ] ab , is differentiable on ( ; ) ab and () () Fa Fb . According to Rolle’s theorem, we have there is a number ( ; ) c ab so that '( ) 0 F c . Joseph Louis Lagrange (1736 - 1813)
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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 ( ) ( )