4. 4
Statement: Let a function f be defined on a closed interval [a,b]. Suppose, further
that
1. f is continuous on the closed interval [a,b]
2. f is derivable in the open interval (a,b); and
3. f(a) = f(b)
Then, Rolle’s theorem states that at least one value , where a b , such
that
1
0f
5. 5
Note: The statement “f is derivable in (a,b)” means that at any point ,x a b ,
right-hand and left-hand derivatives exist and that they are equal. It may be
that at some ,x a b
1 1
0 0f x f x
Or it may be 1 1
0 0f x f x , still the theorem holds.
6. 6
Corollary: If a,b are two roots of the equation f(x) = 0, then the equation.
f1
(x) = 0 will have at least one root between a and b, provided
1. f(x) is continuous in ;a x b and
2. f1
(x) exists in a<x<b
If f(x) be a polynomial, the conditions (1) and (2) are evidently satisfied.
Hence: Between any two zeros of a polynomial f(x) lies at least one zero of
the polynomial f1
(x). (Rolle’s Theorem for Polynomials).
7. 7
Geometric Interpretation:
If the graph of y =f(x) has the ordinates at two points A,B equal, and if the
graph be continuous throughout the interval from A to B and if the curve has a
tangent at every point on it from A to B except possibly at the two extreme
points A and B, then there must exist at least one point on the curve
intervening between A and B where the tangent is parallel to the x-axis.
In this geometrical language, the theorem is almost self-evident. Compare the
Following figures.
9. 9
Observations:
1. The three conditions of Rolle’s theorem are a set of sufficient conditions. They
are by no means necessary. The following illustrations will make the point
clear.
Example-1: 2 2
0,f x x a x in a
Here (a) f(x) is continuous in 0 x a ;
(b)
2 2
1
2 2
2a x
f x
a x
exists in 0<x<a;
(c) 0 0f f a .
All the conditions of Rolle’s theorem are satisfied and as such there exists c
where
f1
(c) = 0 namely, / 2c a where 0<c<a.
10. 10
Example-2: 1,1f x x in
Here (a) f(x) is continuous in 1 1;x
(b) 1 1
1 0 1 1 1 0; 0f x in x in x f does not exist.
f1
(x) does not exist at x = 0
(c) f(1) = f(-1) =1.
Note that f1
(x) nowhere vanishes in [-1, 1] and as such Rolle’s theorem fails.
The failure is explained by the fact that |x| is not derivable in -1<x<1, all other
conditions being satisfied.
11. 11
Example-3: f(x) = sin (1/x) is discontinuous at x=0 in the interval [-1,2] but still
f1
(x) = 0 for an infinite number of points in [-1,2], namely, 2 / 2 1x n .
Where cos(1/x)=0.
Example-4:
1 1
0,1
1
f x in
x x
Here (a) f (x) is continuous in 0<x<1 (not in 0 1x );
(b)
1
2 2
1 1
0 1
1
f x exists in x
xx
(c) 0 1f f , both being undefined.
Thus, all the conditions of Rolle’s theorem do not hold. But yet there exists a,c
where f1
(c) = 0 (namely
1
2
c ) where 0<c<1.
12. 12
The above illustrations lead to the following conclusion:
If f(x) satisfies all the conditions of Rolle’s theorem in [a,b] then the
conclusion f1
(c)=0 where a<c<b is assured, but if any of the conditions are
violated then Rolle’s theorem will not be necessarily true; it may still be true
but the truth is not ensured.
13. 13
2. The theorem asserts the existence of at least one value c, where f1
(c) =0 and
a<c<b. There may be several such values where the derivatives will vanish
e.g.,
(a)
1 1
sin ; , 100 ;
2 2
f x x a b
(b) 0; .f x a x b
In each case f1
(x) =0 for several values in (a,b)
Again according to the theorem, c is strictly within a and b. In special cases
(e.g.,(b) above) a or b themselves may be points where f1
(x) = 0 but then also
there will be other points between a and b where f1
(x) should vanish, if Rolle’s
theorem is to remain valid.
16. 16
Statement: If a function f is
a) continuous in the closed interval [a,b];
b) derivable in the open interval (a,b);
then there exists at least one value of x, say c, such that
1
,
f b f a
f c for a c b
b a
17. 17
Observations:
1. The whole point of the proof is to construct a function x which will satisfy
the hypothesis of Rolle’s Theorem. One could proceed by constructing
x f x f a A x a
Where A is a constant so chosen that 0b . See that a is clearly =0. The
proof of the Mean-Value Theorem (MVT) can then be carried out as above.
18. 18
2. h from of MVT: When b=a+h, a number lying between a and b may be
written as 0 1a h . Then the theorem takes the following form:
Let f be continuous in the closed interval [a,a+h] and have a derivative f1
(x)
whenever a<x<a+h. Then a number , between 0 and 1 for which
1
0 1f a h f a h f a h
The students are advised to construct a function
x f a x f a A x
Where A is a constant so chosen that 0h (then, of course 0 0 )h
and the proof can be carried out independently.
Remember: h form of MVT is usually referred to as Lagrange’s MVT
19. 19
3. Putting h=x, a=0, we have the MVT in the form
1
0 0 1f x f xf x
This form is due to Maclaurin (Maclaurin’s formula as it is often called).
4. The two hypothesis of MVT are only a set of sufficient conditions and by no
means necessary for the conclusion of the theorem.
Take, for instance, 1/ | |, 1, 1f x x a b .
The conditions of the theorem are not satisfied in [-1,1] (Why?). But see that
the conclusion is true iff 1 2b .
20. 20
Geometric Interpretation of MVT:
Let y=f(x) be represented by the curve AB, in Fig. then
tan
f b f a BL
BAL
b a AL
.
it is clear that 1
f is the trigonometrically tangent of the angle, which the
tangent at ,P f makes with OX. The theorem
1f b f a
f
b a
therefore states that at some point P x of the curve between A(x = a) and
B(x = b), the tangent to the curve is parallel to the chord AB.
22. 22
Remember: The fraction
f b f a
b a
measures the mean (or average) rate of
increase of the function in the interval of length b-a. Hence the theorem
expresses the fact that, under the conditions stated, the mean rate of increases
in any interval is equal to the actual rate of increase at some point within the
interval. For instance, the mean velocity of a moving point in any interval of
time is equal to the actual velocity at some instant within the interval. This
justifies the name Mean-Value Theorem.
23. 23
Important Deductions from MVT:
Deduction-I: Suppose that f is continuous on the closed interval [a,b], that f is
derivable in the open interval (a,b) and that f1
(x) = 0 for every ,x a b . Then f
is constant on [a,b]. In fact, we shall prove that , ,f x f a x a b
24. 24
Deduction-II: If F and G are both continuous on [a,b], that they are both derivable
on (a,b) and that 1 1
,F x G x x a b then F(x) and G(x) differ by a constant in
this interval
i.e., F(x) = G(x) + C on [a,b]
Deduction-III: MVT is very useful in estimating the size of function and in making
numerical approximations:
25. 25
Deduction-IV: We recall the following definitions:
Let a functions :f I R be defined over on interval I.
(i) f is said to be monotone increasing or simply, increasing on the interval I if
whenever 1 2,x x I satisfies 1 2x x , then 1 2 ;f x f x f is strictly increasing
on I when 1 2 1 2x x f x f x .
(ii) f is said to be increasing at a given point c I if a neighbourhood (nbd)
of c where f is increasing on this interval
(iii) f is said to be decreasing, strictly decreasing on an interval or f is
decreasing at a point ,c I if f is increasing, strictly increasing on the
interval I or – f is increasing at c I
26. 26
Theorem-I: Let :f I R be a derivable function on the interval I. Then (i) f is
increasing on I if and only if 1
0,f x x I
(ii) f is decreasing on I if and only if 1
0,f x x I
Theorem-II : If f is continuous on [a,b] and f1
(x)>0 in (a,b) then f is strictly
increasing in [a,b].
Theorem-III: If f is continuous in 1
0a x b and f x in a<b then f is
strictly decreasing in a x b (proof is similar and is left out)
27. 27
What about the converses of Theorems I and III ?
The converses may not be true. A strictly increasing (or decreasing) derivable
function may have a derivative that vanishes at certain points [i.e., derivatives
may not be strictly positive (or negative)]
Let f(x) = x3
. This function in strictly increasing on any interval of R but
0x , where f1
(x) = 0
28. 28
Observations:
1. The function f is such that the derivative is strictly positive at a point c of an
interval I where f is defined. The function may not be increasing at the point c.
2 1
2 sin ,
. ., 0
0, 0
x x
e g f x x x
x
See that f1
(0) =1 strictly positive but it can be shown that f is not increasing in
any nbd of x=0.
(In every nbd of 0 f1
(x) takes on both positive and negative values and f is not
monotonic in ay nbd of 0)
2. If 1
0f x in (a,b) and 0f a then f(x) is positive throughout the open
interval a<x<b and if f1
(x)<0 in (a,b) and f(a) <0 then f(x) is negative
throughout the open interval (a,b).
29. 29
Deduction-V: If f1
(x) exists for 0
a x b and f x as x a
, then
1 0
f x as x a
Deduction-VI: If f is continuous at c, and if 1
lim
x c
f x
then f is derivable at c and l
is the derivative there, i.e., f1
(c)=l
Note: This theorem shows that the derived function f1
cannot have a
discontinuity of the first kind.
32. 32
If we apply Lagrange’s mean value theorem to two functions f(x) and g(x),
each of which satisfies the conditions of the theorem in the interval [a,b], we
would obtain
And
1
1 1
1
2 2
, .
,
f b f a b a f c a c b
g b g a b a g c a c b
Dividing one by the other
1
1
1 21
2
, ,
f b f a f c
c c
g b g a g c
being, in general, different
Cauchy takes a step further to make c1 = c2 and establishes a theorem, which
we are going to study now:
33. 33
STATEMENT: It two functions f and g
(a) be both continuous in a x b
(b) are both derivable in a<x<b
(c) g1
(x) does not vanish at any value of x in a <x<b,
Then there exists at least one value of x, say c, such that
1
1
f b f a f c
for a c b
g b g a g c
34. 34
Note : Condition (c) given in the statement of the theorem may be replaced by
(a) f1
(x) and g1
(x) do not both vanish at any value of x in the open interval
(a,b); and
(b) g a g b .
The students should verify that these two conditions and 1
0g x for any x in
(a,b) are equivalent.
h Form: If b = a + h, then at least one such that
1
1
f a h f a f a h
g a h g a g a h
, where 0 1
35. 35
Geometric interpretation of Cauchy’s MVT:
The functions f and g can be considered as determining a curve in the plane
by means of parametric equations x=f(t), y=g(t), where a t b . Cauchy’s
MVT concludes that a point f(c), g(c)) of the curve for some t=c in (a,b)
such that the slope of the line segment joining the end points of the curve
1
1
slope
g c
f c
of the tangent to the curve at t=c.
36. 36
Corollary : If 0f a g a , Cauchy’s result would reduce to
1
1
,
f b f c
a c b
g b g c
Which may be put in this form
If f(x) = 0, g(x) =0 have a common root x=a, then
1
1
,
f x f c
for some c a x
g x g c
37. 37
Note : When g(x) = x, Lagrange’s Mean-Value Theorem becomes a particular
case of Cauchy’s Mean-Value Theorem.
It is to be observed that Rolle’s Theorem can be obtained from Cauchy’s MVT
by letting g(x) = x and f(a) = f(b). However,we could not easily prove this
theorem without using Rolle’s Theorem. Here is a case where we have to use a
special case to prove a general result.
We remind our readers that the conditions of the present theorem are also
sufficient but not necessary.
38. 38
Rolle’s Theorem for polynomials: If f(x) be any polynomial then between
any pair of roots of f(x) = 0 lies a root of f’(x) = 0.
Solution: We have proved Rolle’s theorem for general type of functions. We
give here an algebraical proof of this important theorem valid for polynomials
only. Let and be two consecutive roots, repeated respectively m and n
times, so that . .
m n
f x x x x
Where x is a polynomial which has the same sign, say the positive sign, for
x (since and are two consecutive roots of f(x) = 0, there cannot be
any root of 0x between them)
39. 39
Now
1 1m n
f x x x x
1 1m n n m
m x x n x x x
1 1m n
x x x
,
Where 1
x m x x n x x x x x
Observe that m and n have opposite signs.
Hence x must vanish for some value of x between and
1
f x must vanish for some value of x between and .
Hence the theorem.
40. 40
Example . If a = -1, 1b and 1/ | |f x x , prove that Lagrange’s MVT is not
applicable for f in [a,b]. But check that the conclusion of the theorem is TRUE
if 1 2b .
Solution: The function is not defined at x=0. Even if we define f(0) = A,
where A is some finite quantity, then also
1
0 0
1/ 1 1
0 0 lim lim
h h
h A
f A
h h h
And 1
0 0
1/ 1 1
0 0 lim lim
h h
h A
f A
h h h
i.e., f is not derivable at x = 0
41. 41
The conditions of MVT are not satisfied in [a,b] which always includes the
origin.
Now, the conclusion of the MVT is
1
,
f b f a
f c a c b
b a
This will be true iff
1 1 1
/
d
b a at x c
b a dx x
2
1
c
42. 42
Or,
2 2
11 1
1 1 1 1
b
b b and a
b cc
Or,
2 2
2 2 2 2
2
11
. ., ; sin ,
1 1
bb b b b b
i e c ce c b therefore b
b c b b
Or, 2 2
1 . ., 2 1 0b b b i e b b
i.e.,
2
1 2, . ., 1 2b i e b
Under this condition 1 2b the conclusion of MVT is true but the
conditions for the validity of the theorem are not satisfied.
43. 43
An Important Illustration: If f(x) and x be continuous in a x b and
derivable in a<x<b, then
1
1
,
f a f b f a f c
b a a c b
a b a c
Solution: Let us construct the function
1
1
1
;
f a f x f a f x
F x then F x
a x a x
We observe that F(x) is continuous in a x b and F1
(x) exists in a<x<b.
Hence using MVT, we get
44. 44
1
F b F a b a F c where a c b
1
1
0
f a f b f a f c
b a F a
a b a c
Hence the result
Note . If 1x , we get 1
f b f a b a f c : Lagrange’s MVT. In a
more general way, we have the following result.
45. 45
Example: If ,f x x and x satisfy the first two conditions of Rolle’s
theorem, viz., continuity in a x b and derivability in a<x<b, then there is a
value c between a and b, such that
1 1
0
f a a a
f b b b
f c c c
Solution: Construct the function
f a a a
F x f b b b
f x x x
and proceed as
In the above Ex. Here notice F(a) = 0 = F(b) and so we can at once apply
Rolle’s theorem.
46. 46
Note:
1. Put x a constant k; we have the cauchy’s MVT.
2. Put x x , we get Lagrange’s MVT
3. Put f a f b , the conclusion of Rolle’s theorem is obtained.
47. 47
Example: A twice differentiable function f(x) on a closed interval [a,b] is such
that f(a) = f(b) = 0 and f(x0)>0 where a<x0<b. Prove that at least one value
of x=c (say) between a and b for which f11
(c)<0
Solution: Since f11
(x) exists in [a,b], we conclude that f(x) and f1
(x) both exist
and are continuous on [a,b]
Since a < x0 < b we can apply Lagrange’s MVT to f(x) on two intervals [a, x0]
and [x0, b] and obtain
1
0 0 1 1 0f x f a x a f c when a c x
And 1
0 0 2 0 2f b f x b x f c when x c b
But it is given that 0f a f b .
0 01 1
1 2
0 0
f x f x
f c and f c
x a b x
(1)
48. 48
Where a < c1 < x0 < c2 < b.
Now since f1
(x) is continuous and derivable on [c1, c2] we can again apply
Lagrange’s MVT and obtain
1 1
2 1 11
1 2
2 1
f c f c
f c where c c c
c c
Using (1) we see that
011
2 1 0 0
0
b a f x
f c
c c b x x a
.
49. 49
Example: Show that
tan
when 0
sin 2
x x
x
x x
Solution: To prove
tan
, 0
sin 2
x x
if x
x x
i.e., To prove:
2
sin tan
0 0
sin 2
x x x
for x
x x
Since the Denominator sin 0 0
2
x x for x
it will suffice to prove that
2
sin tan 0 0
2
f x x x x if x
Now
1 2 2
sin sec cos tan 2 sin sin sec 2 .f x x x x x x x x x x
50. 50
The function f1
(x) is continuous and derivable in 0
2
x
11 2 2
cos cos sec 2sin sec tan 2f x x x x x x x
2
cos sec 2 2sec tanx x x x
2
2
cos sec 2 tan secx x x x .
Clearly 11
0 0 / 2f x for x
51. 51
1
f x is an increasing function in that interval. More over 1
0 0f .
Therefore, 1
0 0
2
f x for x
Consequently, f(x) is an increasing function and f(0) = 0.
0 0
2
f x in x
it follows from the previous arguments
tan
0
sin 2
x x
for x
x x
52. 52
Example : Assuming f11
(x) to be continuous on [a,b], prove that
111
2
b c c a
f c f a f b c a c b f
b a b a
Where c and both lies in [a,b].
Solution: To prove b a f c b c f a c a f b
111
2
b a c a c b f
Choose a function
x b a f x b x f a x a f b b a x a x b A
53. 53
Where A is a constant so determined that 0c
i.e,. 0b a f c b c f a c a f b b a c a c b A
or,
b a f c b c f a c a f b
A
b a c a c b
(2)
Clearly, 0a b and satisfies all the conditions of Rolle’s Theorem in
each of the two intervals [a,c] and [c,b] and hence 1 2and in [a,c] and [c,b]
respectively so that 1 1
1 20 0and
54. 54
Again, 1 1
2x b a f x f a f b b a x a b A which is
continuous on
[a, b] and derivable on [a,b] and in particular on 1 2, . Also 1 1
1 2 0 .
Therefore by Rolle’s Theorem 1 2[ , ] such that 11
0 .
But 11 11
x b a f x b a 2A so that
11
2f A b a
Or, 111
2
A f when 1 2a b (3)
From (2) and (3) the result follows.
