Rolle'S Theorem :
According to Rolle’s theorem if a function is continuous in the interval [
a
,
b
] and differentiable in the interval (
a
,
b
) and
f
(
a
) =
f
(
b
), then there exists at least one point c for which
f
' (
c
) = 0 where
c
ε (
a
,
b
)
Or
There exists at least one point c at which slope of tangent is equal to zero.
Here as we can see function is continuous at [
a
,
b
] and differentiable at (
a
,
b
) and
f
(
a
) =
f
(
b
) = 0, so conditions of Rolle’s theorem are satisfied hence there must be a point for which
f’
(
c
) = 0 as below.
In above graph there is a point c for which
f’
(
c
) = 0 or slope of tangent is equal to zero.
Rolle'S Theorem Example 1
Rolle'S Theorem Example 1 :
Verify Rolle’s theorem for the function
f
(
x
) =
x
2
– 2
x
+s 1 on [1, 3]
Sol:
We have
f
(
x
) =
x
2
– 2
x
+ 1 which is a polynomial function and polynomial functions are continuous and differentiable for
x
ε
R
, now to hold Rolle’s there must be c ε (1, 3) for which
f
' (
c
) = 0 holds Let’s check.
f
' (
x
) = 2
x
– 2,
f
' (
x
) = 0
2
x
– 2 = 0
⇒
x
= 1 ε (1, 3)
Hence Rolle’s Theorem verified.
Rolle'S Theorem Example 2
Rolle'S Theorem Example 2 :
Verify Rolle’s theorem for the function
f
(
x
) = |
x
– 6| on [–7, 7]
Sol:
As we can observe |
x
– 6| function is not differentiable at 6 hence Rolle’s theorem is not applicable for the given function.
Rolle'S Theorem Rapid Questions
-
Verify Rolle’s theorem for the function
x
2
-4
on [– 1, 1]?
-
Verify Rolle’s theorem for the function
x-5
+6
on [–3, 6]?
Explanation of Conditions in Rolle’s theorem
Why function must be continuous at [a, b] and differentiable at (a, b)?
If a function is not continuous and differentiable then we can’t generalize the condition
f
’(
c
) = 0 as c is a general point between [a, b]
In above we can see there is no point c for which
f
’(
c
) = 0 exist.
Here
f
’(
c
) = 0 exist,
Above two examples proof’s that Rolle’s theorem conditions are sufficient but not necessary.
Rolle'S Theorem Rapid Questions Example 1
Example 1
: Verify Rolle's theorem for the function
in the interval [0, 5]
Sol:
As we can see in above graph function is not continuous at 1/3 hence Rolle’s theorem not applicable.
Rolle'S Theorem Rapid Questions Example 2
Example 2 :
Verify Rolle’s Theorem for the function |
x
2
– 3
x
+ 2 | in the interval [0, 6].
Sol:
As we can see in above graph function is not differentiable {1, 2} hence Rolle’s theorem not applicable.
Rapid Questions
-
1.
Verify Rolle’s Theorem for the function |
x
2
– 5
x
+ 6 | in the interval [0, 7].
2.
Verify Rolle’s Theorem for the function 15
x
– 3in the interval [0, 3].
Why interval must be [
a
,
b
] for continuity?
We may have three possibilities [
a
,
b
), (
a
,
b
], (
a
,
b
)
Rolle'S Theorem Case-1 (a , b)
Here function is considered as continuous at [
a
,
b
) Let’s make a graph.
In above as we can see function is continuous at [
a
,
b
] and
f
(
a
) =
f
(
b
) also
f
’(
c
) = 0
Exist, Now let’s see second situation.
In above we can see
f
(
a
) =
f
(
b
) also function is continuous and differentiable at [
a
,
b
) and (
a
,
b
) respectively. Still there is no point available [
a
,
b
) such that
f
’(
c
) = 0.
These two situations are possible, because Rolle’s theorem conditions are sufficient but not necessary.
Rolle'S Theorem Case-2 (a , b)
Case-2 [
a
,
b
) :
In similar way [
a
,
b
) can also be explained.
Rolle'S Theorem Case-3 (a , b)
Case-3 (
a
,
b
) :
In above as we can see function is continuous and differentiable at (
a
,
b
),
f
(
a
) =
f
(
b
) and also
f
’(
c
) = 0 exist Now let’s see next situation
In above as we can see function is continuous and differentiable at (
a
,
b
) and also
f
(
a
) =
f
(
b
) but there is no point c exist such that
f
’(
c
) = 0
These two situations are possible, because Rolle’s theorem conditions are sufficient but not necessary.
Why
f
(
a
) =
f
(
b
) ?
As we know f’(c) represents slope of tangent and if starting and ending point has same values as well continuous and differentiable at [a, b] and (
a
,
b
) respectively,
Then function must have to take at least one U turn except the case when line is parallel to x axis (then infinite points would exist).
In above we can see at [
a
,
b
] function is continuous and differentiable at (
a
,
b
) as well as
f
(
a
) =
f
(
b
) hence there are 3 points for which slope of tangent is zero.
In above function is continuous and differentiable at [a, b] and (
a
,
b
) respectively but
f
(
a
) ≠
f
(
b
), still slope of tangent is zero, this is because Rolle’s theorem gives sufficient conditions not necessary.
When function would not take U turn?
If graph is a line parallel to x axis and continuous, differentiable at [a, b] and (a, b) respectively.
Rolle'S Theorem Illustrations
Q. 1.
Verify Rolle’s Theorem for
f
(
x
) = sin2
x
on [0, 2]
Sol :
Since Sin2x is continuous and differentiable function in
and
f
(0)
.
So
f
’(
x
) = 0
⇒
2cos 2
x
= 0, cos2
x
= 0
⇒
x
= 4
Hence verified.
Q. 2.
Verify Rolle’s Theorem for f(x)= 6
x
– 4
x
on [0, 6]
Sol:
Since 6
x
– 4
x
is continuous and differentiable function in [0, 6] and
f
(0) =
f
(6) = 0.
So, f’(
x
) = 0
⇒
6 – 8sin
x
= 0, sin
x
= 34
⇒
x
=.24
Hence verified.
Rapid Questions
-
1.
Verify Rolle’s Theorem for
f
(
x
) =
– sin
on [–1, 0]
2.
Verify Rolle’s Theorem for
f
(
x
) =
x
2
– 5
x
+ 4 on [1, 4]
