3. This property was known in the 12th century in ancient
India. The outstanding Indian astronomer and
mathematician Bhaskara II (1114β1185) mentioned it in
his writings.
In a strict form this theorem was proved in 1691 by the
French mathematician Michel Rolle (1652β1719)
4. Rolleβs theorem
Statement:
Let a function π(π₯) defined on [π, π] be such that
β It is continuous in the interval [π, π]
β It is differentiable in the interval (π, π)
β π π = π(π)
then there exist at least one c β (a, b) such that πβ² π = 0
5. Since π(π₯) is continuous in closed interval π, π
βΉ π(π₯) is bounded I.e. ,π β€ π π₯ β€ π β π₯ β [π, π]
β΄ π π πππ π(π) are either both maxima (minima) or both
are neither maxima nor minima
So there exist at least one point π β π, π , where π π =
π ππ π.
12. Case 1 Case 2 Case 3
In all the above three cases , we observed that all the conditions of Rolleβs theorem are not
satisfied because at least one one of the three conditions is being violated. But still , in each of
the three cases, There exist a point βπ β such that π β (π, π) and πβ² π = 0
13. Verify whether the function π π₯ = sin(π₯)
ππ 0, π satisfies the conditions of Rolleβs theorem and hence
find c as prescribed by the theorem.
Given π π₯ = sin π₯ ππ 0, π .
Here π 0 = 0 = π(π)
Also we know that sin(π₯) is continuous in [0, π] and differentiable in ]0, π[ .
Hence π satisfies all the conditions of the Rolleβs theorem. So there must exist at
least one value of π₯ β ]0, π[ such that πβ² π₯ = 0
Now πβ² π₯ = 0 βΉ cos π₯ = 0 βΉ π₯ =
π
2
β 0, π .
Hence , in Rolleβs theorem , π =
π
2
14. Discuss the applicability of
Rolleβs theorem to π π₯ = 2 + (π₯ β 1)
2
3 ππ 0,2 .
Solution :
Here πβ² π₯ =
2
3
Γ π₯ β 1 β
1
3 = 2/{3 π₯ β 1
1
3}
Which does not exist (i.e is not finite) at π₯ = 1 β]0,2[ , Hence the condition β π(π₯)
is deriveable in]0, 2[ β is not satisfied . Therefore , Rolleβs theorem is not applicable
to π π₯ ππ 0,2 .
15. Example : Discuss the applicability of
Rolleβs theorem to π π₯ = π₯ in [1,-1].
Consider f(x)=|x|
(where |x| is the absolute value of x) on the
closed interval [β1,1].
This function does not have derivative at x=0.
Though f(x) is continuous on the closed
interval [β1,1], there is no point inside the
interval (β1,1) at which the derivative is equal
to zero. The Rolleβs theorem fails here
because f(x) is not differentiable over the
whole interval (β1,1).
16. APPLICATION: If you bike up a hill, then
back down, at some point your elevation
was stationary.
18. let a function π(π₯) is defined on π, π ππ such that it
is
β f(x) is continuous on a closed interval [a,b]
β f(x) is differentiable on the open interval (a,b)
then β at least one π β π, π π€βπππ
19. i.e., where slope of tangent becomes equal To slope of the chord AB.
20.
21. Drawbacks of Lagrange mean
value theorem:
Lagrangeβs mean value theorem fails for the function which does not
satisfy at least one of the two conditions.
Function is discontinuous
at π₯ = π₯1
Function is non-differentiable
at π₯ = π₯1
22. Function is non-differentiable at π₯ = π₯1 still
π β π, π πππ π€βππβ
Function is discontinuous at π₯ = π₯1 still
π β π, π πππ π€βππβ
23. Example : Verify LMVT for the function
π π₯ = π₯ π₯ β 1 π₯ β 2 ππ πππ‘πππ£ππ 0 β€ π₯ β€
1
2
.
Solution:
Here π π₯ = π₯3 β 3π₯2 + 2π₯, being a polynomial , π(π₯) is continuous and derivable in the
closed interval 0 β€ π₯ β€
1
2
.
Hence condition of Lagrangeβs mean value theorem are satisfied. β π β 0 ,
1
2
where
πβ²
π =
π π βπ(π)
πβπ
Now πβ²
π₯ = 3π₯2
β 6π₯ + 2
And
π π βπ(π)
πβπ
=
π
π
Hence
3
4
= 3π₯2
β 6π₯ + 2
π₯ = 1 β
1
6
21 taking minus , sign, we get a value of π₯ i.e.
π₯ = 1 β
1
6
21 β (0 ,
1
2
)
Hence LMVT is verified.
24. If you drive between points A and B,
at some time your speedometer reading was the same as
your average speed over the drive.