mean value theorem

1. Presented by Uzair Ahmad

2. Outlines

3. This property was known in the 12th century in ancient
India. The outstanding Indian astronomer and
mathematician Bhaskara II (1114−1185) mentioned it in
his writings.
In a strict form this theorem was proved in 1691 by the
French mathematician Michel Rolle (1652−1719)

4. Rolle’s theorem
Statement:
Let a function 𝑓(𝑥) defined on [𝑎, 𝑏] be such that
❖ It is continuous in the interval [𝑎, 𝑏]
❖ It is differentiable in the interval (𝑎, 𝑏)
❖ 𝑓 𝑎 = 𝑓(𝑏)
then there exist at least one c ∈ (a, b) such that 𝑓′ 𝑐 = 0

5. Since 𝑓(𝑥) is continuous in closed interval 𝑎, 𝑏
⟹ 𝑓(𝑥) is bounded I.e. ,𝑚 ≤ 𝑓 𝑥 ≤ 𝑀 ∀ 𝑥 ∈ [𝑎, 𝑏]
∴ 𝑓 𝑎 𝑎𝑛𝑑 𝑓(𝑏) are either both maxima (minima) or both
are neither maxima nor minima
So there exist at least one point 𝑐 ∈ 𝑎, 𝑏 , where 𝑓 𝑐 =
𝑚 𝑜𝑟 𝑀.

6. 𝑀 = 𝑚 ⇒ 𝑓(𝑥) is constant ∀ 𝑥 ∈ (𝑎, 𝑏)
⟹ 𝑓 𝑥 = 0 ∀ 𝑥 ∈ (𝑎, 𝑏)

7. 𝑀 ≠ 𝑚 and let there lie a point 𝑐 𝜖 (𝑎, 𝑏), where 𝑓 𝑐 = 𝑀
⟹ 𝑓 𝑐 + ℎ ≤ 𝑓 𝑐 𝑎𝑛𝑑𝑓 𝑐 − ℎ ≤ 𝑓 𝑐
⟹
𝑓 𝑐+ℎ −𝑓(𝑐)
ℎ
≤ 0 and
𝑓 𝑐−ℎ −𝑓(𝑐)
−ℎ
≥ 0
⟹ lim
ℎ⟶0
𝑓 𝑐+ℎ −𝑓(𝑐)
ℎ
≤ 0 𝑎𝑛𝑑 lim
ℎ→0
𝑓 𝑐−ℎ −𝑓(𝑐)
−ℎ
≥ 0
⟹ 𝑓′
𝑐−
≥ 0 𝑎𝑛𝑑𝑓′
𝑐+
≤ 0 but ∵ 𝑓 is
differentiable at c.
∴ 𝑓′
𝑐−
= 𝑓′
𝑐+
⟹ 𝑓′ 𝑐 = 0

8. 𝑀 ≠ 𝑚 and let there lie a point 𝑐 𝜖 (𝑎, 𝑏), where
𝑓 𝑐 = 𝑚
⟹ 𝑓 𝑐 + ℎ ≥ 𝑓 𝑐 𝑎𝑛𝑑𝑓 𝑐 − ℎ ≥ 𝑓 𝑐
⟹
𝑓 𝑐+ℎ −𝑓(𝑐)
ℎ
≥ 0 and
𝑓 𝑐−ℎ −𝑓(𝑐)
−ℎ
≤ 0
⟹ lim
ℎ⟶0
𝑓 𝑐+ℎ −𝑓(𝑐)
ℎ
≥ 0 𝑎𝑛𝑑 lim
ℎ→0
𝑓 𝑐−ℎ −𝑓(𝑐)
−ℎ
≤ 0
⟹ 𝑓′ 𝑐+ ≥ 0 𝑎𝑛𝑑𝑓′ 𝑐− ≤ 0 but ∵ 𝑓 is differentiable at
c.
∴ 𝑓′ 𝑐− = 𝑓′ 𝑐+
⟹ 𝑓′
𝑐 = 0

9. 𝑐1 𝑎𝑛𝑑 𝑐3 𝑎𝑟𝑒 𝑡ℎ𝑒 𝑝𝑜𝑖𝑛𝑡𝑠 𝑜𝑓 𝑙𝑜𝑐𝑎𝑙 𝑚𝑎𝑥𝑖𝑚𝑎 𝑎𝑛𝑑 𝑐2 𝑎𝑛𝑑 𝑐4
𝑎𝑟𝑒 𝑡ℎ𝑒 𝑝𝑜𝑖𝑛𝑡𝑠 𝑜𝑓 𝑙𝑜𝑐𝑎𝑙 𝑚𝑖𝑛𝑖𝑚𝑎.
Hence 𝑓′ 𝑐1 = 𝑓′ 𝑐2 = 𝑓′ 𝑐3 = 𝑓′ 𝑐4 … … … … … = 0

10. Rolle’s theorem fails for the function which does not satisfy
at least one of the three conditions.

11. ➢ 𝑡ℎ𝑒 𝑐𝑜𝑛𝑣𝑒𝑟𝑠𝑒 𝑜𝑓 𝑅𝑜𝑙𝑙𝑒′
𝑠 𝑡ℎ𝑒𝑜𝑟𝑒𝑚 𝑚𝑎𝑦 𝑛𝑜𝑡 𝑏𝑒
𝑡𝑟𝑢𝑒. 𝑖. 𝑒 , 𝑓′
𝑐 𝑚𝑎𝑦 𝑛𝑜𝑡 𝑏𝑒 𝑧𝑒𝑟𝑜 𝑎𝑡 𝑎 𝑝𝑜𝑖𝑛𝑡 𝑖𝑛
𝑎, 𝑏 𝑤𝑖𝑡ℎ𝑜𝑢𝑡 𝑠𝑎𝑡𝑖𝑠𝑓ying 𝑎𝑙𝑙 𝑡ℎ𝑒 𝑡ℎ𝑟𝑒𝑒 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑠

12. Case 1 Case 2 Case 3
In all the above three cases , we observed that all the conditions of Rolle’s 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 ‘𝑐 ‘ such that 𝑐 ∈ (𝑎, 𝑏) and 𝑓′ 𝑐 = 0

13. Verify whether the function 𝑓 𝑥 = sin(𝑥)
𝑖𝑛 0, 𝜋 satisfies the conditions of Rolle’s theorem and hence
find c as prescribed by the theorem.
Given 𝑓 𝑥 = sin 𝑥 𝑖𝑛 0, 𝜋 .
Here 𝑓 0 = 0 = 𝑓(𝜋)
Also we know that sin(𝑥) is continuous in [0, 𝜋] and differentiable in ]0, 𝜋[ .
Hence 𝑓 satisfies all the conditions of the Rolle’s theorem. So there must exist at
least one value of 𝑥 ∈ ]0, 𝜋[ such that 𝑓′ 𝑥 = 0
Now 𝑓′ 𝑥 = 0 ⟹ cos 𝑥 = 0 ⟹ 𝑥 =
𝜋
2
∈ 0, 𝜋 .
Hence , in Rolle’s theorem , 𝑐 =
𝜋
2

14. Discuss the applicability of
Rolle’s theorem to 𝑓 𝑥 = 2 + (𝑥 − 1)
2
3 𝑖𝑛 0,2 .
Solution :
Here 𝑓′ 𝑥 =
2
3
× 𝑥 − 1 −
1
3 = 2/{3 𝑥 − 1
1
3}
Which does not exist (i.e is not finite) at 𝑥 = 1 ∈]0,2[ , Hence the condition “ 𝑓(𝑥)
is deriveable in]0, 2[ ‘ is not satisfied . Therefore , Rolle’s theorem is not applicable
to 𝑓 𝑥 𝑖𝑛 0,2 .

15. Example : Discuss the applicability of
Rolle’s theorem to 𝑓 𝑥 = 𝑥 in [1,-1].
Consider f(x)=|x|
(where |x| is the absolute value of x) on the
closed interval [−1,1].
This function does not have derivative at x=0.
Though f(x) is continuous on the closed
interval [−1,1], there is no point inside the
interval (−1,1) at which the derivative is equal
to zero. The Rolle’s theorem fails here
because f(x) is not differentiable over the
whole interval (−1,1).

16. APPLICATION: If you bike up a hill, then
back down, at some point your elevation
was stationary.

17. By Joseph-Louis Lagrange

18. let a function 𝑓(𝑥) is defined on 𝑎, 𝑏 𝑖𝑠 such that it
is
❑ f(x) is continuous on a closed interval [a,b]
❑ f(x) is differentiable on the open interval (a,b)
then ∃ at least one 𝑐 ∈ 𝑎, 𝑏 𝑤ℎ𝑒𝑟𝑒

19. i.e., where slope of tangent becomes equal To slope of the chord AB.

21. Drawbacks of Lagrange mean
value theorem:
Lagrange’s mean value theorem fails for the function which does not
satisfy at least one of the two conditions.
Function is discontinuous
at 𝑥 = 𝑥1
Function is non-differentiable
at 𝑥 = 𝑥1

22. Function is non-differentiable at 𝑥 = 𝑥1 still
𝑐 ∈ 𝑎, 𝑏 𝑓𝑜𝑟 𝑤ℎ𝑖𝑐ℎ
Function is discontinuous at 𝑥 = 𝑥1 still
𝑐 ∈ 𝑎, 𝑏 𝑓𝑜𝑟 𝑤ℎ𝑖𝑐ℎ

23. Example : Verify LMVT for the function
𝑓 𝑥 = 𝑥 𝑥 − 1 𝑥 − 2 𝑖𝑛 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙 0 ≤ 𝑥 ≤
1
2
.
Solution:
Here 𝑓 𝑥 = 𝑥3 − 3𝑥2 + 2𝑥, being a polynomial , 𝑓(𝑥) is continuous and derivable in the
closed interval 0 ≤ 𝑥 ≤
1
2
.
Hence condition of Lagrange’s mean value theorem are satisfied. ∃ 𝑐 ∈ 0 ,
1
2
where
𝑓′
𝑐 =
𝑓 𝑏 −𝑓(𝑎)
𝑏−𝑎
Now 𝑓′
𝑥 = 3𝑥2
− 6𝑥 + 2
And
𝑓 𝑏 −𝑓(𝑎)
𝑏−𝑎
=
𝟑
𝟒
Hence
3
4
= 3𝑥2
− 6𝑥 + 2
𝑥 = 1 ∓
1
6
21 taking minus , sign, we get a value of 𝑥 i.e.
𝑥 = 1 −
1
6
21 ∈ (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.