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Verify Rolle’s Theorem for the function
The given function,, being a polynomial function, is continuous in [−4, 2] and is differentiable in (−4, 2).
∴ f (−4) = f (2) = 0
⇒ The value of f (x) at −4 and 2 coincides.
Rolle’s Theorem states that there is a point c ∈ (−4, 2) such that
Hence, Rolle’s Theorem is verified for the given function.
Examine if Rolle’s Theorem is applicable to any of the following functions. Can you say some thing about the converse of Rolle’s Theorem from these examples?
(i) 
(ii) 
(iii) 
By Rolle’s Theorem, for a function
, if
(a) f is continuous on [a, b]
(b) f is differentiable on (a, b)
(c) f (a) = f (b)
then, there exists some c ∈ (a, b) such that 
Therefore, Rolle’s Theorem is not applicable to those functions that do not satisfy any of the three conditions of the hypothesis.
(i)
It is evident that the given function f (x) is not continuous at every integral point.
In particular, f(x) is not continuous at x = 5 and x = 9
⇒ f (x) is not continuous in [5, 9].
The differentiability of f in (5, 9) is checked as follows.
Let n be an integer such that n ∈ (5, 9).
Since the left and right hand limits of f at x = n are not equal, f is not differentiable at x = n
∴f is not differentiable in (5, 9).
It is observed that f does not satisfy all the conditions of the hypothesis of Rolle’s Theorem.
Hence, Rolle’s Theorem is not applicable for.
(ii)
It is evident that the given function f (x) is not continuous at every integral point.
In particular, f(x) is not continuous at x = −2 and x = 2
⇒ f (x) is not continuous in [−2, 2].
The differentiability of f in (−2, 2) is checked as follows.
Let n be an integer such that n ∈ (−2, 2).
Since the left and right hand limits of f at x = n are not equal, f is not differentiable at x = n
∴f is not differentiable in (−2, 2).
It is observed that f does not satisfy all the conditions of the hypothesis of Rolle’s Theorem.
Hence, Rolle’s Theorem is not applicable for.
(iii)
It is evident that f, being a polynomial function, is continuous in [1, 2] and is differentiable in (1, 2).
∴f (1) ≠ f (2)
It is observed that f does not satisfy a condition of the hypothesis of Rolle’s Theorem.
Hence, Rolle’s Theorem is not applicable for.
If 
is a differentiable function and if 
does not vanish anywhere, then prove that
.
It is given that is a differentiable function.
Since every differentiable function is a continuous function, we obtain
(a) f is continuous on [−5, 5].
(b) f is differentiable on (−5, 5).
Therefore, by the Mean Value Theorem, there exists c ∈ (−5, 5) such that
It is also given that 
does not vanish anywhere.
Hence, proved.
Verify Mean Value Theorem, if 
in the interval
, where 
and
.
Given function is continuous in closed intervals and differentiable in open intervals f(1) not equal to f(4)
By mean value theorem
where c belongs to (a,b)
f(1)=1-4-3=-6
f(4)=16-16-3=-3
f'(x)= 2x-4
By mvt
C lies between 1 and 4
Hence, Mean Value Theorem is verified for the given function.
Verify Mean Value Theorem, if
in the interval [a, b], where a = 1 and b = 3. Find all 
for which 
The given function f is
f, being a polynomial function, is continuous in [1, 3] and is differentiable in (1, 3) whose derivative is 3x2 − 10x − 3.
Mean Value Theorem states that there exist a point c ∈ (1, 3) such that
Hence, Mean Value Theorem is verified for the given function and 
is the only point for which
Examine the applicability of Mean Value Theorem for all three functions given in the above exercise 2.
Mean Value Theorem states that for a function
, if
(a) f is continuous on [a, b]
(b) f is differentiable on (a, b)
then, there exists some c ∈ (a, b) such that 
Therefore, Mean Value Theorem is not applicable to those functions that do not satisfy any of the two conditions of the hypothesis.
(i) 
It is evident that the given function f (x) is not continuous at every integral point.
In particular, f(x) is not continuous at x = 5 and x = 9
⇒ f (x) is not continuous in [5, 9].
The differentiability of f in (5, 9) is checked as follows.
Let n be an integer such that n ∈ (5, 9).
Since the left and right hand limits of f at x = n are not equal, f is not differentiable at x = n
∴f is not differentiable in (5, 9).
It is observed that f does not satisfy all the conditions of the hypothesis of Mean Value Theorem.
Hence, Mean Value Theorem is not applicable for .
(ii) 
It is evident that the given function f (x) is not continuous at every integral point.
In particular, f(x) is not continuous at x = −2 and x = 2
⇒ f (x) is not continuous in [−2, 2].
The differentiability of f in (−2, 2) is checked as follows.
Let n be an integer such that n ∈ (−2, 2).
Since the left and right hand limits of f at x = n are not equal, f is not differentiable at x = n
∴f is not differentiable in (−2, 2).
It is observed that f does not satisfy all the conditions of the hypothesis of Mean Value Theorem.
Hence, Mean Value Theorem is not applicable for .
(iii) 
It is evident that f, being a polynomial function, is continuous in [1, 2] and is differentiable in (1, 2).
It is observed that f satisfies all the conditions of the hypothesis of Mean Value Theorem.
Hence, Mean Value Theorem is applicable for.
It can be proved as follows.
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