Question 1. Verify Rolle’s theorem for the function f(x) = x2+ 2x – 8, x ∈ [– 4, 2].
Solution:
Now f(x) = x² + 2x – 8 is a polynomial
So, f(x) is continuous in the interval [-4,2] and differentiable in the interval (- 4,2)
f(-4) = (-4)² + 2(-4) – 8 = 16 – 8 – 8 = 0
f(2) = 2² + 4 – 8 = 8 – 8 = 0
f(-4) = f(2)
As Conditions of Rolle’s theorem are satisfied.
Then there exists some c in (-4, 2) such that f′(c) = 0
f'(x) = 2x + 2
f’ (c) = 2c + 2 = 0
c = – 1,
and -1 ∈ [-4,2]
Hence, f’ (c) = 0 at c = – 1.
Question 2. 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 example?
(i) f(x) = [x] for x ∈ [5, 9]
Solution:
In the interval [5, 9],
Now, f (x) = [x] which is neither continuous nor derivable at Integers.
f (x) is neither continuous nor derivable at x = 6,7,8
Hence, Rolle’s theorem is NOT applicable
(ii) f(x) = [x] for x ∈ [– 2, 2]
Solution:
In the interval [– 2, 2],
Now, f (x) = [x] which is neither continuous nor derivable at Integers.
f (x) is neither continuous nor derivable at x = -1,0,1
Hence, Rolle’s theorem is NOT applicable
(iii) f(x) = x2– 1 for x ∈ [1, 2]
Solution:
Now f(x) = x² – 1 is a polynomial
So, f(x) is continuous in the interval [1, 2] and differentiable in the interval (1,2)
f(1) = (1)² – 1 = 0
f(2) = 2² – 1 = 3
f(-4) ≠ f(2)
As Conditions of Rolle’s theorem are NOT satisfied.
Hence, Rolle’s theorem is NOT applicable
Question 3. If f : [– 5, 5] → R is a differentiable function and if f′(x) does not vanish anywhere, then prove that f(– 5) ≠ f(5).
Solution:
For Rolle’s theorem
f is continuous in [a, b] ………(1)
f is derivable in [a, b] ………(2)
f (a) = f (b) ………(3)
then f’ (c)=0, c ∈ (a, b)
So as, f is continuous and derivable
but f ‘(c) ≠ 0
It concludes, f(a) ≠ f(b)
f(-5) ≠ f(5)
Question 4. Verify Mean Value Theorem, if f(x) = x2– 4x – 3 in the interval [a, b], where a = 1 and b = 4.
Solution:
Now f(x) = x² – 4x -3 is a polynomial
So, f(x) is continuous in the interval [1,4] and differentiable in the interval (1,4)
f(1) = (1)² – 4(1) – 3 = -6
f(4) = 4² – 4(4) – 3 = -3
f′(c) = 2c – 4
As Conditions of Mean Value Theorem are satisfied.
Then there exists some c in (1,4) such that
f′(c) = 
= 
= 1
2c – 4 = 1
c = 5/2
and c = 5/2 ∈ (1,4)
Question 5. Verify Mean Value Theorem, if f(x) = x3– 5x2– 3x in the interval [a, b], where a = 1 and b = 3. Find all c ∈ (1, 3) for which f′(c) = 0.
Solution:
Now f(x) = x3– 5x2– 3x is a polynomial
So, f(x) is continuous in the interval [1,3] and differentiable in the interval (1,3)
f(1) = (1)3– 5(1)2– 3(1) = -7
f(3) = 33– 5(3)2– 3(3) = -27
f′(c) = 3c2 – 5(2c) – 3
f′(c) = 3c2 – 10c – 3
As Conditions of Mean Value Theorem are satisfied.
Then there exists some c in (1,3) such that
f′(c) =
= 
= 
= 
3c2 – 10c – 3 = -10
3c2 – 10c + 7 = 0
3c2 – 3c – 7c + 7 = 0
3c (c-1) – 7(c -1) = 0
(3c -7) (c-1) = 0
c = 7/3 or c = 1
As, 1 ∉ (1,3)
So, c = 7/3 ∈ (1,3)
According to the Rolle’s Theorem
As, f(3) ≠ f(1), Then there does not exist some c ∈ (1,3) such that f′(c) = 0
Question 6. Examine the applicability of Mean Value Theorem for all three functions given in the above exercise 2.
(i) f(x) = [x] for x ∈ [5, 9]
Solution:
In the interval [5, 9],
Now, f (x) = [x] which is neither continuous nor derivable at Integers.
f (x) is neither continuous nor derivable at x = 6,7,8
Hence, Mean value theorem is NOT applicable
(ii) f(x) = [x] for x ∈ [– 2, 2]
Solution:
In the interval [– 2, 2],
Now, f (x) = [x] which is neither continuous nor derivable at Integers.
f (x) is neither continuous nor derivable at x = -1,0,1
Hence, Mean value theorem is NOT applicable
(iii) f(x) = x2– 1 for x ∈ [1, 2]
Solution:
Now f(x) = x² – 1 is a polynomial
So, f(x) is continuous in the interval [1,2] and differentiable in the interval (1,2)
f(1) = (1)² – 1 = 0
f(2) = 2² -1 = 3
f′(c) = 2c
As Conditions of Mean Value Theorem are satisfied.
Then there exists some c in (1,2) such that
f′(c) =
= 
= 
= 3
2c = 3
c = 3/2
and c = 3/2 ∈ (1,4)