Mean Value Theorem states that for a function f : [a, b] → R, if

(a) f is continuous on [a, b]

(b) f is differentiable on (a, b)

Then there exists some c ∈ (a, b) such that

If a function does not satisfy any of the above conditions, then Mean Value Theorem is not applicable.

(i) f (x) = [x] for x ∈ [5, 9]

As the given function is a greatest integer function,

(a)f(x) is not continuous in [5, 9]

(b) Let y be an integer such that y ∈ (5, 9)

Left hand limit of f(x) at x = y:

Right hand limit of f(x) at x = y:

Since, left and right hand limits of f(x) at x=y are not equal, f(x) is not differentiable at x=y.

So, f(x) is not differentiable in [5, 9].

Here, f(x) does not satisfy the conditions of Mean Value Theorem.

Mean Value Theorem is not applicable for f(x) = [x] for x ∈ [5, 9].

(ii) f (x) = [x] for x ∈ [– 2, 2]

As the given function is a greatest integer function,

(a)f(x) is not continuous in [-2, 2]

(b) ) Let y be an integer such that y ∈ (-2, 2)

Left hand limit of f(x) at x = y:

Right hand limit of f(x) at x = y:

Since, left and right hand limits of f(x) at x=y are not equal, f(x) is not differentiable at x=y.

So, f(x) is not differentiable in (-2, 2)

Here, f(x) does not satisfy the conditions of Mean Value Theorem.

Mean Value Theorem is not applicable for f(x) = [x] for x ∈ [-2, 2].

(iii) f (x) = x² – 1 for x ∈ [1, 2]

As the given function is a polynomial function,

(a) f(x) is continuous in [1, 2]

(b) f'(x) = 2x

So, f(x) is differentiable in [1, 2].

Here, f(x) satisfies the conditions of Mean Value Theorem.

So, Mean Value Theorem is applicable for f(x).

∴

f(2) = 2² – 1 = 4 -1 = 3

f(1) = 1² – 1 = 1 – 1 = 0

⇒

∴ There is a point c ∈ (1, 2) such that f'(c) = 3

⇒ f'(c) = 3

⇒ 2c = 3

⇒ c = 3/2 where c ∈ (1, 2)

Mean Value Theorem is applicable for f(x) = x² – 1 for x ∈ [1, 2].