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∴ 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 f'(c) = 0
Hence, Rolle’s Theorem is verified for the given function.
(ii)
(iii)
Solution :
By Rolle’s Theorem, for a function f[a,b] →R, 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 f'(c) = 0
Therefore, Rolle’s Theorem is not applicable to those functions that do not satisfy any of the three conditions of the hypothesis.
(i)
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
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
.
It is also given that f'(x) does not vanish anywhere.
Hence, proved.
Mean Value Theorem states that there is a point c ∈ (1, 4) such that f'(c) = 1
Hence, Mean Value Theorem is verified for the given function.
Mean Value Theorem states that there exist a point c ∈ (1, 3) such that f'(c) = - 10
Hence, Mean Value Theorem is verified for the given function and c = 7/3 ∈ (1,3) is the only point for which f'(c) = 0
Therefore, Mean Value Theorem is not applicable to those functions that do not satisfy any of the two conditions of the hypothesis.
(i)
