Rolle’s Theorem is a particular case of the mean value theorem which satisfies certain conditions. At the same time, Lagrange’s mean value theorem is the mean value theorem itself or the first mean value theorem. In general, one can understand mean as the average of the given values. But in the case of integrals, the process of finding the mean value of two different functions is different.
Rolle's Theorem Statement:
Rolle's theorem states that "If a function f is defined in the closed interval [a, b] in such a way that it satisfies the following condition:
i) f is continuous on [a, b], ii) f is differentiable on (a, b), and iii) f (a) = f (b), then there exists at least one value of x, let us assume this value to be c, which lies between a and b i.e. (a < c < b ) in such a way that f‘(c) = 0."
Rolle's Theorem:
(1) f(x) is continuous on [a,b]
(2) f(x) is differentiable on (a,b) and
(3) f(a) = (b)
Then there exists at least on c in(a,b) such that f'(c) = 0
Question: Verify Rolle’s theorem for the function y = x2 + 2, a = –2 and b = 2.
From the definition of Rolle’s theorem, the function y = x2 + 2 is continuous in [– 2, 2] and differentiable in (– 2, 2).
From the given,
f(x) = x2 + 2
f(-2) = (-2)2 + 2 = 4 + 2 = 6
f(2) = (2)2 + 2 = 4 + 2= 6
Thus, f(– 2) = f( 2) = 6
Hence, the value of f(x) at –2 and 2 coincide.
Now, f'(x) = 2x
Rolle’s theorem states that there is a point c ∈ (– 2, 2) such that f′(c) = 0.
At c = 0, f′(c) = 2(0) = 0, where c = 0 ∈ (– 2, 2)
Hence verified.
