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Rolle’s Theorem is considered one the most difficult concept.
Lagrange’s Mean Value Theorem is considered one of the most asked concept.
32 Questions around this concept.
A value of for which conclusion of Mean Value Theorem holds for the function
on the interval
is
Rolle’s Theorem
Let f be a real-valued function defined on the closed interval [a, b] such that
it is continuous on the closed interval [a, b],
it is differentiable on the open interval (a, b), and
f (a) = f (b).
There then exists at least one c ∈ (a, b) such that f ′(c) = 0.
Geometrical interpretation of Rolle’s Theorem:
f (x) be a real-valued function defined on [a, b] such that the curve y = f (x) is a continuous in [a,b] and it is possible to draw a unique tangent at every point on the curve y = f (x) between points A and B. Also, the ordinates at the endpoints of the interval [a, b] are equal. Then there exists at least one point (c, f (c)) lying between A and B on the curve y = f (x) where the tangent is parallel to the x-axis. I.e. f’(c) = 0.
If the function is continuous and differentiable and if f(x) = 0 has two roots α and β, then from the Rolle's theorem there exists at least one root of the equation f’(x) = 0 in the interval (α, β). Similarly, for twice differentiable function y = f(x), between any two roots of the equation f’(x) = 0, there exists at least one root of the equation f”(x) = 0 and so on.
Note:
The converse of Rolle’s theorem need not be true.
For example,
Application of Rolle’s Theorem:
We can use Rolle’s theorem to study and calculate root of the equation as well as locating the roots, The equation must be continuous as well as differentiable.
Let’s look at one example
Consider the equation f(x) = (x - 1)(x - 2)(x - 3)(x - 4). Then how many roots of the equation f’(x) = 0 are positive.
Let f(x) = (x - 1)(x - 2)(x - 3)(x - 4).
Since, f(x) is continuous and differentiable and f(1) = f(2) = f(3) = f(4) = 0
According to Rolle’s theorem there is at least one root of the equation f’(x) = 0 in each of the interval (1, 2), (2, 3) and (3, 4).
Since, f’(x) is a cubic function. So f’(x) = 0 has exactly one root in each interval (1, 2), (2, 3) and (3, 4)
Lagrange’s Mean Value Theorem (LMVT)
Lagrange's Mean Value Theorem generalized Rolle’s theorem by considering functions that do not necessarily have equal value at the endpoints.
Statement
Let f (x) be a function defined on [a, b] such that
it is continuous on [a, b],
it is differentiable on (a, b).
Then there exists at least one real number c (a, b) such that
Geometrical Interpretation:
LMVT states that if f is continuous over the closed interval [a, b] and differentiable over the open interval (a, b), then there exists a point c ∈ (a, b) such that the tangent line to the graph of f at c is parallel to the secant line connecting (a, f(a)) and (b, f(b)) .
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