If any function is defined on the closed intervals [a, b] satisfies the given conditions:
then, there will exist a value x = c in such a way that f'(c) = [f(b) – f(a)]/(b-a).
This theorem is called the first mean value theorem or Lagrange’s mean value theorem.
In the given curve graph, y = f(x) is continuous x = a and x = b and is differentiable within the closed interval [a, b], then according to the mean value theorem, for any function that is differentiable on (a, b) and is continuous on [a, b], then there is some c in the interval (a, b) such that the secant joining the endpoints of the interval given is parallel to the tangent at c.
Verify Mean Value Theorem for the function f(x) = x2 – 4x – 3 in the interval [a, b], where a = 1 and b = 4.
f(x) = x2 – 4x – 3
f'(x) = 2x – 4
a = 1 and b = 4 (given)
f(a) = f(1) = (1)2 – 4(1) – 3 = 1 – 4 – 3 = -6
f(b) = f(4) = (4)2 – 4(4) – 3 = -3
[f(b) – f(a)] / (b – a) = (-3 + 6) / (4 – 1) = 3/3 = 1
According to the mean value theorem statement, there is a point c ∈ (1, 4) such that f'(c) = [f(b) – f(a)]/ (b – a), i.e., f'(c) = 1.
2c – 4 = 1
2c = 5
c = 5/2 ∈ (1, 4)
Verification: f'(c) = 2(5/2) – 4 = 5 – 4 = 1
There is a special case in Lagrange’s theorem according to which, if a function f is defined in the closed interval [a, b] in such a way that it is satisfying the given arguments:
More clearly, if a function is differentiable on the open interval (a, b) and continuous on the closed interval [a, b], then there exists a point x = c in (a, b) such that f'(c) = 0
Mathematically, Rolle’s Theorem is defined as: Let f: [a, b] → R be differentiable on (a, b), and continuous on [a, b] such that f(a) = f(b), where a and b are some real numbers. Then there exists some c in (a, b) such that f′(c) = 0. Normally speaking, if a continuous curve passes through the same y-value (such as the x-axis) twice and has a unique tangent line at every point of the interval, then somewhere in between the endpoints it will have a tangent parallel to the x-axis.