About Mean Value Theorem
In calculus, the Mean Value Theorem is a crucial concept. Parmeshwara, a mathematician from Kerela, India, proposed the earliest form of the mean value theorem in the 14th century. Furthermore, Rolle proposed a simplified version of this in the 17th century: Rolle's Theorem, which was only established for polynomials and was not part of the calculus. Finally, Augustin Louis Cauchy proposed the current version of the Mean Value Theorem in 1823.
The mean value theorem asserts that there is one point on a curve travelling between two points where the tangent is parallel to the secant running through the two points. This mean value theorem gave rise to Rolle's theorem.
What do you mean by Mean Value Theorem?
The mean value theorem asserts that there is at least one point (c, f(c)) on a curve f(x) passing through two provided points (a, f(a)), (b, f(b)) where the tangent is parallel to the secant travelling through the two given points. For a function f(x): [a, b]→ R that is continuous and differentiable across an interval, the mean value theorem is defined in calculus.
Over the interval [a, b], the function f(x) is continuous.
Function f(x) can be differentiated over a range of values (a, b).
In (a, b), there is a point c where f'(c) = [f(b) - f(a)] / (b - a)
The tangent at c is parallel to the secant travelling through the points (a, f(a)), (b, f(b) in this proof. This mean value theorem proves a proposition over a closed interval. The Rolle's theorem is also used to derive the mean value theorem.
Proof of Mean Value Theorem
Statement:The mean value theorem states that if a function f is continuous over the closed interval [a,b] and differentiable over the open interval (a,b), there must be at least one point c in the interval (a,b) where f(c) is the function's average rate of change over [a,b] and parallel to the secant line over [a,b].
Proof: Let g(x) be the secant line connecting (a, f(a)) and (b, f(b)) to f(x). We know that the secant line's equation is y-y1 = m (x- x1).
g(x) - f(a) = [ f(b) - f(a) ] / (b - a) (x-a)
g(x) = [ f(b) - f(a) ] / (b - a) (x-a) + f(a) ----->(1)
Let h(x) be f(x) - g(x)
h(x) = f(x) - [[ f(b) - f(a) ] / (b - a) (x-a) + f(a)] (From (1))
h(a) = h(b) = 0 and h(x) is continuous on [a,b] and differentiable on (a,b).
Using the Rolles theorem, there is some x = c in (a,b) that is equal to h'(c) = 0.
h'(x) = f'(x) - [ f(b) - f(a) ] / (b - a)
For some c in (a,b), h'(c) = 0. Thus
h'(c) = f'(c) - [ f(b) - f(a) ] / (b - a) = 0
f'(c) = [ f(b) - f(a) ] / (b - a)
The mean value theorem is thus established.
Note that if the function is not differentiable, even at a single point in the interval, the result may not hold.
Check out the list of all Maths Formulas on one page prepared by experts.
Download the pdf of Mean Value Theorem
