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Lagrange Mean Value Theorem
Lagrange mean value theorem is a further extension of rolle mean value theorem. The theorem states that for a curve between two points there exists a point where the tangent is parallel to the secant line passing through these two points of the curve.The lagrange mean value theorem is sometimes referred to as only mean value theorem.
Let us learn more about lagrange mean value theorem, proof, its relationship with rolle mean value theorem with examples and faqs.
What Is Lagrange Mean Value Theorem?
Lagrange mean value theorem states that for any two points on the curve there exists a point on the curve such that the tangent drawn at this point is parallel to the secant through the two points on the curve.
Statement of Lagrange Mean Value Theorem: Let a function f be defined such that f : [a, b] → R be a continuous function on [a, b] and differentiable on (a, b). Then there exists some a point c in this interval (a, b) such that the derivative of the function at the point c is equal to the difference of the function values at these points, divided by the difference of the point values.
\[f'(c) = \dfrac{f(b)  f(a)}{b  a}\]
The lagrange mean value theorem can be understood geometrically by presenting the graph of the equation as y = f(x). Here the graph curve of y = f(x) is passing through the points (a, f(a)), (b, f(b)), and there exists a point (c, f(c)) midway between these points and on the curve. The slope of the secant line passing through these points (a, f(a)), (b, f(b)), is \(\dfrac{f(b)  f(a)}{b  a}\). And the slope of the tangent line touching the curve at the point(c, f(c)), is f'(c). Further, the lagrange mean value theorem states that the tangent at a point (c. f(c)) is parallel to the secant line passing through the points (a, f(a)), (b, f(b)), and their slopes are equal. Hence we have \(f'(c) = \dfrac{f(b)  f(a)}{b  a}\).
Proof of Lagrange Mean Value Theorem
Statement: The lagrange 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), then there exists at least one point c in the interval (a,b) such that the slope of the tangent at the point c is equal to the slope of the secant through the endpoints of the curve such that f'(c) = \(\dfrac{f(b)  f(a)}{b  a}\).
Proof: Let g(x) be the secant line to f(x) passing through the points (a, f(a)) and (b, f(b)). We know that the slope of the secant line is m = \(\dfrac{f(b)  f(a)}{b  a}\), and the formula for the secant line is yy\(_1\) = m (x x\(_1\)). Further the equation of the secant line is as follows.
y  f(a) = \(\dfrac{f(b) f(a)}{ba}\) (xa)
y = \(\dfrac{f(b) f(a)}{ba}\) (xa) + f(a)
Since the equation of the sceant line is g(x) = y we have
g(x) = \(\dfrac{f(b) f(a)}{ba}\) (xa) + f(a) >(1)
Let us define a function h(x) which is the difference between the curve f(x), and the secant line g(x) such that we have h(x) = f(x)  g(x).
h(x) = f(x)  g(x)
Here we apply the value of g(x) from the above expression.
h(x) = f(x)  [\(\dfrac{f(b) f(a)}{ba}\) (xa) + f(a)]
Here let us consider the function h(x) is continuous on [a,b] and differentiable on (a,b). Thus applying the Rolles theorem, there is some x = c in (a,b) such that h'(c) = 0.
h'(x) = f'(x)  \(\dfrac{f(b)f(a)}{ba}\)
For some c in (a,b), h'(c) = 0. Thus
h'(c) = f'(c)  \(\dfrac{f(b)f(a)}{ba}\) = 0
f'(c)  \(\dfrac{f(b)f(a)}{ba}\) = 0
f'(c) = \(\dfrac{f(b)f(a)}{ba}\)
Thus the lagrange mean value theorem has been proved.
Lagrange Mean Value Theorem vs Rolle's Mean Value Theorem
The lagrange mean value theorem is a further extension of rolle's mean value theorem. Understanding the rolle;s mean value theorem sets the right foundation for lagrange mean value theorem. Rolle’s mean value theorem defines a function y = f(x), such that the function f : [a, b] → R be continuous on [a, b] and differentiable on (a, b). Here we have the function values equal 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. Geometrically for the graph of the function y = f(x) the line joining (a, f(a)), (b, f(b)) is parallel to the xaxis, and the slope of the tangent at the point (c, f(c))), is equal to zero, f'(c) = 0.
Lagrange mean value theorem has also been defined similar to the rolle's mean value theorem, for a function y = f(x) such that f : [a, b] → R is a function continuous on [a, b] and differentiable on (a, b). Then there exists a point in this interval, c in (a, b) such that the derivative of the function at the point c is equal to the difference of the function values at these points, divided by the difference of the point values \(f'(c) = \dfrac{f(b)  f(a)}{b  a}\). Geometrically the slope of the secant line through the points (a, f(a)), (b, f(b)) is equal to \(\dfrac{f(b)  f(a)}{b  a}\), and is equal to the slope of the tangent through the point(c, f(c)), which is f'(c). Hence we can equalize these two slopes as \(f'(c) = \dfrac{f(b)  f(a)}{b  a}\).
Related Topics
The following topics help in a better understanding of lagrange mean value theorem.
Examples on Lagrange Mean Value Theorem

Example 1: Verify if the function x^{2}+ 2x  8 satisfies lagrange mean value theorem in the interval (4, 4).
Solution:
The given function is f(x) = x^{2}+ 2x  8. The given interval is (4, 4), and is assumed to be continuous in [4, 4].
f'(x) = 2x + 2
f(4) = (4)^{2} + 2(4)  8 = 16 8  8 = 0
f(4) = 4^{2} + 2(4)  8 = 16 + 8  8 = 16
f'(x) = \(\dfrac{f(4)  f(4)}{4  (4)}\) = \(\dfrac{16  0}{4 + 4}\) = 16/8 = 2
f'(c) = 2
2c + 2 = 2
2c = 0
c = 0
Here c = 0 lies in the interval (4, 4).
Answer: Therefore, since c = 0 lies in the interval (4, 4), the function satisfies lagrange mean value theorem.

Example 2: Find if the function f(x) = 2x^{2}  4x + 5 satisfies the lagrange mean value theorem in the interval (2, 5).
Solution:
The given function is f(x) = 2x^{2}  4x + 5. Let us assume that this function is continuous in [2, 5] and is also differentiable in the interval (2, 5).
f'(x) = 4x 4
f(2) = 2.2^{2}  4(2) + 5 = 8  8 + 5 = 5
f(5) = 2.5^{2}  4(5) + 5 = 50  20 + 5 = 30 + 5 = 35
f'(x) = \(\dfrac{f(5)  f(2)}{5  2}\)
= \(\dfrac{35  5}{5  2}\) = 30/3 = 10
f'(c) = 10
4c  4 = 10
4c = 10 + 4
4c = 14
c = 14/4 = 3.5
c = 3.5 belongs to the interval (2, 5)
Answer: Therefore, since c = 3.5 belongs to the interval(2, 5), the function f(x) satisfies the lagrange mean value theorem.
FAQs on Lagrange Mean Value Theorem
What Is Lagrange Mean Value Theorem?
The lagrange 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), then there exists at least one point c in the interval (a,b) such that the slope of the tangent at the point c is equal to the slope of the secant through the endpoints of the curve such that f'(c) = \(\dfrac{f(b)  f(a)}{b  a}\).
What Is the Condition For Lagrange Mean Value Theorem?
The lagrange mean value theorem is defined for a function f, which is continuous over the closed interval [a,b], and differentiable over the open interval (a,b). The condition for lagrange mean value theorem is that there exists a point c in the interval (a, b) such that f'(c) = \(\dfrac{f(b)  f(a)}{b  a}\).
How Do We Know If a Function Satisfies Lagrange Mean Value Theorem?
The function f(x) is said to satisfy lagrange mean value theorem if there exists a point c in the interval (a, b), which satisfies f'(c) = \(\dfrac{f(b)  f(a)}{b  a}\). This has been derived from the slope formula of coordinate geometry, and the slope of the tangent through the point c of the curve f(x) is equal to the slope of the secant line through the end points a and b of the curve f(x).
How Do You Prove Lagrange Mean Value Theorem Using a Graph?
The lagrange mean value theorem can be easily proved with the help of a graph. For a curve f(x) drawn across two points (a, f(a)), (b, f(b)), there exists a point (c, f(c)) on this curve such that the slope of the tangent f'(c) at this point is equal to the slope of the secant line (\(\frac{f(b)  f(a)}{b  a}\)) through the endpoints of the curve.f'(c) = \(\dfrac{f(b)  f(a)}{b  a}\)
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