Overview of the Lagrange’s Mean Value Theorem

In mathematics, the mean value theorem asserts, approximately, that for each planar arc between two endpoints, there is at least one point at which the tangent to the arc is parallel to the secant across the endpoints, and this is known as the mean value of the arc. It is one of the most significant findings in the field of real analysis. When proving claims about a function on an interval, this theorem is used to prove local hypotheses regarding derivatives at locations on the interval, which are then proven. 

Lagrange’s Mean Value Theorem: 

The Lagrange mean value theorem asserts that for any two points on a curve, there exists a point on the curve where the tangent drawn at this point is parallel to the secant drawn through the two points on the curve, which is known as the Lagrange mean value. 

Statement of Lagrange Mean Value Theorem: 

For simplicity, consider the following definition of a function: f: [a, b] → R where f: [a, b] → R is continuous on [a, b] and differentiable on (a, b). When the difference of the function values at these points is divided by the difference of the point values, then there exists some point c in this interval (a, b) such that the derivative of the function at point c equals the difference of the function values at these points divided by the difference of the point values. 

f’(c) = f(b) – f(a)/b-a 

Graph of Lagrange Mean Value Theorem: 

Understanding the lagrange mean value theorem can be accomplished through the use of geometric representations. For example, consider the graph of the equation as y = f(x). In this case, the graph curve of y = f(x) passes through the points (a, f(a)), (b, f(b), and (c, f(c)), and there is a point (c, f(c)) in the middle of the curve and halfway between these places. When the secant line passes through the points (a, f(a)), (b, f(b)), and (c, f(c)), the slope of the secant line is f(b)-f(a)/b-a. Also known as f’(c), the slope of a tangent line intersecting the curve at the point(c,f(c)) is the slope of the curve (c). A further result of the lagrange mean value theorem is that, for any location (c. F(c)), the tangent at that point is parallel to the secant line that passes through the points (a, f(a)), and (c, f(b), and their slopes are equal. As a result, f’(c) = f(b)-f(a)/b-a. 

Proof of Lagrange Mean Value Theorem: 

Statement: 

It is stated in the lagrange mean value theorem, which 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, at the point c, the slope of the tangent is equal to the slope of a secant through the endpoints of the curve such that f’(c) = f(b)-f(a)/b-a. 

Proof: 

Take the secant line to f(x) running through the points (a, f(a)) and (b, f(b)) as an example of g(x). We already know that the secant line has a slope of m = f(b)-f(a)/b-a. Furthermore, the secant line’s formula is given by the equation y-y1 = m(x- x1). The following is the equation for the secant line in more detail. 

y – f(a) = [f(b)-f(a)/b-a](x-a) 

y = [f(b)-f(a)/b-a](x-a)+f(a) 

Because the equation of the secant line is g(x) = y, we have the following equation: 

g(x) = [f(b)-f(a)/b-a](x-a)+f(a) ……….. (1) 

Define a function h(x) that is the difference between the curve f(x) and the secant line g(x), such that the equation h(x) = f(x) – g(x). 

h(x) = f(x)-g(x) 

In this case, we use the value of g(x) from the previous expression. 

h(x) = f(x)-{[f(b)-f(a)/b-a](x-a)+f(a)} 

In this section, we will investigate the function h(x), which is continuous on [a,b] and differentiable on (a,b). As a result, using Rolle’s theorem, we may conclude that there is some x = c in (a,b) such that h’(c) = 0. 

h’(x) = f’(x) – [f(b)-f(a)/b-a] 

For some c in (a,b), h’(c) = 0. As a result, 

h’(c) = f’(c) – [f(b)-f(a)/b-a] = 0 

f’(c) – [f(b)-f(a)/b-a] = 0 

f’(c) = [f(b)-f(a)/b-a] 

As a result, the Lagrange mean value theorem has been established. 

Lagrange’s Mean Value Theorem examples: 

Example: Verify that the function x²+ 2x – 8 is consistent with the lagrange mean value theorem throughout the interval (-4, 4). 

Solution: The function f(x) = x2+ 2x – 8 is presented as an example. The specified interval is (-4, 4) and it is believed 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) = f(4)-f(-4)/4-(-4) = 16-0/4+4 = 16/8 = 2 

f’(c) = 2 

2c+2 = 2 

2c = 0 

c = 0 

As a result, c = 0 lies in the interval (-4,4). 

As a result, the function satisfies the lagrange mean value theorem because c = 0 is located within the interval (-4, 4). 

Conclusion: 

When we look at this theorem from an algebraic perspective, it tells us that, given a polynomial function f (x) as its representation in x and two roots of the equation f(x)=0 as the values a and b, there exists at least one root of the equation f’(x) = 0 that lies between these values.