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Lagrange's Mean Value Theorem

Lagrange's mean value theorem is also known as the mean value theorem or MVT or LMVT. It states that if a function f (x) is a continuous in a close interval [a, b] where (a≤x ≤b) and differentiable in open interval [a, b] where (a < x< b), then there is at least one point x = c on this interval, given as

f(b) - f (a) = f' (c) (b-a)

The above theorem is called first mean value theorem. It enables to express the increment of a function in the given interval through the derivative value at an intermediate point of the segment.

Proof of Lagrange's mean value theorem

Consider the auxiliary function given below

F (x) = f (x) + ?x

Here, we will select a number ? so that the condition F (a) = F (b) is satisfied. Then

Lagrange's Mean Value Theorem

The function F (x) is continuous in the closed interval (a≤x ≤b), differentiable in the open interval (a < x< b), and takes equal values at the endpoints of the interval. So, it satisfies all the conditions of Rolle's theorem. Then, there is a point c exist in the interval (a,b) given as

F' (c) = 0.

It follows that

Lagrange's Mean Value Theorem

Let's understand this concept with the help of an example

Example 1:

Verify mean value theorem for f(x) = x2 in interval [4,6]

Solution: Here, you need first to check out the function is continuous or not in the given closed interval; in our case, it is continuous. Then, you need to check the function is differentiable or not in the open interval (4,6); in our case, it is differentiable.

F' (x) = 2x

f(4) = 8

and f(6) = 36

Lagrange's Mean Value Theorem

Mean value theorem states that there is a point c ε (4,6) such that

f' (c) = 14.

But,

f' (x) = 2x

which implies that c = 7. Thus at c = 7 ε (4,6), we have

f' (c) = 14

Geometric Interpretation

The geometrical meaning of Lagrange's mean value theorem states that when the chord passing through the points of the graph corresponding to the ends of the segment a and b has the slope equals to

Lagrange's Mean Value Theorem

then there is a point x = c located inside the interval [a,b], where the tangent to the graph is parallel to the chord.

Lagrange's Mean Value Theorem

Physical Interpretation

Lagrange's mean value theorem has a very clear physical interpretation. Let's consider that f(p) represents the position of an object moving along a line, depending on the time p. The ratio of Lagrange's Mean Value Theorem is the average velocity of the object in the period b - a. Since f'(p) is the instantaneous velocity, this theorem means that there exists a moment of time c, in which the instantaneous speed is equal to the average speed.

Lagrange's mean value theorem has broad applications in mathematical analysis, computational mathematics and other fields. Let us consider two different results.

Output 1

In a specific case when the values of the function f (x) at the endpoints of the segment [a,b] are equal that is f(a) = f (b), then the mean value theorem implies that there is a point c ε (a,b) such that f'(c) - Lagrange's Mean Value Theorem = 0

We get Rolle's theorem, which can be considered a special case of Lagrange's mean value theorem.

Output 2

If the derivative f' (x) is null at all points in the interval [a,b], the function f (x) is constant on this interval. Undoubtably, for any two points p1 and p2 in the interval [a,b], there exists a point c ε (a,b), such that

f(p2)- f (p1) = f' (c) (p2 - p1) = 0. (x2-x1) = 0

Consequently

F(p1) = f (p2)

Questions based on Lagrange's mean value theorem

Example 1:

A thermometer was taken from the deep freezer and placed in hot water. It took 30 sec for the thermometer to rise from -5 degrees Celsius to 100 degrees Celsius. Find the average rate of temperature change.

Solution:

Given;

T (t1) = -5

T (t2) = 100

And t2 - t1 = 30 0C

As we know,

The average rate of temperature change Lagrange's Mean Value Theorem is described by the right-hand side of the formula given by Lagrange's mean value theorem.

Lagrange's Mean Value Theorem

Example 2:

Check the validity of Lagrange's mean value theorem for the function

f(x) = (x2 - 2x + 3)

on the interval [1, 2]. If the theorem holds, determine a point x satisfying the conditions of the theorem.

Solution:

The given quadratic equation is continuous and differentiable on the entire set of real numbers. Therefore, we can apply Lagrange's mean value theorem in the given equation. The derivative of the function has the form

F'(x) = (x2 - 2x + 3) = 2x - 2

Determine the coordinates of the point c:

Lagrange's Mean Value Theorem

You can see that the point c = 2,3 lies in the interval (1,2)

Example 3:

The position of an object is given by the function s (t) = t2 +5t - 6. Determine the time t = c in the interval 0 ≤t ≤6 when the object's instantaneous velocity is equal to its average velocity in this interval.

Solution:

The given function s (t) satisfies the mean value theorem principle, so we can write

Lagrange's Mean Value Theorem

where, s = 0, and b = 6

taking the derivative:

s'(x) = (t2 +5t -6)' = 2t + 5

substituting this equation we get,

Lagrange's Mean Value Theorem
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