﻿ Average or Lagrange value theorem. Exercises solved step by step.

# Average or Lagrange value theorem. Exercises solved step by step.

Next I’m going to explain what it says and how to interpret the average value theorem, also known as Lagrange or finite increment theorem. This theorem is explained in the 2nd year of high school when the applications of the deviradas are studied.

We will see what it means step by step and we will apply it in several solved exercises.

## Average value theorem

The theorem of the average value says so:

If we have a continuous f(x) function in the closed interval [a,b] (it has to be continuous in x=a and x=b) and derivable in the open interval (a,b) (it doesn’t have to be derivable either in x=a or in x=b), then there is at least one point c, belonging to the open interval (a,b), such that at that point it is verified: Also f(a) and f(b) have to be different.

Symbolically, we can express it this way: And what does this mean?

We have a f(x) function that is continuous in [a,b], derivable in (a,b), like this one: The points x=a and x=b, belong to the function and we also see that the point x=a, has a value of the function f(a) and the point b has a value of the function f(b) that is different from f(a).

Therefore, this function fulfills the conditions for the theorem of the average value to be fulfilled.

If we draw a line through points A and B: The slope of that line has the following formula: That corresponds to the slope of a line passing through two points.

What the theorem of the average value says is that if all the previous conditions are fulfilled, which we have seen yes, then there is at least one point c, in which the tangent line at that point is parallel to the line that passes through points A and B: The equation of the slope of the tangent line at a point is equal to that derived from the function at that point. Therefore, at point c, the equation of the slope of the tangent line will be: When two lines are parallel, it means that they have the same slope, so the slope of the tangent line at point c and the slope of the line through A and B are equal and therefore: The average value theorem says that there is at least one point c, which verifies all of the above, or in other words, that there can be more than one point.

In this case, as we see in the graph of the function, we have another point d where the line tangent to the function is parallel to the line passing through A and B: Therefore, this point is also fulfilled: ## How to apply the theorem of the average value. Exercises solved

Let us now see some examples of how to apply the average value theorem and calculate the c point of the theorem.

### Example 1

Calculate the point c that satisfies the average value theorem for the next function in the interval [0.1]: First of all, we must check if the conditions are fulfilled so that the theorem of the average value can be applied. We must check if the equation is continuous in [0,1] and derivable in (0,1).

Continuity:

The function is continuous in all R, being a polynomial function, so it will also be continuous in the interval [0,1].

Derivability:

The function is derivable in (0,1) if its derivative is continuous in that interval.

The derivative of the function is: That is continuous in all R being a polynomial function, therefore f(x) is derivable.

It is continuous in [0,1] and derivable in (0,1), therefore, there is a value of c in that interval such that: Let’s go on to calculate point c of the theorem.

We calculate what the function is worth at the extremes of the interval:  And we calculate f'(c): On the other hand, we calculate f'(c) from f'(x): Substituting x for c: We match both results of f'(c) and we are left with an equation that depends on c and where we can clear it and find the value of c they are asking for: ### Example 2

Calculates the c point that satisfies the average value theorem for the next function in the interval [0.4]: We have to check that the function is continuous and devirable in that interval. We have a critical point at point x=2, so we are going to study continuity and derivability at that point (both sections are continuous and derivable because they are polynomials).

Continuity:

To see if the function is continuous in x=2, we have to check that its lateral limits and the value of the function in x=2 coincide.

The limit on the left of x=2 is: The limit on the right: And the value of the function: The lateral limits and the value of the function in x=2 coincide: So the function is continuous at x=2

Now let’s see if the function is derivable in x=2

For this, we obtain the derivative of f(x): And now we check if f'(x) is continuous at x=2.

The limit on the left is: The limit on the right: And the value of f'(x) in x=2 is: The lateral limits and the value of f'(x) coincide: Therefore f'(x) is continuous for x=2 and f(x) is derivable for x=2.

They fulfill the two obligatory conditions, then the theorem of the average value can be applied and there will be a point c in the interval [0,4] such that: We calculate the value of the function at the extremes:  And we calculate the value of f'(c): On the other hand, we get f'(c), from f'(x), replacing x with c: In the first stretch we get no value of c, but in the second stretch, it depends on c, which we equal to the value of f'(c) previously calculated and we get what c is worth: ### Example 3

Find a and b for f(x) to meet the conditions of the average value theorem in [0,2] and calculate the c point that satisfies the average value theorem for that interval: For the conditions of the average value theorem to be met, the function must be continuous and derivable at point x=1.

For it to be continuous, the lateral limits of x=1 must coincide.

The limit on the left is: And the limit on the right is: As they have to be equal, we match both results, obtaining the following equation For the function to be derivable, its derivative must be continuous. Therefore, we obtain the derivative of the function: And we check its continuity in x=1. For it the lateral limits must coincide:  Matching both results, we obtain directly the value of b: This value of b, we substitute it in the equation obtained previously: From where we clear the value of a: Once the values of a and b have been calculated, we substitute them in the function And we are going to calculate point c of the theorem of the average value.

We already know that the function is continuous and derivable, so we now calculate the value of the function at the extremes of the interval:  And now we calculate the value of f'(c): On the other hand, we obtain f'(c), from f'(x), replacing x with c: The first section depends on c, so we equal it to the value of f'(c) that we have obtained before and we clear the value of c: 