# Calculus/Mean Value Theorem

**Mean Value Theorem**

If is continuous on the closed interval and differentiable on the open interval , there exists a number, , in the open interval such that

- .

## Examples[edit]

What does this mean? As usual, let us utilize an example to grasp the concept. Visualize (or graph) the function . Choose an interval (anything will work), but for the sake of simplicity, [0,2]. Draw a line going from point (0,0) to (2,8). Between the points *x* = 0 and *x* = 2 exists a number *x* = *c*, where the derivative of at point *c* is equal to the slope of the line you drew.

**Solution:**

1: Using the definition of the mean value theorem

insert values. Our chosen interval is [0,2]. So, we have

2: By the definition of the mean value theorem, we know that somewhere in the interval exists a point that has the same slope as that point. Thus, let us take the derivative to find this point *x* = *c*.

Now, we know that the slope of the point is 4. So, the derivative at this point *c* is 4. Thus, . The square root of 4/3 is the point.

**Example 2:** Find the point that satisifes the mean value theorem on the function and the interval [0,].

**Solution:**

1: Always start with the definition:

so,

(Remember, and are both 0.)

2: Now that we have the slope of the line, we must find the point *x* = *c* that has the same slope. We must now get the derivative!

The cosine function is 0 at (where *n* is an integer). Remember, we are bound by the interval [], so is the point *c* that satisfies the Mean Value Theorem.

## Differentials[edit]

Assume a function that is differentiable in the open interval (a,b) that contains x.

The "Differential of x" is the . This is an approximate change in x and can be considered "equivalent" to . The same holds true for y. What is this saying? One can approximate a change in y by knowing a change in x and a change in x at a point very nearby. Let us view an example.

Example: A schoolteacher has asked her students to discover what is. The students, bereft of their calculators, are too lazy to multiply this out by hand or in their head and desire to utilize calculus. How can they approximate this?

1: Set up a function that mimics the procedure. What are they doing? They are taking a number (Call it x) and they are squaring it to get a new number (call it y). Thus, y = x^2 Write yourself a small chart. Make notes of values for x, y, , , and . We are seeking what y really is, but we need the change in y first.

2: Choose a number close by that is easy to work with. Four is very close to 4.1, so write that down as x. Your is .1 (This is the "change" in x from the approximation point to the point you chose.)

3: Take the derivative of your function.

. Now, "split" this up (This is not really what is happening, but to keep things simple, assume you are "multiplying" over.)

3b. Now you have . We are assuming and are approximately the same as the change in x, thus we can use and y.

3c. Insert values: . Thus, .

4: To find , take to get an approximation. 16 + .8 = 16.8; This approximation is nearly exact (The real answer is 16.81. This is only one hundredth off!)

## Definition of Derivative[edit]

The exact value of the derivative at a point is the rate of change over an infinitely small distance, approaching zero. Therefore, if h approaches 0 and the function is f(x):

If h approaches 0, then:

## Cauchy's Mean Value Theorem[edit]

**Cauchy's Mean Value Theorem**

If , are continuous on the closed interval and differentiable on the open interval , and , then there exists a number, , in the open interval such that

- .