Calculus/Mean Value Theorem

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Mean Value Theorem
Mean Value Theorem

If  f(x) \ is continuous on the closed interval  [a, b] \ and differentiable on the open interval  (a,b) \ , there exists a number,  c \ , in the open interval  (a,b) \ such that

 f'(c) = \frac{f(b) - f(a)}{b - a} .


Lagrange mean value theorem.svg

What does this mean? As usual, let us utilize an example to grasp the concept. Visualize (or graph) the function  f(x) = x^3. 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  f at point c is equal to the slope of the line you drew.


1: Using the definition of the mean value theorem

f(b) - f(a) \over b - a

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

 \frac{f(2) - f(0)}{2 - 0} = \frac{8}{2} = 4

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.

\frac{dy}{dx} = 3x^2

Now, we know that the slope of the point is 4. So, the derivative at this point c is 4. Thus,  4 = 3x^2 . The square root of 4/3 is the point.

Example 2: Find the point that satisifes the mean value theorem on the function f(x) = \sin(x) and the interval [0,\pi].


1: Always start with the definition:

f(b) - f(a) \over b - a


\frac{\sin(\pi) - \sin(0)}{\pi - 0} = 0

(Remember,  \sin(\pi) and  \sin(0) 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!

 \frac{d\sin(x)}{dx} = \cos(x) = 0

The cosine function is 0 at  \pi/2 + \pi n (where n is an integer). Remember, we are bound by the interval [0, \pi], so  \pi/2 is the point c that satisfies the Mean Value Theorem.


Assume a function  y = f(x) that is differentiable in the open interval (a,b) that contains x.  \Delta y =  dy \over dx  \cdot \Delta x

The "Differential of x" is the  \Delta x . This is an approximate change in x and can be considered "equivalent" to  dx . 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  4.1^2 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,  \Delta x,  \Delta y , and  dy \over dx. 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  \delta x is .1 (This is the "change" in x from the approximation point to the point you chose.)

3: Take the derivative of your function.

 dy \over dx  = 2x. Now, "split" this up (This is not really what is happening, but to keep things simple, assume you are "multiplying"  dx over.)

3b. Now you have  dy = 2x \cdot dx . We are assuming  dy and  dx are approximately the same as the change in x, thus we can use  \Delta x and y.

3c. Insert values:  dy  = 2 \cdot 4 \cdot .1 . Thus,  dy = .8 .

4: To find  F(4.1), take  F(4) + dy 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):

 f'(x) = \frac{f(x+h)-f(x)}{(x+h)-x}

If h approaches 0, then:

 f'(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h}

Cauchy's Mean Value Theorem[edit]

Cauchy's Mean Value Theorem

If  f(x) \ ,  g(x) \ are continuous on the closed interval  [a, b] \ and differentiable on the open interval  (a,b) \ ,  g(a) \ne g(b) \ and  g'(x) \ne 0 \ , then there exists a number,  c \ , in the open interval  (a,b) \ such that

 \frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{g(b) - g(a)} .