Calculus/Mean Value Theorem

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Mean Value Theorem
Mean Value Theorem

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


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   and   exists a number   , where the derivative of   at point   is equal to the slope of the line you drew.


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   .


Now, we know that the slope of the point is 4. So, the derivative at this point   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   .


1: Always start with the definition:




(Remember,   and   are both 0.)

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


The cosine function is 0 at   , where   is an integer. Remember, we are bound by the interval   , so   is the point   that satisfies the Mean Value Theorem.


Assume a function   that is differentiable in the open interval   that contains   .  

The "Differential of  " is the   . This is an approximate change in   and can be considered "equivalent" to   . The same holds true for   . What is this saying? One can approximate a change in   by knowing a change in   and a change in   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  ) and they are squaring it to get a new number (call it  ). Thus,   Write yourself a small chart. Make notes of values for   . We are seeking what   really is, but we need the change in   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   . Your   is .1 (This is the "change" in   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   are approximately the same as the change in   , thus we can use   and   .

3c. Insert values:   . Thus,   .

4: To find   , take   to get an approximation. 16 + 0.8 = 16.8; This approximation is nearly exact (The real answer is 16.81. This is only one hundredth off!)

Definition of DerivativeEdit

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


If h approaches 0, then:


Cauchy's Mean Value TheoremEdit

Cauchy's Mean Value Theorem

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