The Mean Value Theorem

mean value theorem, derivative, interval, f prime

If f(x) is a continuous differentiable function on the interval [a, b], we have a "c" in [a, b] such that f'(c)=f(b)-f(a)/(b-a).

This is fairly intuitive.  Since the derivative on the interval [a, b] is the instantaneous slope at each point between a, and b, the mean value theorem is saying that the average slope in [a,b] has to equal one of the instantaneous slopes where "c" denotes one of these slopes.

On a graph, the direct straight line that we get by connecting the value of "f" at "a" and its value at "b" represents the average slope and this is what is represented by f(b)-f(a)/(b-a).  The same old slope that we know from doing linear functions.  

Now, f'(c) is the slope at one of the points in the interval [a,b] and surely, this slope for one point "c" has to equal the average slope.  


Example.  Suppose we have the function x² in the interval [1,3], then, we are saying that for some number "c" between 1 and 3, the derivative has to equal the average slope.

We find the average slope as follows:

f(3)-f(1)
----------
  3-1

which is equivalent to

3²-1²
-----
  2

which is (9-1)/2=(8/2)=4.   So for some "c" in [1,3], the derivative must equal 4.  

Now, we know the derivative through the shortcut method is 2x for x².  So, f'(x)=2x and so, f'(c)=2c, and we are saying that 2c=4 for some c in [1,3].  Solving for C in in 2c=4, dividing both sides by 2 tells us that c=2.  Indeed, f'(2)=2(2)=4 and as desired, 2 is between 1 and 3.  This is very logical and this is a profound statement that we call the Mean value Theorem.

mean value theorem, derivative, interval, f prime