The Mean Value Theorem

If a function is continues and differentiable within the closed intervals [a, b] then:
f'(c)= [f(b)- f(a)]/ b-a
IN other words this mean that there is a point between the closed intervals of which the instantaneous slope is equal to the average slope of the closed intervals.
For Example:


f(x)=X^2, [1, 4] (its both differentiable and continues between the closed intervals)
the average slope OR in other word the secant line is 4x
To find the secant line you have to use the formula [f(b)- f(a)]/ b-a it give you the slope which is 4. Then you use the point-slope form y-y1=m(x-x1)using either one of the close interval points (1,1) or (4, 16).

According to the mean value theorem there is a point between the close intervals [1,4] of which the instantaneous slope or in other word the tangent line is parallel to the secant line.
Now you use the slope you got and set it equal to f'(c)
f'(x)= 2x
4=2x
x=2
Now you got to find the tangent line at x=2
you got the slope m=4 (parallel lines have the same slope) and the point (2,4)
use the point-slope form to find the tangent line
y=4x-4

As you can see the secant line of the closed intervals and the tangent line of which is parallel is within the closed intervals making the Mean Value Theorem TRUE

The Theorem FAILS when the function between the interval is not differentiable or continuous
Example (not differentiable):
f(x)= abs (x-6)+3 , [2, 10]
the secant line is y=7
the tangent line that is parallel is when x=3 but at that point the function is not differentiable therefore contradicting the theorum (f'(c)= [f(b)- f(a)]/ b-a)






Example (when discontinuous):
f(X)= tan (x) [0, 3.14]
The secant line is y=0
the tangent line that is parallel to the closed intervals does not exist because sec^2(X)can't equal 0