Mean Value Theorem Edits :&

Well I only put the corrections here, and the rest of the post is still on the original! (:

1. ----------------------
f'(c)= [f(b)-f(a)]/(b-a)

• This means that the slope of the secant line at 2 points A and B, on a continuous and differentiable function between the intervals [a,b], is guaranteed to be equal to the slope of the tangent line at one point, c. On the graph, c may be the midpoint of [a,b].
  
• Because the slopes of the tangent and secant lines are equal, the lines will be PARALLEL to each other.


• Lets try another function that ISN'T f(x)=x^2, because Ms. Hwang doesn't like it's simplicity (:


• So this time, how about... f(x)=e^x

As you can see, the slope of the tangent line at c=0 is equal to the slope of the secant line between the interval (-1,1), so the lines are PARALLEL.

**I don't like how that one turned out, since the secant line looks weird...,so I shall use another example.. f(x)= -sin(x+pi)-1 between the interval (pi, 2pi).

The graph shows that between the interval (pi, 2pi), there is a tangent line at point c=3pi/2 with the equation y=-2 and a secant line, y=-1. The function, being both differentiable and continuous, allows the slope of the tangent line to be equal to the slope of the secant line, therefore allowing the Mean Value Theorem to apply. Both slopes are equal to 0.

2. -----------------------

It only works for continuous and differentiable functions because if between the interval [a,b] the function is discontinuous(because of a jump or asymptote,) or not differentiable (because of a jump, asymptote, cusp,corner...) then there may not be a tangent line at point c (or there may not even be a point c on the function if it is discontinuous where point c should be) that is parallel to the secant line going through points A and B.

• Since I had an equation involving abs of x, (f(x)=abs (x +1)+3), and im not sure whether I have to re-do this one, I will anyway.


• So... ANOTHER example of a CONTINUOUS but NOT DIFFERENTIABLE function is f(x)=x^(2/3)+1 on (-1,1).

There should be a tangent line (y=1) parallel to the secant line (y=2) at point c=0, but it is not possible for such a tangent line to exist due to the fact that the function is not differentiable at point c.(the slope of the line before c=-1 and the slope of the line after c=-1 are not equal) despite the continuity. Whenever there is a cusp at point c, the mean value theorem does not apply because f ' (c) does not exist.

Ohhhh AND to answer your question ms Hwang, related to my LAST post, the secant line of the second equation is y=1

A Very Mean Value Theorem! xD

f'(c)= [f(b)-f(a)]/(b-a)

1. Explain what this means GRAPHICALLY by showing a good example!

• This means that the slope of the secant line at 2 points A and B, on a continuous and differentiable function between the intervals [a,b], is guaranteed to be equal to the slope of the tangent line at one point, c. On the graph, c may be the midpoint of [a,b].
  
• Because the slopes of the tangent and secant lines are equal, the lines will be PARALLEL to each other.
  
• This is proven through the graph of the function f(x)=x^2

The secant line (green line) goes through the points (0,0) and (2,4), where a=0 and b=2. It's slope should be equal to at least one slope of a tangent line(blue line) at a point c. In our case c=1, which is the midpoint between a=0 and b=2.

slope of secant line=[f(b)-f(a)]/(b-a)
=[f(2)-f(0)]/(2-0) ----use(f(x)=x^2)
=(4-0)/2
=4/2
=2

You know that c=1 because

f'(c)= 2 (2 is the slope of the secant line)
2c=2 (slope of tangent is 2x[or 2c],found by taking derivative of f(x)= x^2)
c= 1

f'(c)=2c
f'(1)=2(1)
f'(1)=2 (Slope of tangent line)

2=2
Slope of tangent line=slope of secant line (PARALLEL)


2. Explain why this only works for continuous and differentiable functions.

• It only works for continuous and differentiable functions because if between the interval [a,b] the function is discontinuous(because of a jump or asymptote,) or not differentiable (because of a jump, asymptote, cusp,corner...) then there may not be a tangent line at point c (or there may not even be a point c on the function if it is discontinuous where point c should be) that is parallel to the secant line going through points A and B.
  
• An example of a discontinuous function that fails the Mean Value Theorem is 1/abs(x).
  
  
  
  A point c does not exist between the interval [-2,2] where f'(c)=[f(b)-f(a)]/b-a.
  
  (NOTE: A function may be discontinuous between an interval [a,b] and STILL have a tangent line at a point c that is equal to [f(b)-f(a)]/b-a]!!! Because of the discontinuity, there is just not a GUARANTEED point c, since the Mean Value Theorem fails).
  
• An example of a continuous but not differentiable function that fails the Mean Value Theorem is abs(x+1)+3 on (-2,0).
  
  

There should be a tangent line parallel to the secant line (green line y=4) at point c=-1, as shown in the picture (blue line), but it is not possible for the tangent line to actually exist because the function is not differentiable at the point c (the slope of the line before c=-1 and the slope of the line after c=-1 are not equal)despite the continuity.