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.