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1. The Mean Value Theorem

The Mean Value Theorem states the following:

• Given that f(x) is a continuous and differentiable function, there is a guaranteed point where the slope of the tangent line at point c, is equal to the slope of the secant line passing through the two end points of an interval. If the interval is [a,b], the secant line passes through the points a & b.
• The secant line and tangent line are parallel, for lines with the same slope are always parallel. They will never intersect!

For example...

The graph of f(x)= sin(x) + 2 on the interval [0,3], proves the Mean Value Theorem. The function is continuous and differentiable on the given interval, a requirement of the Mean Value Theorem. Also, the slope of the function's tangent line (y=3) is parallel to its secant line (y=2). Both of these lines have a slope of 0. The tangent line touches the function sin(x) at point c (3). The secant line passes through points x=0 (point a) and x=3 (point b), the two endpoints on the interval (0,3) [a,b0].







2. Why does the Mean Value Theorem ONLY work for continuous and differentiable functions ?

Okay, let's get down to the basics.

A continuous function is defined as a function for which small changes in the input result in small changes in the output.
Continuous functions have the following properties:

• For all the points on a continuous graph, the limit from the right equals the limit from the left. Therefore, the limit at each point exists.
• [The limit as x approaches a number is the actual output (the y value) at that x].
  

• Also, on a closed interval, every continuous has an absolute maximum and absolute minimum.
• A function may be continuous, but have a point where it is not differentiable [a point where the slope is undefined due to a corner, cusp, jump discontinuity, oscillating discontinuity, or infinite discontinuity (asymptote)].

A function f(x) is differentiable at a point if its derivative exists at that point.
Differentiable functions have the following qualities:

• The slope must be defined at a particular point in the function

• If f is differentiable at a point c, then f is continuous at that point c. (but NOT vice versa).

Continuous but not DIFFERENTIABLE:
Consider the function f(x) = x^(7/9) + 1 on the interval [-1,1]. It is continuous at every point in this interval and also differentiable at every point except x=0, where there is a cusp. At x=0, the slope is undefined. This function fails the Mean Value Theorem first and foremost because there is no tangent line parallel to the secant line y=2. Technically, the line y=1 does not exist, since as i said before, x=0 is undifferentiable. However, if the slope at x=0 existed, it would be 0 and be parallel to the secant line x=2.
This is why a graph must be differentiable and continuous to have a guaranteed point c.









Differentiable but not CONTINUOUS:

Consider the graph of f(x)=(sin(x)/x on the interval [-2,2]. This graph is differentiable at every point except x=0, where it is also discontinuous. This graph fails the Mean Value Theorem because it does not have a guaranteed point c. Point c in this case would the tangent line at y=0, but since the function is neither differentiable nor continuous there, the "tangent line" does not exist. Only the secant line that goes through both -2 and 2 exists.

A very mean value theorem !

The Mean Value Theorem !

The Mean Value Theorem relates the slope of a secant line to the slope of a tangent line. According to the Mean Value Theorem, if a function, f, is continuous on and differentiable on , there is a guaranteed point c in in which

In other words, there is a guaranteed point on the interval [a,b] where the slope of the tangent line through point c (f'(c)) is equal to the average slope throughout the interval [a,b] [f(b) - f(a)]/(b - a). This also means that the secant line and tangent line are parallel, for lines with the same slope are always parallel.

This can be shown graphically by utilizing the function f(x)= x^2.

This function is both continuous and differentiable in the interval [0,2], as shown to the left.

So, to prove that the Mean Value Theorem applies to this function, we must find the slope of the secant line by using the part of the theorem that reads [f(b) - f(a)]/(b - a). Let's use the interval [0,2], where a=0 and b=2.

[f(2) - f(0)]/[2 - 0]

= (4-0)/2

= 2

This is shown to the left, where the green line represents the secant line of the function f(x)=x^2.

The slope of the secant line is 2. According to the Mean Value Theorem, we must be able to find a point between x=0 and x=2 where the slope is 2. To do so, we must find the slope of the function f(x).

Take the derivative of f(x)=x^2.

f'(x)=2x.

Now we have found f'(x) and [f(b) - f(a)]/(b - a), the slope of the tangent line and slope of the secant line, respectively. Now, in accordance to the Mean Value Theorem, we must set them equal to one another to find point c, where the slope of the tangent line is equal to that of the secant line.

2x=2

x=1.

This means that at x=1, the slope of the tangent line is equal to the slope of the secant line. At x=1, both the tangent line and secant line are parallel to one another. Since x=1 lies within the interval [0,2], this proves the Mean Value Theorem.

Now, we need to find the equation of the tangent line. We know the slope is 2 and x=1.

We need to find a point, so we can plug in 1 into the original function f(x)= x^2.

f(1)=1

so the point is (1,1)

use y-y1=m(x-x1)

y-1=2(x-1)

y=2x-2+1

y=2x-1.

The equation of the tangent line is y=2x-1.

The Mean Value Theorem only applies to continuous, differentiable functions!

The Mean Value Theorem only works for differentiable and continuous functions. This is so because a discontinuous function or one that is not differentiable at one or more points does not guarantee a point c where the slope of the tangent line is equal to the slope of the tangent line. In fact, the place where that point c would have been may not be continuous due to an asymptote, cusp, jump, or corner. The point where that c would have been may also not be differentiable, meaning the slope of the tangent line does not exist. Sometimes, the point where the c would have been is neither continuous nor differentiable. For these reasons, a function must be continuous AND differentiable in order for the Mean Value Theorem to apply.

An example of a function in which there is no GUARANTEED point c is f(x)= [sin(x)]/x.

The table in the lower right hand corner suggests the function is not differentiable at x=0.

[sin(x)]/x FAILS the Mean Value Theorem because it is discontinuous at the point x=0.

Another good counterexample to the Mean Value Theorem would be f(x)=| x |.

The Mean Value Theorem FAILS because f(x)=| x | is not differentiable at x=0, where there is a corner.