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.