The Mean Value Theorem

Suppose that you drive 100 miles in 2 hours. Was there any time during your trip at which you were going 50 MPH?

Suppose the rate at which a well produces water varies between 2 and 10 gallons per minute. If you run the well for an hour, can conclude that the total water pumped is between 120 and 600 gallons?

If f '(x) > 0 at every point of an interval, is f(x) necessarily increasing on that interval?

The answers are "yes", "of course", and "yes" (this is the first derivative test), but ALL of these answers depend on a theorem called the Mean Value Theorem (MVT). Let's see what the Mean Value Theorem is about.

The Statement of the Mean Value Theorem

Consider a function which is continuous on a closed interval [a,b] and differentiable on the open interval (a,b).

If we connect the point (a, f(a)) to the point (b, f(b)), we produce a line-segment whose slope is the average rate of change of f(x) over the interval (a,b). The derivative of f(x) at any point c is the instantaneous rate of change of f(x) at c.

Two points c such that f '(c) is the slope of the line joining (a,f(a)) to (b,f(b))

The Mean Value Theorem says that there is a point c in (a,b) at which the function's instantaneous rate of change is the same as its average rate of change over the entire interval [a,b].

The Mean Value Theorem

Let f(x) be continuous on the closed interval [a,b] and differentiable on the open interval (a,b). Then there is a point c in the interval (a,b) where

	f(b) - f(a)
f '(c) =
	b - a

Example

y = cos(x)

Statement: There is a point in the interval [0, 3

/2] where the tangent to the curve y = cos(x) has slope -2/3

Discussion: cos(0) = 1 and cos(3/2) = 0. Hence by the MVT there is some point c in the interval [0, 3/2] where

	cos(3/2) - cos(0)		0 - 1
f '(c) =		=		= -2/3
	3/2 - 0		3/2 - 0

Comment: Since the derivative of cos(x) is continuous, we could try applying the Intermediate Value Theorem to the derivative. But D_xcos(x) = -sin(x), and -sin(0) = 0 while - sin(3/2) = 1, so the IVT allows us to conclude that there are points in the interval [0, 3/2] where the tangent to the curve y = cos(x) has slope between 0 and 1, but not that there is a point with slope -2/3.

Alternative Form

An alternative form of the MVT is this:

The Mean Value Theorem

Let f(x) be continuous on the closed interval [a,b] and differentiable on the open interval (a,b). Then there is a point c in the interval (a,b) such that

f(b) = f(a) + f '(c)(b - a)

This form comes from simplifying

	f(b) - f(a)
f '(c) =
	b - a

but has a different interpretation. It says there is a linear function L(x) = f(a) + m(x - a) which

passes through the point (a,f(a)),
has the same value as f(x) at x = b, and
has slope equal to the slope of the graph of y = f(x) at some point between a and b.

Green line is the graph of the linear function L(x).

Example

Statement: There is a number c between 0 and .5 such that sin(.5) = cos(c)(.5).

Discussion: The derivative of sin(x) is cos(x). At x = 0, the value of the derivative is 1. The value of sin(x) at x = 0 is 0. The MVT says there is point c between 0 and .5 such that

sin(.5) = 0 + cos(c)(.5 - 0) = cos(x)(.5).

Comment: In Stage 4 we saw that sin(x) is approximately equal to x for small x. The MVT implies that for any x > 0,

sin(x) = cos(c)^.x for some c with 0 < c < x.
If x is small, then c is near zero, and cos(c) is near 1, leading to the same approximation of sin(x) by x, but in a more controlled fashion.

The next page contains more sample consequences of the MVT.

/Stage7/Lesson/MVT.html

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