**The Mean Value Theorem (MVT)**

If f is a continuous function on [a, b] and is differentiable on (a, b), then there exists

at least one number c in the interval (a, b) such that

'

=

__Note__: This says that the slope of the tangent line to f at c equals the slope of the

secant line through the points (a, f(a)) and (b, f(b)).

__Simple example__: Suppose that we have a profit function P(x) which is differentiable

on the interval (0, 50) and continuous on the interval [0, 50]. We also know that

P(2) = 600 and P(10) = 600. What does the MVT tell us about the function P(x)?

We let a = 2 and b = 10. The MVT then says that there is at least one number c,

with 2 < c < 10, such that

'

=

'

Thus

=

=

=

In other words, our function P(x) has at least one critical point between 2 and 10

(a place where P might have a relative maximum or relative minimum).

__A more interesting example__: We can use the MVT to prove the following useful

theorem:

If g is continuous on [a, b] and differentiable on (a, b), while g'(x) > 0 for all x in

(a, b), then g is increasing on [a, b].

__Proof__: Choose any numbers

and

so that

<=

<

<=

Since these 2 numbers are arbitrary numbers in [a, b] with the first number to the

left of the second, to show that g is increasing it suffices to show that

<

Apply the MVT to function g(x) on the interval

The MVT says that there exists at least one number c with

<

<

'

and

=

'

Thus

=

'

By our hypothesis above,

>

and by the choice of our two numbers

>

Hence

>

which clearly gives us that

<

as desired.

Thus g is an increasing function on the interval [a, b].