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].