Calculus 1 Lecture Notes
Section 2.9
Page 1 of 6
Section 2.9: The Mean Value Theorem
Big Idea: For any differentiable function on a closed interval, there will always be a point in the
interval where the derivative of the function equals the slope of the line between the endpoints. This
result will be critical for proving the relationship between differentiation and integration.
Big Skill: You should be able to verify the conclusion of the Mean Value Theorem for given functions,
to prove results about the number of zeros of a function, and to find antiderivatives of simple
functions.
In section 1.3, the Intermediate Value Theorem told us that for a continuous function, there
would always be at least one x value between the endpoints of an interval at which the value of the
function would take on any y value between the y values of the endpoints. In this section, the Mean
Value Theorem tells us that for a differentiable function, there will always be at least one point in an
interval at which the derivative is equal to the slope between the endpoints.
Intermediate Value Theorem
Mean Value Theorem
Graph of y = f(x) = 2x3 – 2x
Graph of y = f(x) = 2x3 – 2x
Since f(0.5) = -0.75 and f(1.1) = +0.462, and
-0.75 < 0 < 0.462, there must be at least one x
value in the interval x  [0.5, 1.1] such that
f(x) = 0. It turns out that f(x) = 0 when x = 1.
Since the slope of the line between the points
(0.5, -0.75) and (1.1, 0.462) is m = 2.02, there
must be at least one x value in the interval
x  [0.5, 1.1] such that f (x) = 2.02 (i.e., the
derivative is equal to the slope of the line through
the endpoints). It turns out that
2.01
f (x) = 2.02 when x 
 0.819 .
3
Calculus 1 Lecture Notes
Section 2.9
Page 2 of 6
To prove the Mean Value Theorem, we need a “helper” theorem:
Theorem 9.1 (Rolle’s Theorem): (The “What goes up must come down” Theorem)
If f(x) is continuous on the interval [a, b], differentiable on the interval (a, b), and f(a) = f(b), then there
is a number c  (a, b) such that f  (c) = 0.
Picture:
Proof:
1) Case #1: f(x) is constant on [a, b].
Then f  (c) = 0, because the derivative of a constant is zero.
2) Case #2: f(x) is not constant on [a, b].
It must be the case then that f(x) increases to the right of a,
or that f(x) decreases to the right of a.
a) Subcase #2a: f(x) increases to the right of a.
There must be a point at which the f(x) begins to decrease so that it may return to its original
value.
Call the x value of this point x = c.
Since f  (c) exists, it must be the case that f  (c) > 0, f  (c) = 0, or f  (c) < 0.
i) Subcase #2ai: Show that it is not the case that f  (c) > 0 using a proof by contradiction.
f  x  f c
 0.
Assuming f  (c) > 0, we have f   c   lim
x c
xc
f  x  f c
0.
So, for every x sufficiently close to c, (think delta-epsilon),
xc
If x > c, then x – c > 0, and the above inequality then implies that f(x) – f(c) > 0, or f(x) >
f(c).
Calculus 1 Lecture Notes
Section 2.9
Page 3 of 6
This is a contradiction of the fact that the function has started to decrease to the right of c,
and thus it can not be true that f  (c) > 0.
ii) Subcase #2aii: Show that it is not the case that f  (c) < 0 using a proof by contradiction…
Therefore, f  (c) = 0 at some x = c for the case when f(x) increases to the right of a.
b) Subcase #2b: f(x) decreases to the right of a.
There must be a point at which the f(x) begins to increase so that it may return to its original
value.
Call the x value of this point x = c.
Since f  (c) exists, it must be the case that f  (c) > 0, f  (c) = 0, or f  (c) < 0.
i) Subcase #2bi: Show that it is not the case that f  (c) > 0 using a proof by contradiction…
ii) Subcase #2bii: Show that it is not the case that f  (c) < 0 using a proof by contradiction.
Therefore, f  (c) = 0 at some x = c for the case when f(x) decreases to the right of a.
 there is a number c  (a, b) such that f  (c) = 0.
Practice:
Find a value of c satisfying the conclusion of Rolle’s Theorem for f(x) = x3 – 3x2 + 2x + 2 on the
interval [0, 1]. Note: finding a c value is not the point of Rolle’s Theorem; Rolle’s Theorem is
important because it can be used in the proof of the Mean Value Theorem.
y = x^3
- 3x^2
+ 2xy + 2







x

Calculus 1 Lecture Notes
Section 2.9
Page 4 of 6
Theorem 9.2 (Corollary #1 to Rolle’s Theorem):
If f(x) is continuous on the interval [a, b], differentiable on the interval (a, b), and f(x) = 0 has two
solutions in [a, b], then f  (x) = 0 has at least one solution in (a, b).
Picture:
Theorem 9.3 (Corollary #2 to Rolle’s Theorem):
For any integer n > 0, if f(x) is continuous on the interval [a, b], differentiable on the interval (a, b),
and f(x) = 0 has n solutions in [a, b], then f ‘ (x) = 0 has at least n-1 solutions in (a, b).
Picture:
Practice:
1. Determine how many zeros the function 2x3 – 3x2 + 4x – 5 has.
2. Prove that a quadratic function has at most two zeros.
Calculus 1 Lecture Notes
Section 2.9
Page 5 of 6
Theorem 9.4 (Mean Value Theorem):
If f(x) is continuous on the interval [a, b] and differentiable on the interval (a, b), then there is a
number c  (a, b) such that
f (b)  f (a )
f ' (c ) 
ba
Picture:
Proof:
The proof essentially uses the idea that you can perform a linear transformation of skewing the graph
until f(a) and f(b) lie on a horizontal line, and thus Rolle’s Theorem applies.
Define a new function g(x) that “tilts” the function f(x) until the endpoints of f on the interval are
horizontal:
g  x   f  x    m  x  a   f  a   , where y  m  x  a   f  a  is the equation of the secant line
f (b)  f (a )
). Notice that now
ba
g b   f b    m b  a   f  a 
between  a, f  a   and  b, f  b   (which means that the slope m 
g  a   f  a    m  a  a   f  a 
 f b   f  a 

 f b  
b  a   f  a  
ba


0
0
Thus, by Rolle’s Theorem, there exists x = c such that g(c) = 0.
d
d
g  x    f  x    m  x  a   f  a   
dx
dx

g  x  f  x  m
 f a  0  f a
and
g c  f  c  m  0
f c  m
f c 
f b  f  a 
ba
Calculus 1 Lecture Notes
Section 2.9
Page 6 of 6
Practice:
Find a value of c satisfying the conclusion of the Mean Value Theorem f(x) = x3 – x2 – x +1 on the
interval [0, 2].
Theorem 9.5 (Corollary #1 to the Mean Value Theorem):
If f  (x) = 0 for all x in some open interval I, then f (x) is constant on I.
Picture:
Theorem 9.6 (Corollary #2 to the Mean Value Theorem):
If f  (x) = g  (x) for all x in some open interval I, then for some constant c,
f (x) = g(x) + c for all x  I.
Picture:
Practice:
1. Find all functions that have a derivative equal to 3x2 + 1.
2. Find all functions that have a derivative equal to x4 + 3x.