Differential Calculus 201-NYA-05
Vincent Carrier
Mean Value Theorem
The Mean Value Theorem is one of the most important theorems of calculus. It has farreaching applications, most notably in the proof of the Fundamental Theorem of Calculus.
It is a generalization of the following theorem, discovered by Michel Rolle (1652-1719).
Rolle’s Theorem: Let f be a function satisfying the following three conditions:
1) continuous on [a, b];
2) differentiable on (a, b);
3) f (a) = f (b).
Then there is a c ∈ (a, b) such that f 0 (c) = 0.
y 6
r
-
0
a
c
b
x
Proof:
Let us consider two cases:
1) The function is constant on [a, b]. Then
f 0 (x) = 0
for all x ∈ (a, b)
so c is any number in (a, b).
2) The function is not constant on [a, b]. Then, by the Extreme Value Theorem, it has
an absolute maximum or an absolute minimum in (a, b) at x = c (say). Since the
absolute maximum or minimum is also a local maximum or minimum, it follows by
a theorem seen previously that f 0 (c) = 0.
Example: Consider the function f (x) = x3 + 2x2 − x − 1 on [−1, 1]. Show that Rolle’s
theorem applies and find all values of c ∈ (−1, 1) that satisfy f 0 (c) = 0.
It is clear that f is continuous on [−1, 1] and differentiable on (−1, 1). Furthermore,
f (−1) = −1 + 2 + 1 − 1 = 1
f (1) = 1 + 2 − 1 − 1 = 1.
and
Thus, Rolle’s Theorem applies. The equation
f 0 (x) = 3x2 + 4x − 1 = 0
has two solutions, namely
√
√
√
7
−4 ± 2 7
2
−4 ± 28
=
= − ±
6
6
3
3
of which only the first one belongs to the interval [−1, 1]. Thus,
√
2
7
c = − +
.
3
3
The Mean Value Theorem is a generalization of Rolle’s Theorem.
Mean Value Theorem: Let f be a function satisfying the following two conditions:
1) continuous on [a, b];
2) differentiable on (a, b).
Then there is a c ∈ (a, b) such that
f 0 (c) =
f (b) − f (a)
.
b−a
The expression on the right represents the slope of the dotted line below.
y 6
f (b)
r
f (a)
0
-
a
c
b
x
Proof:
Let
x−a
[f (b) − f (a)].
b−a
g(x) = f (x) −
The function g is continuous on [a, b] and differentiable on (a, b). Furthermore,
g(a) = f (a) −
a−a
[f (b) − f (a)] = f (a),
b−a
g(b) = f (b) −
b−a
[f (b) − f (a)] = f (a).
b−a
Thus, g satisfies the three conditions of Rolle’s Theorem. It means that there is a c ∈ (a, b)
such that
f (b) − f (a)
g 0 (c) = f 0 (c) −
= 0
b−a
which amounts to
f 0 (c) =
f (b) − f (a)
.
b−a
Examples: Say why the Mean Value Theorem applies and find the value(s) c whose
existence is predicted by the theorem.
a) f (x) = 3e2x
[a, b] = [−1, 1].
The function f (x) = 3e2x is
1) continuous on [−1, 1];
2) differentiable on (−1, 1).
f 0 (x) = 6e2x =
6e2x =
e2x =
3(e2 − e−2 )
3(e4 − 1)
3e2 − 3e−2
=
=
.
1 − (−1)
2
2e2
3(e4 − 1)
2e2
e4 − 1
4e2
2x = ln
x =
Thus,
c=
1
e4 − 1
ln
.
2
4e2
e4 − 1
4e2
1
e4 − 1
ln
.
2
4e2
b) f (x) = arcsin x
[a, b] = [1/2, 1].
The function f (x) = arcsin x is
1) continuous on [1/2, 1];
2) differentiable on (1/2, 1).
1
arcsin 1 − arcsin(1/2)
π/2 − π/6
π/3
2π
f 0 (x) = √
=
=
=
=
.
1 − 1/2
1/2
1/2
3
1 − x2
√
1 − x2 =
1 − x2 =
3
2π
r
x = ±
9
4π 2
√
= ±
2
x2 =
4π − 9
4π 2
Thus,
√
c=
c) f (x) = arctan x
4π 2 − 9
4π 2
4π 2 − 9
.
2π
4π 2 − 9
.
2π
√
[a, b] = [1, 3].
The function f (x) = arctan x is
√
1) continuous on [1, 3];
√
2) differentiable on (1, 3).
√
1
arctan 3 − arctan 1
π/3 − π/4
π
√
√
√
f (x) =
=
=
=
.
1 + x2
3−1
3−1
12( 3 − 1)
0
1+x
2
x2
√
12( 3 − 1)
=
π
√
12( 3 − 1) − π
=
π
Thus,
s
c=
s
x = ±
√
12( 3 − 1) − π
.
π
√
12( 3 − 1) − π
.
π