The Mean Value Theorem
Theorem 1 (The Mean Value Theorem) Let a, b ∈ IR with a < b. If f : [a, b] → IR
is continuous on [a, b] and differentiable on (a, b), then there is a c ∈ (a, b) such that
f (b) = f (a) + f ′ (c)(b − a)
(∗)
Proof: Define L to be chord joining a, f (a) and b, f (b) and F (x) to be the distance
(a)
from x, f (x) to L. Notice that c is such that (∗) is true if and only if f ′ (c) = f (b)−f
.
b−a
That is, if and only if the slope of the tangent line to y = f (x) at x = c is the same as
the slope of L, so that L and the tangent line are parallel. This happens when c is a local
minimum or maximum for F (x). See the figure on the left below.
(b,f (b))
(c,f (c))
θ
(b,f (b))
(x,f (x))
(x,f (x))
F (x)
β f (b)−f (a)
F (x)
θ
α
(a,f (a))
x−a
b−a
(a,f (a))
We now derive an analytic expression for F (x). Use the notation in the figure on the
b−a
6 0. By similar
=
right above. In particular, the angle θ obeys sin θ = √
2
2
(b−a) +(f (b)−f (a))
triangles
α
x−a
=
f (b)−f (a)
b−a
=⇒ α =
f (b)−f (a)
(x
b−a
− a)
Since α + β = f (x) − f (a),
h
F (x) = β sin θ = [f (x) − f (a) − α] sin θ = f (x) − f (a) −
f (b)−f (a)
(x
b−a
i
− a) sin θ
Actually, everything
i motivation. I could have just said
h up to this point has just been
f (b)−f (a)
“Consider F (x) = f (x) − f (a) − b−a (x − a) sin θ.”
The function F (x) is continuous on the compact set [a, b] and hence achieves its
maximum and minimum. Furthermore F (x) is differentiable with derivative
h
i
(a)
F ′ (x) = f ′ (x) − f (a) − f (b)−f
sin θ
b−a
c Joel Feldman.
2014. All rights reserved.
November 26, 2014
The Mean Value Theorem
(a)
as desired.
If F has a max or min at some c ∈ (a, b), then F ′ (c) = 0 and f ′ (c) = f (b)−f
b−a
In fact, I claim that F must have a max or min in (a, b). Because otherwise, both the
max and min must occur at a or b. As F (a) = F (b) = 0 this forces both the max and min
to be 0 and F (x) to be identically 0. In this case, every x ∈ (a, b) is a max and a min,
contradicting the hypothesis that F has no max or min in (a, b).
Theorem 2 (Generalized Mean–Value Theorem) Let the functions F (x) and G(x)
both be defined and continuous on a ≤ x ≤ b and both be differentiable on a < x < b.
Then, there is a number c obeying a < c < b such that
[F (b) − F (a)]G′ (c) = [G(b) − G(a)]F ′ (c)
Proof:
Define
h(x) = F (b) − F (a) G(x) − G(b) − G(a) F (x)
First observe that h(x) is continuous on [a, b] and is also differentiable on (a, b) and has
derivative
h′ (x) = F (b) − F (a) G′ (x) − G(b) − G(a) F ′ (x)
Our goal is to find a c ∈ (a, b) with h′ (c) = 0. To do so, it suffices to find a local minimum
or maximum of h(x) somewhere in (a, b). Because h(x) is a continuous function on the
compact set [a, b], it achieves its maximum and minimum values somewhere in [a, b], but
we have to show that at least one of them is in (a, b). Now
h(a) = h(b) = F (b)G(a) − F (a)G(b)
Either
◦ h(x) is a constant, in which case h′ (c) = 0 for all a < c < b, or
◦ h(x) is not a constant, in which case F (b)G(a)−F (a)G(b) cannot be both its minimum
and maximum values. So at least one of the minimum and maximum has to be
achieved somewhere in (a, b).
Corollary 3 (Consequences of the Mean–Value Theorem) Assume that the function
f : (a, b) → IR is differentiable on (a, b).
(a) If f ′ (x) ≥ 0 for all a < x < b, then f (x) is monotonically increasing.
(b) If f ′ (x) > 0 for all a < x < b, then f (x) is strictly monotonically increasing.
(c) If f ′ (x) = 0 for all a < x < b, then f (x) is constant.
(d) If f ′ (x) ≤ 0 for all a < x < b, then f (x) is monotonically decreasing.
(e) If f ′ (x) < 0 for all a < x < b, then f (x) is strictly monotonically decreasing.
(f ) If x, y ∈ (a, b) and f ′ (x) < λ < f ′ (y) then there is a c ∈ (a, b) such that f ′ (c) = λ.
c Joel Feldman.
2014. All rights reserved.
November 26, 2014
The Mean Value Theorem
Proof: (a)–(e) all follow immediately from the observation that if a < x < y < b, there
is a x < c < y such that f (y) − f (x) = f ′ (c) (y − x).
(f) Define g(t) = f (t) − λt. Suppose that x < y. (If y < x, just exchange x and y in what
follows.) We wish to find a c ∈ (x, y) with g ′ (c) = 0. Now
◦ Since g ′ (x) = f ′ (x) − λ < 0, there must exist a t1 > x with g(t1 ) < g(x). (Otherwise
as t1 tends to x would give g ′ (x) ≥ 0.) Thus t = x is not
taking the limit of g(t1t1)−g(x)
−x
a minimum for g(t).
◦ Since g ′ (y) = f ′ (y) − λ > 0, there must exist a t2 < y with g(t2 ) < g(y). (Otherwise
taking the limit of g(t2t2)−g(y)
as t2 tends to y would give g ′ (y) ≤ 0.) Thus t = y is not
−y
a minimum for g(t).
This tells us that g(t) must have a local minimum at some c ∈ (x, y) so that g ′ (c) = 0
Remark 4
(a) Corollary 3.f is an Intermediate Value Theorem for derivatives.
(b) Beware. As a consequence of Corollary 3.f, we have that there is no function f (x) that
is differentiable on (−1, 1) with
f ′ (x) =
n
1
0
if x > 0
if x < 0
(c) Beware. Corollary 3.f does not imply that the derivatives of differentiable functions
have to be continuous. For example, the function
f (x) =
x2 sin x12
0
if x 6= 0
if x = 0
is differentiable at every x ∈ IR with
′
f (x) =
− x2 cos x12 + 2x sin x12
0
if x 6= 0
if x = 0
But f ′ (x) is wildly discontinuous at x = 0.
c Joel Feldman.
2014. All rights reserved.
November 26, 2014
The Mean Value Theorem