Mean Value Theorem
MATH 161 Calculus I
J. Robert Buchanan
Department of Mathematics
Spring 2015
Preview
I
The Mean Value Theorem (MVT) is of fundamental
importance in calculus.
Preview
I
The Mean Value Theorem (MVT) is of fundamental
importance in calculus.
I
From the MVT many important other concepts will be
derived.
Preview
I
The Mean Value Theorem (MVT) is of fundamental
importance in calculus.
I
From the MVT many important other concepts will be
derived.
I
We begin with an elementary version of the MVT called
Rolle’s Theorem.
Rolle’s Theorem
Theorem (Rolle’s Theorem)
Suppose f is continuous on interval [a, b] and differentiable on
interval (a, b) and f (a) = f (b). Then there is a number
c ∈ (a, b) such that f 0 (c) = 0.
y
f HaL= f HbL
a
c
b
x
Example
Suppose f (x) = 1 + sin x on the interval [−π, π]. Find the
number c referred to in Rolle’s Theorem.
Example
Suppose f (x) = 1 + sin x on the interval [−π, π]. Find the
number c referred to in Rolle’s Theorem.
We note that f (−π) = 1 = f (π) and that f (x) is continuous on
[−π, π] and differentiable on (−π, π). Thus according to Rolle’s
Theorem there exists −π < c < π for which
f 0 (c) = 0
cos c = 0
π
c =
.
2
Example
Suppose f (x) = 1 + sin x on the interval [−π, π]. Find the
number c referred to in Rolle’s Theorem.
We note that f (−π) = 1 = f (π) and that f (x) is continuous on
[−π, π] and differentiable on (−π, π). Thus according to Rolle’s
Theorem there exists −π < c < π for which
f 0 (c) = 0
cos c = 0
π
c =
.
2
The number c = −π/2 would also solve the equation.
Consequences of Rolle’s Theorem (1 of 2)
Theorem
If f is continuous on [a, b], differentiable on (a, b) and f (x) = 0
has two solutions in [a, b], then f 0 (x) = 0 has at least one
solution in (a, b).
Proof.
Let the two solutions of f (x) = 0 be x = s and x = t where
s < t. Apply Rolle’s Theorem on the interval [s, t].
Consequences of Rolle’s Theorem (2 of 2)
Theorem
For any integer n > 0, if f is continuous on [a, b], differentiable
on (a, b) and f (x) = 0 has n solutions in [a, b], then f 0 (x) = 0
has at least (n − 1) solutions in (a, b).
Proof.
Let the n solutions of f (x) = 0 be {x1 , x2 , . . . , xn }. Apply Rolle’s
Theorem on the intervals [xi , xi+1 ] for i = 1, 2, . . . , n − 1.
Example
Prove that x 3 + 4x − 1 = 0 has just one real number solution.
Example
Prove that x 3 + 4x − 1 = 0 has just one real number solution.
Hint: show that it has at least one solution, then suppose the
equation has two solutions and reach a contradiction.
Solution
I
Let f (x) = x 3 + 4x − 1. Since f is a polynomial, it is
continuous and differentiable at all real numbers.
Solution
I
Let f (x) = x 3 + 4x − 1. Since f is a polynomial, it is
continuous and differentiable at all real numbers.
I
Since f (0) = −1 < 0 < 4 = f (1) then according to the
Intermediate Value Theorem (IVT) there is at least one
solution to the equation f (x) = 0 in the interval (0, 1).
Solution
I
Let f (x) = x 3 + 4x − 1. Since f is a polynomial, it is
continuous and differentiable at all real numbers.
I
Since f (0) = −1 < 0 < 4 = f (1) then according to the
Intermediate Value Theorem (IVT) there is at least one
solution to the equation f (x) = 0 in the interval (0, 1).
I
Suppose there are two solutions, in other words there exist
numbers x1 < x2 such that f (x1 ) = 0 = f (x2 ). By Rolle’s
Theorem there exists a number x1 < c < x2 for which
f 0 (c) = 0
3c 2 + 4 = 0
Solution
I
Let f (x) = x 3 + 4x − 1. Since f is a polynomial, it is
continuous and differentiable at all real numbers.
I
Since f (0) = −1 < 0 < 4 = f (1) then according to the
Intermediate Value Theorem (IVT) there is at least one
solution to the equation f (x) = 0 in the interval (0, 1).
I
Suppose there are two solutions, in other words there exist
numbers x1 < x2 such that f (x1 ) = 0 = f (x2 ). By Rolle’s
Theorem there exists a number x1 < c < x2 for which
f 0 (c) = 0
3c 2 + 4 = 0
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Since there are no real number solutions to this equation,
then this contradicts the assumption that f (x) = 0 has two
different solutions. Therefore it can have only one solution.
Mean Value Theorem (1 of 2)
Theorem (Mean Value Theorem)
Suppose f is continuous on [a, b] and differentiable on (a, b).
There exists a number c ∈ (a, b) such that
f 0 (c) =
f (b) − f (a)
.
b−a
Remark: the MVT says the slope of the tangent line
somewhere in (a, b) equals the slope of the secant line across
[a, b].
Mean Value Theorem (1 of 2)
Theorem (Mean Value Theorem)
Suppose f is continuous on [a, b] and differentiable on (a, b).
There exists a number c ∈ (a, b) such that
f 0 (c) =
f (b) − f (a)
.
b−a
Remark: the MVT says the slope of the tangent line
somewhere in (a, b) equals the slope of the secant line across
[a, b].
Proof.
f (b) − f (a)
(x − a) + f (a)
b−a
(a)
Define s(x) = f (x) − f (b)−f
b−a (x − a) − f (a)
Secant line: y =
Mean Value Theorem (2 of 2)
s(a) = s(b)
=⇒
s0 (c) = 0
(by Rolle’s Theorem)
y
f HaL
f HbL
a
c
b
x
Example
Consider the function f (x) = x 3 + x 2 + 1 on the interval [−1, 1].
Find the value of c guaranteed by the Mean Value Theorem
such that
f 0 (c) =
f (b) − f (a)
.
b−a
Example
Consider the function f (x) = x 3 + x 2 + 1 on the interval [−1, 1].
Find the value of c guaranteed by the Mean Value Theorem
such that
f (b) − f (a)
.
b−a
f (1) − f (−1)
3c 2 + 2c =
1 − (−1)
3−1
3c 2 + 2c =
2
3c 2 + 2c = 1
f 0 (c) =
Example
Consider the function f (x) = x 3 + x 2 + 1 on the interval [−1, 1].
Find the value of c guaranteed by the Mean Value Theorem
such that
f (b) − f (a)
.
b−a
f (1) − f (−1)
3c 2 + 2c =
1 − (−1)
3−1
3c 2 + 2c =
2
3c 2 + 2c = 1
f 0 (c) =
3c 2 + 2c − 1 = 0
(3c − 1)(c + 1) = 0
1
c =
3
Illustration
f (x) = x 3 + x 2 + 1
y
3.0
2.5
2.0
1.5
1.0
0.5
-1.0
-0.5
0.5
1.0
x
Constants and the MVT
Recall: if c is a constant then
d
dx c
= 0.
Constants and the MVT
Recall: if c is a constant then
d
dx c
= 0.
Question: are there any non-constant, differentiable functions
such that f 0 (x) = 0 on an interval?
Constants and the MVT
Recall: if c is a constant then
d
dx c
= 0.
Question: are there any non-constant, differentiable functions
such that f 0 (x) = 0 on an interval?
Theorem
Suppose f 0 (x) = 0 for all x in an open interval I, then f (x) is
constant on I.
Proof.
f (s) − f (t) = f 0 (c)(s − t)
f (s) − f (t) = 0
f (x) = f (t)
Derivatives that Agree
Note: If f (x) = x 3 + 2x + 1 and g(x) = x 3 + 2x then
f 0 (x) = g 0 (x) for all x even though f (x) 6= g(x) for all x.
Derivatives that Agree
Note: If f (x) = x 3 + 2x + 1 and g(x) = x 3 + 2x then
f 0 (x) = g 0 (x) for all x even though f (x) 6= g(x) for all x.
Question: suppose two functions have the same derivative,
what is the relationship between the functions?
Derivatives that Agree
Note: If f (x) = x 3 + 2x + 1 and g(x) = x 3 + 2x then
f 0 (x) = g 0 (x) for all x even though f (x) 6= g(x) for all x.
Question: suppose two functions have the same derivative,
what is the relationship between the functions?
Corollary
Suppose f (x) and g(x) are functions such that f 0 (x) = g 0 (x) for
all x in some open interval I, then there exists a constant c
such that f (x) = g(x) + c for all x in I.
Example
Find all the functions f (x) such that f 0 (x) = 9x 4 + 3x + 1.
Example
Find all the functions f (x) such that f 0 (x) = 9x 4 + 3x + 1.
Remark: this process is known as antidifferentiation.
Example
Find all the functions f (x) such that f 0 (x) = 9x 4 + 3x + 1.
Remark: this process is known as antidifferentiation.
We need to raise each power of x by 1 and divide the
coefficient by the new power.
f (x) =
9
3
9
3
x 4+1 +
x 1+1 + x + C = x 5 + x 2 + x + C
4+1
1+1
5
2
where C is an arbitrary constant.
Inequalities and the MVT
Use the MVT to prove the inequality
| cos u − cos v | ≤ |u − v |.
Inequalities and the MVT
Use the MVT to prove the inequality
| cos u − cos v | ≤ |u − v |.
Let f (x) = cos x, then by the MVT
f (u) − f (v )
= f 0 (c) (c between u and v )
u−v
cos u − cos v
= − sin c
u−v
cos u − cos v = | − sin c| ≤ 1
u−v
| cos u − cos v |
≤ 1
|u − v |
| cos u − cos v | ≤ |u − v |.
Increasing and Decreasing Functions
Definition
Function f (x) is increasing on the interval I if for all a and b in I
with a < b then f (a) < f (b). Function f (x) is decreasing on the
interval I if for all a and b in I with a < b then f (a) > f (b).
Homework
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Read Section 2.10
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Exercises: 1–45 odd