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The Mean Value Theorem
Related Topics
Wize University Calculus 1 Textbook > Applications of Differentiation
The Mean Value Theorem
3 Activities
If
f
(
x
)
f(x)
f
(
x
)
is continuous and differentiable on
[
6
,
15
]
[6,15]
[
6
,
15
]
with
f
(
6
)
=
−
2
f(6)=-2
f
(
6
)
=
−
2
and
f
′
(
x
)
≤
10
f^{\prime}(x)\leq 10
f
′
(
x
)
≤
10
, what is the largest possible value for
f
(
15
)
f(15)
f
(
15
)
?
≥
88
\ge 88
≥
88
≤
88
\leq88
≤
88
≤
90
\leq 90
≤
90
≥
90
\ge90
≥
90
I don't know
Check Submission
More The Mean Value Theorem Questions:
Practice: MVT
For the function
g
(
x
)
=
(
x
+
1
)
3
g(x)=(x+1)^{3}
g
(
x
)
=
(
x
+
1
)
3
on
[
−
1
,
1
]
[-1,1]
[
−
1
,
1
]
, find the number
c
c
c
guaranteed by the Mean Value Theorem.
Mean Value Theorem with Maximum Value
If
f
(
x
)
f(x)
f
(
x
)
is continuous and differentiable on
[
6
,
15
]
[6,15]
[
6
,
15
]
with
f
(
6
)
=
−
2
f(6)=-2
f
(
6
)
=
−
2
and
f
′
(
x
)
≤
10
f^{\prime}(x)\leq 10
f
′
(
x
)
≤
10
, what is the largest possible value for
f
(
15
)
f(15)
f
(
15
)
?
Mean Value Theorem with Maximum Value
If
f
(
x
)
f(x)
f
(
x
)
is continuous and differentiable on
[
6
,
15
]
[6,15]
[
6
,
15
]
with
f
(
6
)
=
−
2
f(6)=-2
f
(
6
)
=
−
2
and
f
′
(
x
)
≤
10
f^{\prime}(x)\leq 10
f
′
(
x
)
≤
10
, what is the largest possible value for
f
(
15
)
f(15)
f
(
15
)
?
The Mean Value Theorem
For
g
(
x
)
=
−
x
2
g(x)=-x^{2}
g
(
x
)
=
−
x
2
on
[
−
1
,
2
]
[-1,2]
[
−
1
,
2
]
, find the number
c
c
c
guaranteed by the Mean Value Theorem.
Suppose that
f
f
f
is a differentiable function such that
f
′
(
x
)
≥
3
f'(x)\ge3
f
′
(
x
)
≥
3
for all
x
x
x
. If
f
(
2
)
=
3
,
f(2)=3,
f
(
2
)
=
3
,
what is the minimum possible value of
f
(
4
)
f(4)
f
(
4
)
?
Mean Value Theorem: Speed
In 2 hours, a car driver covered 150 km on a road with speed limit 65 km per hour. Did the car go over the speed limit? Answer using the mean-value theorem.
Mean Value Theorem with Maximum Value
If
f
(
x
)
f(x)
f
(
x
)
is continuous and differentiable on
[
6
,
15
]
[6,15]
[
6
,
15
]
with
f
(
6
)
=
−
2
f(6)=-2
f
(
6
)
=
−
2
and
f
′
(
x
)
≤
10
f^{\prime}(x)\leq 10
f
′
(
x
)
≤
10
, what is the largest possible value for
f
(
15
)
f(15)
f
(
15
)
?
Practice: MVT
Q:
\textbf{Q:}
Q:
For the function
g
(
x
)
=
(
x
+
1
)
3
g(x)=(x+1)^{3}
g
(
x
)
=
(
x
+
1
)
3
on
[
−
1
,
1
]
[-1,1]
[
−
1
,
1
]
, find the number
c
c
c
guaranteed by the Mean Value Theorem.
The Mean Value Theorem
Let
f
(
x
)
f(x)
f
(
x
)
be a function so that
f
(
x
)
,
f
′
(
x
)
,
f
′
′
(
x
)
f(x), f'(x), f''(x)
f
(
x
)
,
f
′
(
x
)
,
f
′′
(
x
)
exist and are continuous for all
x
x
x
, and
∣
f
(
x
)
−
sin
(
x
)
∣
≤
1
/
4
|f(x) - \sin(x)| \leq 1/4
∣
f
(
x
)
−
sin
(
x
)
∣
≤
1/4
for all
x
x
x
.
The Mean Value Theorem
Let
f
(
x
)
f(x)
f
(
x
)
be a continuous function defined for all real numbers
x
x
x
. Suppose
f
(
x
)
f(x)
f
(
x
)
is increasing on the intervals
(
−
∞
,
−
2
)
(-\infty, -2)
(
−
∞
,
−
2
)
and
(
4
,
∞
)
(4, \infty)
(
4
,
∞
)
, decreasing on
(
−
2
,
4
)
(-2, 4)
(
−
2
,
4
)
,
f
(
−
2
)
=
5
f(-2) = 5
f
(
−
2
)
=
5
and
f
(
4
)
=
1
f(4) = 1
f
(
4
)
=
1
. How many zeros does
f
(
x
)
f(x)
f
(
x
)
have?
The Mean Value Theorem
Why doesn't the function
f
(
x
)
=
∣
x
∣
f(x) = \left|x\right|
f
(
x
)
=
∣
x
∣
on the interval
[
−
1
,
1
]
[-1,1]
[
−
1
,
1
]
violate the Mean Value Theorem?
In other words, how is it that
f
(
1
)
−
f
(
−
1
)
1
−
(
−
1
)
=
0
\dfrac{f(1)-f(-1)}{1-(-1)}=0
1
−
(
−
1
)
f
(
1
)
−
f
(
−
1
)
=
0
, but there does not exist a value of
c
c
c
such that
f
′
(
c
)
=
0
f'(c) = 0
f
′
(
c
)
=
0
?
Given
f
(
x
)
=
x
2
−
3
x
+
2
f(x) = x^2 - 3x + 2
f
(
x
)
=
x
2
−
3
x
+
2
on the interval
x
∈
[
−
1
,
3
]
x \in [-1,3]
x
∈
[
−
1
,
3
]
, what value of
c
c
c
does the mean value theorem guarantee exists?
Suppose that
f
f
f
is a differentiable function such that
f
′
(
x
)
≥
3
f'(x)\ge3
f
′
(
x
)
≥
3
for all
x
x
x
. If
f
(
2
)
=
3
,
f(2)=3,
f
(
2
)
=
3
,
what is the minimum possible value of
f
(
4
)
f(4)
f
(
4
)
?
Given
f
(
x
)
=
x
2
−
3
x
+
2
f(x) = x^2 - 3x + 2
f
(
x
)
=
x
2
−
3
x
+
2
on the interval
x
∈
[
−
1
,
3
]
x \in [-1,3]
x
∈
[
−
1
,
3
]
, what value of
c
c
c
does the mean value theorem guarantee exists?
tan question
Given the function
f
(
x
)
=
tan
x
f(x)=\tan x
f
(
x
)
=
tan
x
. By the Mean Value Theorem, there exists a
c
c
c
in the interval
(
1
,
2
)
(1,2)
(
1
,
2
)
such that
f
′
(
c
)
f'(c)
f
′
(
c
)
is equal to
f
(
2
)
−
f
(
1
)
2
−
1
\dfrac{f(2)-f(1)}{2-1}
2
−
1
f
(
2
)
−
f
(
1
)
. This statement is:
Practice: Mean Value Theorem
Practice: Mean Value Theorem
Given
f
(
x
)
=
x
2
−
3
x
+
2
f(x) = x^2 - 3x + 2
f
(
x
)
=
x
2
−
3
x
+
2
on the interval
x
∈
[
−
1
,
3
]
x \in [-1,3]
x
∈
[
−
1
,
3
]
, what value of
c
c
c
does the mean value theorem guarantee exists?
Practice: Mean Value Theorem
Practice: Mean Value Theorem
If
f
(
x
)
f(x)
f
(
x
)
is continuous and differentiable on
[
6
,
15
]
[6,15]
[
6
,
15
]
with
f
(
6
)
=
−
2
f(6)=-2
f
(
6
)
=
−
2
and
f
′
(
x
)
≤
10
f^{\prime}(x)\leq 10
f
′
(
x
)
≤
10
, what is the largest possible value for
f
(
15
)
f(15)
f
(
15
)
?
Given
f
(
x
)
=
x
2
−
3
x
+
2
f(x) = x^2 - 3x + 2
f
(
x
)
=
x
2
−
3
x
+
2
on the interval
x
∈
[
−
1
,
3
]
x \in [-1,3]
x
∈
[
−
1
,
3
]
, what value of
c
c
c
does the mean value theorem guarantee exists?
Mean Value Theorem: Speed
In 2 hours, a car driver covered 150 km on a road with speed limit 65 km per hour. Did the car go over the speed limit? Answer using the mean-value theorem.
Practice: MVT
For the function
g
(
x
)
=
(
x
+
1
)
3
g(x)=(x+1)^{3}
g
(
x
)
=
(
x
+
1
)
3
on
[
−
1
,
1
]
[-1,1]
[
−
1
,
1
]
, find the number
c
c
c
guaranteed by the Mean Value Theorem.
Mean Value Theorem with Maximum Value
If
f
(
x
)
f(x)
f
(
x
)
is continuous and differentiable on
[
6
,
15
]
[6,15]
[
6
,
15
]
with
f
(
6
)
=
−
2
f(6)=-2
f
(
6
)
=
−
2
and
f
′
(
x
)
≤
10
f^{\prime}(x)\leq 10
f
′
(
x
)
≤
10
, what is the largest possible value for
f
(
15
)
f(15)
f
(
15
)
?
The Mean Value Theorem
For
g
(
x
)
=
−
x
2
g(x)=-x^{2}
g
(
x
)
=
−
x
2
on
[
−
1
,
2
]
[-1,2]
[
−
1
,
2
]
, find the number
c
c
c
guaranteed by the Mean Value Theorem.
Example: Mean Value Theorem
Is there a constant
c
c
c
in
f
(
x
)
=
∣
x
−
1
∣
f(x)=| x-1|
f
(
x
)
=
∣
x
−
1∣
where
f
(
3
)
−
f
(
0
)
=
f
′
(
c
)
(
3
−
0
)
f(3)-f(0)=f'(c)(3-0)
f
(
3
)
−
f
(
0
)
=
f
′
(
c
)
(
3
−
0
)
? Why does this not contradict the Mean Value Theorem?
Exercise
Suppose
f
f
f
is a differentiable function such that
f
′
(
x
)
≤
5
f'(x)\leq 5
f
′
(
x
)
≤
5
for all
x
∈
[
1
,
5
]
x\in[1,5]
x
∈
[
1
,
5
]
and
f
(
1
)
=
3
f(1)=3
f
(
1
)
=
3
. What is an upper bound on
f
(
5
)
f(5)
f
(
5
)
?
Suppose that
f
f
f
is a differentiable function such that
f
′
(
x
)
≥
3
f'(x)\geq 3
f
′
(
x
)
≥
3
for all
x
x
x
. If
f
(
2
)
=
3
f(2)=3
f
(
2
)
=
3
, what is the minimum possible value of
f
(
4
)
f(4)
f
(
4
)
?
Given
f
(
x
)
=
x
2
−
3
x
+
2
f(x) = x^2 - 3x + 2
f
(
x
)
=
x
2
−
3
x
+
2
on the interval
x
∈
[
−
1
,
3
]
x \in [-1,3]
x
∈
[
−
1
,
3
]
, what value of
c
c
c
does the mean value theorem guarantee exists?