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Professor Switkes

Professor Switkes

Mean Value Theorem and Rolle's Theorem

Slide Duration:

Table of Contents

I. Overview of Functions
Review of Functions

26m 29s

Intro
0:00
What is a Function
0:10
Domain and Range
0:21
Vertical Line Test
0:31
Example: Vertical Line Test
0:47
Function Examples
1:57
Example: Squared
2:10
Example: Natural Log
2:41
Example: Exponential
3:21
Example: Not Function
3:54
Odd and Even Functions
4:39
Example: Even Function
5:10
Example: Odd Function
5:53
Odd and Even Examples
6:48
Odd Function
6:55
Even Function
8:43
Increasing and Decreasing Functions
10:15
Example: Increasing
10:42
Example: Decreasing
10:55
Increasing and Decreasing Examples
11:41
Example: Increasing
11:48
Example: Decreasing
12:33
Types of Functions
13:32
Polynomials
13:45
Powers
14:06
Trigonometric
14:34
Rational
14:50
Exponential
15:13
Logarithmic
15:29
Lecture Example 1
15:55
Lecture Example 2
17:51
Additional Example 3
-1
Additional Example 4
-2
Compositions of Functions

12m 29s

Intro
0:00
Compositions
0:09
Alternative Notation
0:32
Three Functions
0:47
Lecture Example 1
1:19
Lecture Example 2
3:25
Lecture Example 3
6:45
Additional Example 4
-1
Additional Example 5
-2
II. Limits
Average and Instantaneous Rates of Change

20m 59s

Intro
0:00
Rates of Change
0:11
Average Rate of Change
0:21
Instantaneous Rate of Change
0:33
Slope of the Secant Line
0:46
Slope of the Tangent Line
1:00
Lecture Example 1
1:14
Lecture Example 2
6:36
Lecture Example 3
11:30
Additional Example 4
-1
Additional Example 5
-2
Limit Investigations

22m 37s

Intro
0:00
What is a Limit?
0:10
Lecture Example 1
0:56
Lecture Example 2
5:28
Lecture Example 3
9:27
Additional Example 4
-1
Additional Example 5
-2
Algebraic Evaluation of Limits

28m 19s

Intro
0:00
Evaluating Limits
0:09
Lecture Example 1
1:06
Lecture Example 2
5:16
Lecture Example 3
8:15
Lecture Example 4
12:58
Additional Example 5
-1
Additional Example 6
-2
Formal Definition of a Limit

23m 39s

Intro
0:00
Formal Definition
0:13
Template
0:55
Epsilon and Delta
1:24
Lecture Example 1
1:40
Lecture Example 2
9:20
Additional Example 3
-1
Additional Example 4
-2
Continuity and the Intermediate Value Theorem

19m 9s

Intro
0:00
Continuity
0:13
Continuous
0:16
Discontinuous
0:37
Intermediate Value Theorem
0:52
Example
1:22
Lecture Example 1
2:58
Lecture Example 2
9:02
Additional Example 3
-1
Additional Example 4
-2
III. Derivatives, part 1
Limit Definition of the Derivative

22m 52s

Intro
0:00
Limit Definition of the Derivative
0:11
Three Versions
0:13
Lecture Example 1
1:02
Lecture Example 2
4:33
Lecture Example 3
6:49
Lecture Example 4
10:11
Additional Example 5
-1
Additional Example 6
-2
The Power Rule

26m 1s

Intro
0:00
Power Rule of Differentiation
0:14
Power Rule with Constant
0:41
Sum/Difference
1:15
Lecture Example 1
1:59
Lecture Example 2
6:48
Lecture Example 3
11:22
Additional Example 4
-1
Additional Example 5
-2
The Product Rule

14m 54s

Statement of the Product Rule
0:08
Lecture Example 1
0:41
Lecture Example 2
2:27
Lecture Example 3
5:03
Additional Example 4
-1
Additional Example 5
-2
The Quotient Rule

19m 17s

Intro
0:00
Statement of the Quotient Rule
0:07
Carrying out the Differentiation
0:23
Quotient Rule in Words
1:00
Lecture Example 1
1:19
Lecture Example 2
4:23
Lecture Example 3
8:00
Additional Example 4
-1
Additional Example 5
-2
Applications of Rates of Change

17m 43s

Intro
0:00
Rates of Change
0:11
Lecture Example 1
0:44
Lecture Example 2
5:16
Lecture Example 3
7:38
Additional Example 4
-1
Additional Example 5
-2
Trigonometric Derivatives

26m 58s

Intro
0:00
Six Basic Trigonometric Functions
0:11
Patterns
0:47
Lecture Example 1
1:18
Lecture Example 2
7:38
Lecture Example 3
12:15
Lecture Example 4
14:25
Additional Example 5
-1
Additional Example 6
-2
The Chain Rule

23m 47s

Intro
0:00
Statement of the Chain Rule
0:09
Chain Rule for Three Functions
0:27
Lecture Example 1
1:00
Lecture Example 2
4:34
Lecture Example 3
7:23
Additional Example 4
-1
Additional Example 5
-2
Inverse Trigonometric Functions

27m 5s

Intro
0:00
Six Basic Inverse Trigonometric Functions
0:10
Lecture Example 1
1:11
Lecture Example 2
8:53
Lecture Example 3
12:37
Additional Example 4
-1
Additional Example 5
-2
Equation of a Tangent Line

15m 52s

Intro
0:00
Point Slope Form
0:10
Lecture Example 1
0:47
Lecture Example 2
3:15
Lecture Example 3
6:10
Additional Example 4
-1
Additional Example 5
-2
IV. Derivatives, part 2
Implicit Differentiation

30m 5s

Intro
0:00
Purpose
0:09
Implicit Function
0:20
Lecture Example 1
0:32
Lecture Example 2
7:14
Lecture Example 3
11:22
Lecture Example 4
16:43
Additional Example 5
-1
Additional Example 6
-2
Higher Derivatives

13m 16s

Intro
0:00
Notation
0:08
First Type
0:19
Second Type
0:54
Lecture Example 1
1:41
Lecture Example 2
3:15
Lecture Example 3
4:57
Additional Example 4
-1
Additional Example 5
-2
Logarithmic and Exponential Function Derivatives

17m 42s

Intro
0:00
Essential Equations
0:12
Lecture Example 1
1:34
Lecture Example 2
2:48
Lecture Example 3
5:54
Additional Example 4
-1
Additional Example 5
-2
Hyperbolic Trigonometric Function Derivatives

14m 30s

Intro
0:00
Essential Equations
0:15
Six Basic Hyperbolic Trigc Functions
0:32
Six Basic Inverse Hyperbolic Trig Functions
1:21
Lecture Example 1
1:48
Lecture Example 2
3:45
Lecture Example 3
7:09
Additional Example 4
-1
Additional Example 5
-2
Related Rates

29m 5s

Intro
0:00
What Are Related Rates?
0:08
Lecture Example 1
0:35
Lecture Example 2
5:25
Lecture Example 3
11:54
Additional Example 4
-1
Additional Example 5
-2
Linear Approximation

23m 52s

Intro
0:00
Essential Equations
0:09
Linear Approximation (Tangent Line)
0:18
Example: Graph
1:18
Differential (df)
2:06
Delta F
5:10
Lecture Example 1
6:38
Lecture Example 2
11:53
Lecture Example 3
15:54
Additional Example 4
-1
Additional Example 5
-2
V. Application of Derivatives
Absolute Minima and Maxima

18m 57s

Intro
0:00
Minimums and Maximums
0:09
Absolute Minima and Maxima (Extrema)
0:53
Critical Points
1:25
Lecture Example 1
2:58
Lecture Example 2
6:57
Lecture Example 3
10:02
Additional Example 4
-1
Additional Example 5
-2
Mean Value Theorem and Rolle's Theorem

20m

Intro
0:00
Theorems
0:09
Mean Value Theorem
0:13
Graphical Explanation
0:36
Rolle's Theorem
2:06
Graphical Explanation
2:28
Lecture Example 1
3:36
Lecture Example 2
6:33
Lecture Example 3
9:32
Additional Example 4
-1
Additional Example 5
-2
First Derivative Test, Second Derivative Test

27m 11s

Intro
0:00
Local Minimum and Local Maximum
0:14
Example
1:01
First and Second Derivative Test
1:26
First Derivative Test
1:36
Example
2:00
Second Derivative Test (Concavity)
2:58
Example: Concave Down
3:15
Example: Concave Up
3:54
Inconclusive
4:19
Lecture Example 1
5:23
Lecture Example 2
12:03
Lecture Example 3
15:54
Additional Example 4
-1
Additional Example 5
-2
L'Hopital's Rule

23m 9s

Intro
0:00
Using L'Hopital's Rule
0:09
Informal Definition
0:34
Lecture Example 1
1:27
Lecture Example 2
4:00
Lecture Example 3
5:40
Lecture Example 4
9:38
Additional Example 5
-1
Additional Example 6
-2
Curve Sketching

40m 16s

Intro
0:00
Collecting Information
0:15
Domain and Range
0:17
Intercepts
0:21
Symmetry Properties (Even/Odd/Periodic)
0:33
Asymptotes (Vertical/Horizontal/Slant)
0:45
Critical Points
1:15
Increasing/Decreasing Intervals
1:24
Inflection Points
1:38
Concave Up/Down
1:52
Maxima/Minima
2:03
Lecture Example 1
2:58
Lecture Example 2
10:52
Lecture Example 3
17:55
Additional Example 4
-1
Additional Example 5
-2
Applied Optimization

25m 37s

Intro
0:00
Real World Problems
0:08
Sketch
0:11
Interval
0:20
Rewrite in One Variable
0:26
Maximum or Minimum
0:34
Critical Points
0:42
Optimal Result
0:52
Lecture Example 1
1:05
Lecture Example 2
6:12
Lecture Example 3
13:31
Additional Example 4
-1
Additional Example 5
-2
Newton's Method

25m 13s

Intro
0:00
Approximating Using Newton's Method
0:10
Good Guesses for Convergence
0:32
Lecture Example 1
0:49
Lecture Example 2
4:21
Lecture Example 3
7:59
Additional Example 4
-1
Additional Example 5
-2
VI. Integrals
Approximating Areas and Distances

36m 50s

Intro
0:00
Three Approximations
0:12
Right Endpoint, Left Endpoint, Midpoint
0:22
Formulas
1:05
Velocity and Distance
1:35
Lecture Example 1
2:28
Lecture Example 2
12:10
Lecture Example 3
19:43
Additional Example 4
-1
Additional Example 5
-2
Riemann Sums, Definite Integrals, Fundamental Theorem of Calculus

22m 2s

Intro
0:00
Important Equations
0:22
Riemann Sum
0:28
Integral
1:58
Integrand
2:35
Limits of Integration (Upper Limit, Lower Limit)
2:43
Other Equations
3:05
Fundamental Theorem of Calculus
4:00
Lecture Example 1
5:04
Lecture Example 2
10:43
Lecture Example 3
13:52
Additional Example 4
-1
Additional Example 5
-2
Substitution Method for Integration

23m 19s

Intro
0:00
U-Substitution
0:13
Important Equations
0:30
Purpose
0:36
Lecture Example 1
1:30
Lecture Example 2
6:17
Lecture Example 3
9:00
Lecture Example 4
11:24
Additional Example 5
-1
Additional Example 6
-2
VII. Application of Integrals, part 1
Area Between Curves

19m 59s

Intro
0:00
Area Between Two Curves
0:12
Graphic Description
0:34
Lecture Example 1
1:44
Lecture Example 2
5:39
Lecture Example 3
8:45
Additional Example 4
-1
Additional Example 5
-2
Volume by Method of Disks and Washers

24m 22s

Intro
0:00
Important Equations
0:16
Equation 1: Rotation about x-axis (disks)
0:27
Equation 2: Two curves about x-axis (washers)
3:38
Equation 3: Rotation about y-axis
5:31
Lecture Example 1
6:05
Lecture Example 2
8:28
Lecture Example 3
11:55
Additional Example 4
-1
Additional Example 5
-2
Volume by Method of Cylindrical Shells

30m 29s

Intro
0:00
Important Equations
0:50
Equation 1: Rotation about y-axis
1:04
Equation 2: Rotation about y-axis (2 curves)
7:34
Equation 3: Rotation about x-axis
8:15
Lecture Example 1
8:57
Lecture Example 2
14:26
Lecture Example 3
18:15
Additional Example 4
-1
Additional Example 5
-2
Average Value of a Function

16m 31s

Intro
0:00
Important Equations
0:11
Origin of Formula
0:34
Lecture Example 1
2:51
Lecture Example 2
5:30
Lecture Example 3
8:13
Additional Example 4
-1
Additional Example 5
-2
VIII. Extra
Graphs of f, f', f''

23m 58s

Intro
0:00
Slope Function of f(x)
0:41
Slope is Zero
0:53
Slope is Positive
1:03
Slope is Negative
1:13
Slope Function of f'(x)
1:31
Slope is Zero
1:42
Slope is Positive
1:48
Slope is Negative
1:54
Lecture Example 1
2:23
Lecture Example 2
8:06
Lecture Example 3
12:36
Additional Example 4
-1
Additional Example 5
-2
Slope Fields for Differential Equations

18m 32s

Intro
0:00
Things to Remember
0:13
Graphic Description
0:42
Lecture Example 1
1:44
Lecture Example 2
6:59
Lecture Example 3
9:46
Additional Example 4
-1
Additional Example 5
-2
Separable Differential Equations

17m 4s

Intro
0:00
Differential Equations
0:10
Focus on Exponential Growth/Decay
0:27
Separating Variables
0:47
Lecture Example 1
1:35
Lecture Example 2
6:41
Lecture Example 3
9:36
Additional Example 4
-1
Additional Example 5
-2
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Lecture Comments (4)

1 answer

Last reply by: Cheng Jiang
Wed Oct 30, 2013 5:37 PM

Post by Shehan Gunasekara on May 1, 2012

Thanks!! helped alot with uni

0 answers

Post by Real Schiran on October 29, 2011

:-)

0 answers

Post by Ahmed Shiran on October 29, 2011

Cool !

Mean Value Theorem and Rolle's Theorem

  • You should memorize the Mean Value Theorem and Rolle’s Theorem – including the continuity and differentiability hypotheses.
  • If you are using one of these theorems, do check that the continuity and differentiability hypotheses are satisfied.
  • If a problem asks you to verify a conclusion of one of these theorems, use algebra to do so.
  • Note that Rolle’s Theorem is simply a special case of the Mean Value Theorem.

Mean Value Theorem and Rolle's Theorem

Use Rolle's Theorem to show that f′(c) = 0 for some c in the interval [0, 4] with f(x) = 4x2 − 16x + 4
  • Polynomial, so it's continuous
  • f′(x) = 8x − 16 defined for all x
  • 4x2 − 16x + 4 = 0
  • x = [(16 ±√{(−16)2 − 4(4)(4)})/2(4)]
  • x = [(16 ±√{256 − 64})/8]
  • x = 2 ±[(√{192})/8]
  • x = 2 ±[(√{64 * 3})/8]
  • x = 2 ±√3 Both solutions are within the interval
Yes. There exists some c in the interval [0,4] such that f′(c) = 0.
Find the value of x such that f′(x) = 0 for f(x) = 4x2 − 16x + 4
  • f′(x) = 8x − 16
  • 8x − 16 = 0
  • 8x = 16
  • x = [16/8]
f′(2) = 0
Use Rolle's Theorem to show that f′(c) = 0 for some c in the interval [−1, 1] with f(x) = |x| − 1
  • We can rewrite this as f(x) = √{x2} − 1
  • The function is continuous on the interval
  • f′(x) = [1/2] [1/(√{x2})] [d/dx] x2
  • f′(x) = [x/(√{x2})] Removable discontinuity at x = 0
  • limx → 0+ f′(x) ≠ limx → 0 f′(x)
The function is not differentiable at x = 0, which is in the interval. Rolle's Theorem cannot be applied.
Confirm the Rolle's Theorem conditions on the interval [−1, 1] with f(x) = x − 1
  • Polynomial, so it's continuous
  • f′(x) = 1
  • Differentiable on that interval
  • Let's check the endpoints
  • f(−1) = −2
  • f(1) = 0
  • The endpoints don't match. They don't necessarily have to equal zero, but there must be two equal values of y within that interval. Rolle's Theorem is only a special case of the Mean Value Theorem, which is covered in the next lesson
The conditions for Rolle's Theorem are not met.
Use Rolle's Theorem to show that f′(c) = 0 for some c in the interval [0, π] with f(x) = sinx
  • f(x) = sinx
  • sinx is defined for all x. Continuous
  • f′(x) = cosx
  • cosx is defined for all x and has no edges or kinks in its graph. Differentiable.
  • sinx = 0
  • x = 0, π
Yes. There exists some c in the interval [0, π] such that f′(c) = 0.
Find the value of x in the interval [0, π] such that f′(x) = 0 for f(x) = sinx
  • f′(x) = cosx
  • f′([(π)/2]) = 0
f′([(π)/2]) = 0
Use Rolle's Theorem to show that f′(c) = 0 for some c in the interval [0, 4] with f(x) = x3 − 12x
  • Polynomial function, so it's continuous everywhere
  • f′(x) = 3x2 − 12 continuous everywhere
  • x3 − 12x = 0
  • x(x2 − 12) = 0
  • x(x + √{12})(x − √{12}) = 0
  • Three solutions, x = 0, −√{12}, √{12}, and two of these are within the interval
Yes. There exists some c in the interval [0, 4] such that f′(c) = 0.
Use Rolle's Theorem to show that f′(c) = 0 for some c in the interval [−2, 2] with f(x) = ex
  • ex is a continuous function so it's continuous function. In fact, it can be expressed as a polynomial function
  • f′(x) = ex defined everywhere. In fact, it can be expressed as a polynomial function
  • ex≠ 0
  • Let's test the end-points
  • f(−2) = e−2
  • f(2) = e2
  • f(2) ≠ f(−2)
Rolle's Theorem conditions not met.
Find x such that f′(x) = 0 where f(x) = x2
  • f′(x) = 2x
  • f′(0) = 0
When x = 0, f′(x) = 0. Rolle's Theorem can be used to prove that a solution in an interval exists, but it doesn't necessarily prove there is no solution. In truth, the same
Use Rolle's Theorem to show that f′(c) = 0 for some c in the interval [0, 1] with f(x) = x4 − x2
  • Polynomial function, so it's continuous
  • f′(x) = 4x3 − 2x is defined everywhere
  • x4 − x2 = 0
  • x2(x2 − 1) = 0
  • x = −1, 0, 1
Yes. There exists some c in the interval [0, 1] such that f′(c) = 0.
Find the value of c that satisfies the the Mean Value Theorem conditions for f(x) = ex on the interval [0, ln2]
  • f′(x) = ex
  • Continuous and differentiable everywhere
  • f′(c) = [(f(ln2) − f(0))/(ln2 − 0)]
  • f′(c) = [(eln2 − e0)/ln2]
  • f′(c) = [(2 − 1)/ln2] = [1/ln2]
  • Rolle's Theorem conditions are not met (f(a) ≠ f(b)), but the Mean Value conditions are met.
  • f′(c) = ec = [1/ln2]
  • lnec = ln[1/ln2]
c = ln[1/ln2]
Find the value of c that satisfies the the Mean Value Theorem conditions for f(x) = sinx on the interval [0, 2π]
  • f′(x) = cosx
  • Continuous and differentiable everywhere
  • f′(c) = [(cos2π− cos0)/(2π− 0)]
  • f′(c) = [(1 − 1)/(2π)]
  • f′(c) = 0
  • cosc = 0
  • There are many solutions, but only two in the interval
c = [(π)/2], [(3π)/2]
Find the value of c that satisfies the the Mean Value Theorem conditions for f(x) = x3 − x2 + 1 on the interval [0, 1]
  • f′(x) = 3x2 − 2x
  • f′(c) = [(13 − 12 + 1 − 0)/(1 − 0)]
  • f′(c) = 1
  • 3c2 − 2c = 1
  • 3c2 − 2c − 1 = 0
  • c = [(2 ±√{4 − 4(3)(−1)})/(−6)]
  • c = [(2 ±4)/(−6)]
  • c = 1, [(−1)/3]
  • Only one of these values is in the interval
c = 1
Find the mean value of f′(x) using the Mean Value Theorem for f(x) = 3x2 on the interval [1, 2]
  • Polynomial. Continuous and differentiable.
  • f′(x) = 6x
  • f′(c) = [(3(22) − 3(12))/(2 − 1)]
f′(c) = 9
Find the value of c that satisfies the the Mean Value Theorem conditions for f(x) = [1/x] on the interval [−1,1]
  • f(x) is undefined at x = 0 and that is within the interval
  • f(x) is not continuous on the interval
Mean Value Theorem cannot be applied
Find the value of c that satisfies the the Mean Value Theorem conditions for f(x) = [1/x] on the interval [1, 2]
  • f′(x) = −[1/(x2)]
  • f′(c) = [([1/2] − [1/1])/(2 − 1)]
  • f′(c) = −[1/2]
  • −[1/2] = −[1/(c2)]
  • [1/2] = [1/(c2)]
  • c2 = 2
  • c = ±√2
  • Only one solution in the interval
c = √2
f(x) is continuous and differentiable on [a,b] and f(a) = f(b) and a ≠ b. Show that f′(x) has a root in the interval
  • In order for some c to be a root of f′(x), f′(c) = 0
  • Continuous and differentiable, so Mean Value Theorem can be used
  • f′(c) = [(f(b) − f(a))/(b − a)]
  • f′(c) = [(f(b) − f(b))/(b − a)]
  • f′(c) = [0/(b − a)]
  • f′(c) = 0
f′(c) = 0, so there is a root at x = c
Find at least one root of f′(x) on the interval [−1, 1] given f(x) = x2
  • Polynomial. Continuous and differentiable everywhere
  • f′(x) = 2x
  • f′(c) = [(12 − (−12))/(1 − (−1))]
  • f′(c) = 0
  • 2c = 0
  • c = 0
There's a root of f′(0)
f(x) is continuous and differentiable on [0,5]. Also, f(0) = 1 and f′(x) ≤ 3 in the interval. Find the largest possible value of f(5).
  • f′(c) = [(f(5) − f(0))/(5 − 0)]
  • f′(c) = [(f(5) − 1)/5]
  • 5 f′(c) = f(5) − 1
  • f(5) = 5 f′(c) + 1
  • Plug in the largest value
  • f(5) = 5(3) + 1
The largest possible value of f(5) is 16
Find the value of c that satisfies the the Mean Value Theorem conditions for f(x) = √x on the interval [1, 4]
  • f(x) is continuous so long as x ≥ 0
  • f′(x) = [1/(2 √x)]
  • Differentiable so long as x > 0
  • Continuous and differentiable in the interval
  • f′(c) = [(√4 − √1)/(4 − 1)]
  • f′(c) = [1/3]
  • [1/(2 √c)] = [1/3]
  • 2√c = 3
  • √c = [3/2]
c = [9/4]
Find the average value, f(c), of of the function f(x) = sinx on the interval [0, π] using the Mean Value Theorem for Integrals
  • f(c) = [1/(π− 0)] ∫0π sinx  dx
  • f(c) = [1/(π)] (−cosx) |0π
  • f(c) = [1/(π)] (− cosπ+ cos0)
  • f(c) = [1/(π)] (1 + 1)
f(c) = [2/(π)]
Find the average value, f(c) of of the function f(x) = sinx on the interval [0, 2π] using the Mean Value Theorem for Integrals
  • f(c) = [1/(2π− 0)] ∫0 sinx  dx
  • f(c) = [1/(2π)] (−cosx) |0
  • f(c) = [1/(2π)] (− cos2π+ cos0)
  • f(c) = [1/(2π)] (−1 + 1)
f(c) = 0
Find the average value,f(c), of of the function f(x) = x2 + 7 on the interval [0, 3] using the Mean Value Theorem for Integrals
  • f(c) = [1/(3 − 0)] ∫03 x2 + 7  dx
  • f(c) = [1/3] ([(x3)/3] + 7x) |03
  • f(c) = [1/3] ([(33)/3] + 21 − (0 + 0))
  • f(c) = [1/3] 30
f(c) = 10
Find the average value,f(c), of of the function f(x) = x3 + 2x + 1 on the interval [0, 4] using the Mean Value Theorem for Integrals
  • f(c) = [1/(4 − 0)] ∫04 x3 + 2x + 1  dx
  • f(c) = [1/4] ([(x4)/4] + x2 + x) |04
  • f(c) = [1/4] ([(44)/4] + 42 + 4 − 0 − 0 − 0)
  • f(c) = 42 + 4 + 1
f(c) = 21
Find the average value,f(c), of of the function f(x) = [1/x] on the interval [−1, 1] using the Mean Value Theorem for Integrals
  • f(x) has a discontinuity at x = 0 and that is in the interval
Mean Value Theorem for Integrals cannot be applied
Find the average value,f(c), of of the function f(x) = [1/x] on the interval [1, e] using the Mean Value Theorem for Integrals
  • f(x) is continuous for x > 0
  • f(c) = [1/(e − 1)] ∫1e [1/x] dx
  • f(c) = [1/(e − 1)] lnx |1e
  • f(c) = [1/(e − 1)] (lne − ln1)
  • f(c) = [1/(e − 1)] (1)
f(c) = [1/(e − 1)]
Find the average value,f(c), of of the function f(x) = cosx + 4 on the interval [0, 2π] using the Mean Value Theorem for Integrals
  • f(c) = [1/(2π− 0)] ∫0 cosx + 4  dx
  • f(c) = [1/(2π)] (sinx + 4x) |0
  • f(c) = [1/(2π)] (sin2π+ 8 π− sin0 − 0)
  • f(c) = [1/(2π)] (8 π)
f(c) = 4
Find the average value,f(c), of of the function f(x) = [1/(√{4 − x2})] on the interval [0, 1] using the Mean Value Theorem for Integrals
  • f(x) is undefined for x ≤ −2 and x ≥ 2
  • The function is defined on the interval [0, 1]
  • f(c) = [1/(1 − 0)] ∫01 [1/(√{4 − x2})] dx
  • f(c) = ∫01 [1/(√{22 − x2})] dx
  • f(c) = sin−1 [x/2] |01
  • f(c) = sin−1 [1/2] − sin−1 0
  • f(c) = [(π)/6] − 0
f(c) = 0
Find the average value,f(c), of of the function f(x) = 4x on the interval [−5, 5] using the Mean Value Theorem for Integrals
  • f(c) = [1/(5 − (−5))] ∫−55 4x  dx
  • f(c) = [1/10] 2x2 |−55
  • f(c) = [1/10] (2(5)2 − 2(−5)2)
  • f(c) = [1/10] (0)
f(c) = 0
Find the average value,f(c), of of the function f(x) = tan[x/4] on the interval [0, π] using the Mean Value Theorem for Integrals
  • f(x) has discontinuities whenever cos[x/4] = 0
  • So near zero, f(x) is discontinuous at −2π and 2π. Our interval does not include a discontinuity
  • f(c) = [1/(π− 0)] ∫0π tan[x/4] dx
  • u = [x/4]
  • du = [1/4] dx
  • f(c) = [1/(π)] 4 ∫0π tanu  du
  • f(c) = [4/(π)] ∫0π [sinu/cosu] du
  • v = cosu
  • dv = −sinu  du
  • f(c) = [4/(π)] (−1) ∫0π [1/v] dv
  • f(c) = − [4/(π)] lnv |0π
  • f(c) = − [4/(π)] lncosu |0π
  • f(c) = − [4/(π)] lncos[x/4] |0π
  • f(c) = − [4/(π)] (lncos[(π)/4] − lncos0)
  • f(c) = − [4/(π)] (ln[1/(√2)] − ln1)
  • f(c) = − [4/(π)] (ln[1/(√2)] − 0)
  • f(c) = − [4/(π)] (ln1 − ln√2)
  • f(c) = − [4/(π)] (0 − ln2[1/2])
  • f(c) = − [4/(π)] (− [1/2] ln2)
f(c) = [2 ln2/(π)]

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Mean Value Theorem and Rolle's Theorem

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Theorems 0:09
    • Mean Value Theorem
    • Graphical Explanation
    • Rolle's Theorem
    • Graphical Explanation
  • Lecture Example 1 3:36
  • Lecture Example 2 6:33
  • Lecture Example 3 9:32
  • Additional Example 4
  • Additional Example 5
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