Professor Switkes

Mean Value Theorem and Rolle's Theorem

Slide Duration:

Section 1: 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
-1
-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
-1
-2
Section 2: 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
-1
-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
-1
-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
-1
-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
-1
-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
-1
-2
Section 3: 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
-1
-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
-1
-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
-1
-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
-1
-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
-1
-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
-1
-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
-1
-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
-1
-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
-1
-2
Section 4: 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
-1
-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
-1
-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
-1
-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
-1
-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
-1
-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
-1
-2
Section 5: 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
-1
-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
-1
-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
-1
-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
-1
-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
-1
-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
-1
-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
-1
-2
Section 6: 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
-1
-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
-1
-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
-1
-2
Section 7: 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
-1
-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
5:31
Lecture Example 1
6:05
Lecture Example 2
8:28
Lecture Example 3
11:55
-1
-2
Volume by Method of Cylindrical Shells

30m 29s

Intro
0:00
Important Equations
0:50
1:04
Equation 2: Rotation about y-axis (2 curves)
7:34
8:15
Lecture Example 1
8:57
Lecture Example 2
14:26
Lecture Example 3
18:15
-1
-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
-1
-2
Section 8: 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
-1
-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
-1
-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
-1
-2
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### 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.

### Mean Value Theorem and Rolle's Theorem

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• 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

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