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Calculus AB Mean Value Theorem

Section 4: Applications of Derivatives: Lecture 7 | 9:39 min

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Lecture Comments (10)

	0 answers Post by Angela Zhang on December 7, 2014 John, you are very cute! I like you <3
	0 answers Post by Yu Han on July 24, 2014 For example 1, how can you tell if it met these two conditions?
	0 answers Post by Justin Powell on February 10, 2014 Just for clarity: The Mean Value Theorem says that if a function f(x) is continuous and differentiable between two intervals x=a and x=b, then solving the function for these two values will give the coordinates {a,f(a)} and {b,f(b)}. Now if you draw a line between these two points, the slope will be: (f(b)-f(a))/(b-a) which is the rise over the run. The Mean Value Theoem states that at some point, {c,f(c)}, on the graph of the function between the intervals x=a an x=b, the derivative or slope must be equal to (f(b)-f(a))/(b-a) at least once. So the derivative of f(x) at x=c is equal to: f'(c)=(f(b)-f(a))/(b-a) In the last example: the slope between the coordinates {-1,-1} and {5,125} equals 126/6 = 21=f'(c)=3c^2 c=sqrt(7). At f(sqrt7) the slope of the curve of the graph is parallel to the line between {-1,-1} and {5,125}. also for the MVT, f'(c)=0 is only true when f(a) and f(b) are equal (Rolles theorem). Sorry for overclarifying, but the video left out some key points.
	1 answer Last reply by: Andrew Mu Mon Jan 6, 2014 8:30 AM Post by Willie Wang on January 13, 2013 AT Example 4, why c cannot be negative sqr(7)ï¼Ÿ
	1 answer Last reply by: Johnny Zamora Fri Jan 10, 2014 2:58 AM Post by James Xie on December 17, 2012 What exactly does c stand for? Is it the Instant Rate Of Change (IROC?
	0 answers Post by Ryan Menezes on November 18, 2012 That's exactly what i want to know......
	1 answer Last reply by: Siyun Liu Sat Jan 24, 2015 6:57 PM Post by Steve Denton on October 18, 2012 At 5:00, what happened to the 1/2 in front of (c-1)^-1/2? rewritten then 1/ (2 sq rt (c-1))?????

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Mean Value Theorem

There exists a point c, between aand b where the slope is defined by change in f(x)divided by change in x.
Applying Mean Value Theorem

Check conditions
Take derivative
Set derivative function =
Replace xwith c
Solve for c value

Mean Value Theorem

Find the value of c that satisfies the the Mean Value Theorem conditions for f(x) = e^x on the interval [0, ln2]

f′(x) = e^x
Continuous and differentiable everywhere
f′(c) = [(f(ln2) − f(0))/(ln2 − 0)]
f′(c) = [(e^ln2 − 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) = e^c = [1/ln2]
lne^c = 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]

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Answer

Mean Value 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

Mean Value Theorem

Example 1

Example 2

Example 3

Example 4

Intro 0:00
Mean Value Theorem 0:06

Rolle's Theorem
Mean Value Theorem Conditions
Mean Value Theorem Definition

Example 1 0:56
Example 2 2:44
Example 3 5:28
Example 4 7:15

Mathematics: AP Calculus AB

Section 1: Functions
	Definitions & Properties of Functions	11:26
	Graphing	13:58
	Inverse Functions	6:47
	Polynomial Functions	5:04
	Trigonometric Functions	6:45
	Inverse Trigonometric Functions	5:58
	Trigonometric Identities	17:42
	Exponential Functions	5:53
	Logarithmic Functions	7:08
	Rational Functions	15:36
	Conic Sections	14:58
Section 2: Limits and Continuity
	Limit Definition & Properties	7:15
	Solving Limits with Algebra	8:01
	Rational Limit Rules	3:16
	One Sided Limits	9:57
	Special Trigonometric Limits	8:28
	Limits & Continuity	10:14
	Limits: Multiple Choice Practice	6:16
Section 3: Derivatives
	Derivative Definition & Properties	4:11
	Basic Rules of Differentiation	7:07
	Power Rule	7:14
	Trigonometric Rules	7:53
	Product Rule	11:11
	Quotient Rule	16:50
	Chain Rule	19:48
	Higher Order Derivatives	15:00
	Derivatives of Exponential Functions	13:14
	Derivatives of Logarithmic Functions	11:30
	Derivatives of Inverse Trigonometric Functions	16:54
	Implicit Differentiation	16:53
	Multiple Choice Practice: Derivatives	11:07
Section 4: Applications of Derivatives
	Tangent & Normal Lines	22:36
	Position Velocity & Acceleration	18:42
	Related Rates	26:22
	Minimum & Maximum	12:22
	Concavity	11:43
	Rolles Theorem	8:28
	Mean Value Theorem	9:39
	Differentials	12:25
	Applications of Derivatives: Multiple Choice Practice	13:21
	Applications of Derivatives: Free Response Practice	10:22
Section 5: Integrals
	Definition of Integrals	1:08
	Integrals of Power Rule	8:50
	Integrals Basic Rules of Integration	9:43
	Trigonometric Rules of Integrals	8:58
	Chain Rule	13:59
	Integrals of Exponential Functions	12:52
	Integrals of Natural Logarithmic Functions	13:00
	Integrals of Inverse Trigonometric Functions	8:29
	Integrals: Multiple Choice Practice	15:37
Section 6: Applications of Integrals
	Fundamental Theorem of Calculus	15:55
	Area Under A Curve	18:34
	Reimann Sums	10:35
	Trapezoid Rule	12:46
	Mean Value Theorem	11:22
	Second Fundamental Theorem of Calculus	4:44
	Area Between Curves	16:39
	Revolving Solids Washer Disk Methods	21:09
	Revolving Solids Cylindrical Shells Method	26:46
	Revolving Solids Known Cross Sections	27:41
	Differential Equations Eulers Method	17:54
	Differential Equations Slope Fields	16:30
	Application of Integrals: Multiple Choice Practice	14:19
	Application of Integrals: Free Response Practice	9:14
Section 7: Sample AP Test
	AP Calculus AB Practice test: Section 1: Multiple Choice Part 1	17:50
	AP Calculus AB Practice test: Section 1: Multiple Choice Part 2	17:32
	AP Calculus AB Practice test: Section 1: Multiple Choice Part 3	22:14
	AP Calculus AB Practice test: Section 1: Multiple Choice Part 4	19:35
	AP Calculus AB Practice test: Section 1: Multiple Choice Part 5	25:43
	AP Calculus AB Practice Test: Section 2: Free Response Part 1	16:50
	AP Calculus AB Practice Test: Section 2: Free Response Part 2	21:36
	AP Calculus AB Practice Test: Section 2: Free Response Part 3	13:39

Related Books

	Cracking the AP Calculus AB Exam, 2015 Edition Author: Princeton Review ISBN: 804124809 Publisher: Princeton Review Year: 2014 The book includes 3 full length practice tests with detailed explanations, a review of all the key concepts, and targeted strateeies to ace th exam for your highest score.
	Cliffs AP: Calculus AB & BC, 3rd Edition Authors: Dale W Johnson, Kerry J King ISBN: 764586831 Publisher: Cliffs Notes Year: 2001 The book has a topic-by topic breakdown and lots of problem approach suggestions for both free response and multiple choice calculus questions. The book also includes two full length tests.

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John Zhu

Mean Value Theorem

Slide Duration:

Table of Contents

Section 1: Functions

Definitions & Properties of Functions

11m 26s

Intro