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Calculus AB Rolles Theorem

Section 4: Applications of Derivatives: Lecture 6 | 8:28 min

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Lecture Comments (13)

	1 answer Last reply by: Marcelo Quijano Fri Nov 7, 2014 8:08 PM Post by Marcelo Quijano on November 7, 2014 Can someone explain to me how he messed up on example 2?
	2 answers Last reply by: Chonglin Xu Thu Jan 1, 2015 1:12 AM Post by BAILIN ZHANG on February 5, 2014 Example 1 is not correct.
	0 answers Post by Jimmy Wu on October 20, 2013 Remember, both points are a straight line. It can either be a Slope of 0/1 or it can also be of the same value.
	0 answers Post by Jimmy Wu on October 20, 2013 Example 3 can be applied. A must equal to B which does.
	0 answers Post by Sahil Gupta on January 20, 2013 You messed up in example 1 and 2.
	4 answers Last reply by: Akshay Tiwary Sat Apr 27, 2013 10:17 AM Post by abbas esmailzadeh on November 3, 2012 example 1:f(x)=x^2-x how f'(x)=4x-4

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Rolles Theorem

There exists a point c, between aand b where the slope is 0.
Applying Rolle’s Theorem

Check conditions
Take derivative of function
Set derivative function = 0
Solve for cvalue

Most often straight forward. Just check the conditions and apply the method

Rolles 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
lim_{x → 0⁺} f′(x) ≠ lim_{x → 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) = e^x

e^x is a continuous function so it's continuous function. In fact, it can be expressed as a polynomial function
f′(x) = e^x defined everywhere. In fact, it can be expressed as a polynomial function
e^x≠ 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.

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Answer

Rolles 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

Rolle's Theorem

Example 1

Example 2

Example 3

Intro 0:00
Rolle's Theorem 0:07

Conditions
Summary

Example 1 1:09
Example 2 3:08
Example 3 4:48

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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Rolles Theorem

Slide Duration:

Table of Contents

Section 1: Functions

Definitions & Properties of Functions

11m 26s

Intro