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College Calculus: Level I Mean Value Theorem and Rolle's Theorem

Section 5: Application of Derivatives: Lecture 2 | 20:00 min

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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

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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
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.

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]

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)] ∫₀^π sinx dx
f(c) = [1/(π)] (−cosx) |₀^π
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)] ∫₀^2π sinx dx
f(c) = [1/(2π)] (−cosx) |₀^2π
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)] ∫₁^e [1/x] dx
f(c) = [1/(e − 1)] lnx |₁^e
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)] ∫₀^2π cosx + 4 dx
f(c) = [1/(2π)] (sinx + 4x) |₀^2π
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)] ∫₀^π tan[x/4] dx
u = [x/4]
du = [1/4] dx
f(c) = [1/(π)] 4 ∫₀^π tanu du
f(c) = [4/(π)] ∫₀^π [sinu/cosu] du
v = cosu
dv = −sinu du
f(c) = [4/(π)] (−1) ∫₀^π [1/v] dv
f(c) = − [4/(π)] lnv |₀^π
f(c) = − [4/(π)] lncosu |₀^π
f(c) = − [4/(π)] lncos[x/4] |₀^π
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)]

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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

Theorems

Lecture Example 1

Lecture Example 2

Lecture Example 3

Additional Example 4

Additional Example 5

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

College Calculus 1 Online Course

Section 1: Overview of Functions
	Review of Functions	26:29
	Compositions of Functions	12:29
Section 2: Limits
	Average and Instantaneous Rates of Change	20:59
	Limit Investigations	22:37
	Algebraic Evaluation of Limits	28:19
	Formal Definition of a Limit	23:39
	Continuity and the Intermediate Value Theorem	19:09
Section 3: Derivatives, part 1
	Limit Definition of the Derivative	22:52
	The Power Rule	26:01
	The Product Rule	14:54
	The Quotient Rule	19:17
	Applications of Rates of Change	17:43
	Trigonometric Derivatives	26:58
	The Chain Rule	23:47
	Inverse Trigonometric Functions	27:05
	Equation of a Tangent Line	15:52
Section 4: Derivatives, part 2
	Implicit Differentiation	30:05
	Higher Derivatives	13:16
	Logarithmic and Exponential Function Derivatives	17:42
	Hyperbolic Trigonometric Function Derivatives	14:30
	Related Rates	29:05
	Linear Approximation	23:52
Section 5: Application of Derivatives
	Absolute Minima and Maxima	18:57
	Mean Value Theorem and Rolle's Theorem	20:00
	First Derivative Test, Second Derivative Test	27:11
	L'Hopital's Rule	23:09
	Curve Sketching	40:16
	Applied Optimization	25:37
	Newton's Method	25:13
Section 6: Integrals
	Approximating Areas and Distances	36:50
	Riemann Sums, Definite Integrals, Fundamental Theorem of Calculus	22:02
	Substitution Method for Integration	23:19
Section 7: Application of Integrals, part 1
	Area Between Curves	19:59
	Volume by Method of Disks and Washers	24:22
	Volume by Method of Cylindrical Shells	30:29
	Average Value of a Function	16:31
Section 8: Extra
	Graphs of f, f', f''	23:58
	Slope Fields for Differential Equations	18:32
	Separable Differential Equations	17:04

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	Calculus by Larson, 9th Edition Authors: Ron Larson, Bruce H. Edwards ISBN: 0547167024 Publisher: Brooks Cole Year: 2009 The book is filled with examples and detailed explanations. Capstone exercises are also included for each section and synthesize the main concepts into a single example.
	Calculus by Larson, 9th Edition, Solutions Manual, Volume 1 Author: Ron Larson ISBN: 547213093 Publisher: Brooks Cole Year: 2008 This solutions manual includes worked out solutions to every odd-numbered exercise in Calculus of a Single Variable, 9th edition.
	Calculus by Larson, 9th Edition, Solutions Manual, Volume 2 Authors: Ron Larson, Bruce H. Edwards ISBN: 547213107 Publisher: Brooks Cole Year: 2009 This solutions manual includes worked out solutions to every odd-numbered exercise in Multivariable Calculus, 9th edition.

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Professor Switkes

Mean Value Theorem and Rolle's Theorem

Slide Duration:

Table of Contents

Section 1: Overview of Functions

Review of Functions

26m 29s

Intro