Rolle’s theorem and Mean Value Theorem ( Section 4.2)

1 / 17

# Rolle’s theorem and Mean Value Theorem ( Section 4.2) - PowerPoint PPT Presentation

Rolle’s theorem and Mean Value Theorem ( Section 4.2). Alex Karassev. Rolle’s Theorem. y. y. f ′ (c) = 0. y = f(x). y = f(x). f(a) = f(b). x. x. c 3. b. c 2. a. c 1. a. b. c. Example 1. Example 2. Rolles Theorem. Suppose f is a function such that f is continuous on [a,b]

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about 'Rolle’s theorem and Mean Value Theorem ( Section 4.2)' - rachel-puckett

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

### Rolle’s theorem and Mean Value Theorem(Section 4.2)

Alex Karassev

Rolle’s Theorem

y

y

f ′ (c) = 0

y = f(x)

y = f(x)

f(a) = f(b)

x

x

c3

b

c2

a

c1

a

b

c

Example 1

Example 2

Rolles Theorem
• Suppose f is a function such that
• f is continuous on [a,b]
• differentiable at least in (a,b)
• If f(a) = f(b) then there exists at least one c in (a,b) such that f′(c) = 0

y

f ′ (c) = 0

y = f(x)

x

a

b

c

Applications of Rolle’s Theorem
• It is used in the prove of theMean Value Theorem
• Together with the Intermediate Value Theorem, it helps to determine exact number of roots of an equation

y = cos x

Example
• Prove that the equation cos x = 2x has exactly one solution

y = 2x

Solution
• cos x = 2x has exactly one solution means two things:
• It has a solution – can be proved using the IVT
• It has no more than one solution – can be proved using Rolle's theorem
cos x = 2x has a solution:

Proof using the IVT

• cos x = 2x is equivalent to cos x – 2x = 0
• Let f(x) = cos x – 2x
• f is continuous for all x, so the IVT can be used
• f(0) = cos 0 – 2∙0 = 1 > 0
• f(π/2) = cos π/2 – 2 ∙π/2 = 0 – π = – π < 0
• Thus by the IVT there exists c in [0, π/2] such thatf(c) = 0, so the equation has a solution

Proof by contradiction using Rolle's theorem

f(x) = cos x – 2x = 0 has no more than one solution:
• Assume the opposite: the equation has at least two solutions
• Then there exist two numbers a and b s.t. a ≠ b andf(a) = f(b) = 0
• In particular, f(a) = f(b)
• f(x) is differentiable for all x, and hence Rolle's theorem is applicable
• By Rolle's theorem, there exists c in [a,b] such that f′(c) = 0
• Find the derivative: f ′ (x) = (cos x – 2x) ′ = – sin x – 2
• So if we had f ′ (c) = 0 it would mean that –sin c – 2 = 0 orsin c = – 2, which is impossible!
• Thus we obtained a contradiction
• All our steps were logically correct so the fact that we obtained a contradiction means that our original assumption "the equation has at least two solutions" was wrong
• Thus, the equation has no more than one solution!

If it had two roots, then there would exist a ≠ b such thatf(a) = f(b) = 0

f ′ (c) = 0

y = f(x)

b

a

c

But f ′ (x) = (cos x – 2x) ′ = – sin x – 2

So if we had f ′ (c) = 0 it would mean that –sin c – 2 = 0 orsin c = – 2, which is impossible

y

y

y = f(x)

y = f(x)

x

x

b

a

b

a

c

Example 2

Example 1

Mean Value Theorem

There exists at least one point on the graphat which tangent line is parallel to the secant line

f(b) – f(a) (b-a)

f′(c) =

Mean Value Theorem
• Slope of secant line is the slope of line through the points (a,f(a)) and (b,f(b)), so it is
• Slope of tangent line is f′(c)

y

y = f(x)

x

b

a

c

f(b) – f(a) (b-a)

f′(c) =

MVT: exact statement
• Suppose f is continuous on [a,b] and differentiable on (a,b)
• Then there exists at least one point c in (a,b) such that

y

y = f(x)

x

b

a

c

Interpretation of the MVTusing rate of change
• Average rate of changeis equal to the instantaneous rate of changef′(c) at some moment c
• Example: suppose the cities A and B are connected by a straight road and the distance between them is 360 km. You departed from A at 1pm and arrived to B at 5:30pm. Then MVT implies that at some moment your velocity v(t) = s′(t) was:(s(5.5) – s(1)) / (5.5 – 1) = 360 / 4.5 = 80 km / h
Application of MVT
• Estimation of functions
• Connection between the sign of derivative and behavior of the function:

if f ′ > 0 function is increasing

if f ′ < 0 function is decreasing

• Error bounds for Taylor polynomials
Example
• Suppose that f is differentiable for all x
• If f (5) = 10 and f ′ (x)≤ 3 for all x, how small can f(-1) be?
Solution
• MVT: f(b) – f(a) = f ′(c) (b – a) for some c in (a,b)
• Applying MVT to the interval [ –1, 5], we get:
• f(5) – f(–1) = f ′(c) (5 – (– 1)) = 6 f ′(c) ≤ 6∙3 = 18
• Thus f(5) – f(-1) ≤ 18
• Therefore f(-1) ≥ f(5) – 18 = 10 – 18 = – 8