Section 4.2 Mean Value Theorem

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Section 4.2 Mean Value Theorem What you’ll learn Mean Value Theorem Physical Interpretation Increasing and Decreasing Functions Other Consequences Why? The Mean Value Theorem is an important theoretical tool to connect the average and instantaneous rate of change.

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Section 4.2 Mean Value Theorem. What you’ll learn Mean Value Theorem Physical Interpretation Increasing and Decreasing Functions Other Consequences Why? The Mean Value Theorem is an important theoretical tool to connect the average and instantaneous rate of change. Warm Up. - PowerPoint PPT Presentation

Transcript of Section 4.2 Mean Value Theorem

Page 1: Section 4.2  Mean Value Theorem

Section 4.2 Mean Value Theorem

What you’ll learnMean Value TheoremPhysical InterpretationIncreasing and Decreasing FunctionsOther Consequences

Why?The Mean Value Theorem is an important theoretical tool to connect the average and instantaneous rate of change.

Page 2: Section 4.2  Mean Value Theorem

Warm Up

Group Activity:

AP Test Prep Series Pages 131 – 133

Due in 15 minutes

Correct Homework

Page 3: Section 4.2  Mean Value Theorem

This means…….

IFy = f(x) is continuousy = f(x) is on a closed interval [a,b]y = f(x) is differentiable at every point in its interior

(a,b)

THENSomewhere between points A and B on a

differentiable curve, there is at least one tangent line parallel to chord AB.

http://justmathtutoring.com/ Mean Value Thm.

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Example 1: Exploring the Mean Value Theorem

Show that the function f(x) = x2 satisfies the hypothesis of

the Mean Value Theorem on the interval [0,2]. Then find a solution c to the equation

Consider f(x) = x2. Is it continuous? Closed interval? Differentiable?

If so, by the MVT we are guaranteed a point c in the interval [0,2] for which ab

afbfcf

)()(

)('

ab

afbfcf

)()(

)('

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Example 1 continued

Given f(x) = x2

Interval [0,2]

To use the MVT Find the slope of the chord

with endpoints (0, f(0)) and (2, f(2)).

Find f’

Set f’ equal to the slope of the chord, solve for c

Find c

ab

afbfcf

)()(

)('

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Interpret your answer

The slope of the tangent line to f(x) = x2 at

x = 1 is equal to the slope of the chord AB.

OR The tangent line at x = 1 is parallel to chord AB.

Write equations for line AB and the tangent line of y = x2 at x=1.

Graph and investigate. The lines should be parallel.

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Homework

Lesson 4.2

Quick Review Ex 1-10

Watch justmathtutoring

Mean Value Theorem

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Warm Up

Page 195

Exercise 51 a-c

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Theorem 3: Mean Value Theorem for Derivatives If y = f(x) is continuous at every point on the

closed interval [a, b] and differentiable at every point of its interior (a,b), then there is at least one point c in (a,b) at which

(The derivative at some point c = the slope of the chord.)

ab

afbfcf

)()(

)('

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Show that the function satisfies the hypothesis of the MVT on the interval [0,1]. Then find c

1. Is it continuous? Closed interval? Differentiable?

2. Find c

3. Interpret your findings

12)( 2 xxxf

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Example 2: Further Exploration of the MVT

Explain why each of the following functions fails to satisfy the conditions of the Mean Value Theorem on the interval [-1,1].

You try:

1)( 2 xxf1,1

1,3{)(

2

3

xx

xxxf

3

1

)( xxf

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Example 3: Applying the Mean Value Theorem

Find a tangent to f in the interval (-1,1) that is parallel to the secant AB.

Given: , A = (-1,f(-1)) and B = (1, f(1))

1) Find the slope of AB and f’(c)

2) Apply MVT to find c

3) Evaluate f(x) at c, use that point and the slope from step 1 to find the equation of the tangent line

21)( xxf

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You Try -

1) Find slope of chord AB that connects endpoints

2) Find f ’ and apply MVT formula to find c.

3) Evaluate f(x) at c, use that point and the slope from step 1 to find the equation of the tangent line. Graph & check.

31 x1)( xxf

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Physical Interpretation of the Mean Value TheoremThe MVT says the instantaneous change at

some interior point must equal the average change over the entire interval.

f’(x) = instantaneous change at a point

= average change over the

interval

ab

afbf

)()(

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Example 4: Interpreting the MVTIf a car accelerating from zero takes 8 sec to go

352 ft, its average velocity for the 8-second interval is 352 / 8 = 44 ft/sec, or 30 mph.

Can we cite the driver for speeding if he / she is in a residential area with a speed limit of 25 mph?

hour

miles

hourft

mileft 30

1

sec3600

5280

1

sec

44

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Homework

Page 202

Exercises 1-11 Odds, 12-14

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Today’s Agenda

Present Homework on board

Page 202 Exercises 1-11 Odds, 12-14

4.2 Power Point

Examples 5-8

Start today’s homework

Page 203 Exercises 15-33 odds, 37, 43, 45

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4.2 Continued:Increasing Functions / Decreasing Functions

Let f be a function defined on an interval I and

let x1 and x2 be any two points on I

f increases on I if x1 < x2 => f(x1) < f(x2)

f decreases on I if x1 < x2 => f(x1) > f(x2)

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Corollaries to the Mean Value TheoremCorollary 1: Increasing & Decreasing Functions

Let f be continuous on [a,b] and differentiable on (a,b).1) If f ’ > 0 at each point of (a,b), then f increases on [a,b]2) If f ’< 0 at each point of (a,b), then f decreases on [a,b]

Corollary 2: Functions with f’ = 0 are ConstantIf f ’(x) = 0 at each point of an interval I, then there is a constant C

for which f(x) = C for all x in I

Corollary 3: Functions with the same derivative differ by a constant.If f ’(x) = g ’(x) at each point of an interval I, then there is a constant

C such that f(x) = g(x) + C for all x in I.

http://justmathtutoring.com/ Using 1st derivative to find where f is increasing or decreasing, find max or min points

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Example 5 Determining where graphs rise or fall

Use corollary 1 to determine where the graph of

f(x) = x2 – 3x is increasing and decreasing.

1) Find f’ and set it equal to zero to find critical points.

2) Where f ’ > 0, f is increasing.3) Where f ’ < 0, f is decreasing.4) Any maximum or minimum values?

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Example 6 Determining where graphs rise or fall

Where is the function increasing and where is it decreasing?

Graphically: Use window [-5,5] by [-5,5]Confirm Analytically: Find f ’, evaluate f ’ = 0 to find critical points.

Where f ’ > 0, f is increasing. Where f ’ < 0, f is decreasing.

xxxf 4)( 3

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Example 7 Applying Corollary 3

Find the function f(x) whose derivative is sin x and whose graph passes through the point (0,2).

Write f(x) = antiderivative function + C Use (x,y) = (0, 2) in equation, solve for C.Write f(x), the antiderivative of sin x through (0, 2)

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Definition: Antiderivative

A function F(x) is an antiderivative of a function f(x) if F’(x) = f(x) for all x in the domain of f. The process of finding an antiderivative is antidifferentiation.

If you are given a derivative function, “Think Backwards” to get the original using all of the chapter 3 differentiation relationships.

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Find the Antiderivative!

Find the antiderivative of

f(x) = 6x2.

Reverse the power rule Add 1 to the exponent Divide the coefficient by

(exponent + 1) Add C You have found the

antiderivative, F(x)

Find the antiderivative of

f(x) = cos x

Think Backwards: What function has a derivative of cos x?

Add C

You have found the antiderivative, F(x).

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Find each antiderivative – don’t forget C!1) f’(x) = -sin x

2) f’(x) = 2x + 6

3) f’(x) = 2x2 + 4x - 3

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Specific Antiderivatives

Find the antiderivative of

through the point (0, 1).

Write equation for antiderivative + c Insert point (0,1) for (x, y) and solve for c Write specific antiderivative.

3

1

3

2)( xxf

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Example 8 Finding Velocity and Position

We can use antidifferentiation to find the velocity and position functions of a body falling freely from a height of 0 meters under each of the following sets of conditions.

a) The acceleration is 9.8 m/sec2 and the body falls from rest.

b) The acceleration is 9.8 m/sec2 and the body is propelled downward with an initial velocity of 1 m/sec2.

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The acceleration is 9.8 m/sec2 and the body falls from rest.

Work backwards

Velocity function + c, P(0,0) (why?)

Position function + c (from velocity function)

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The acceleration is 9.8 m/sec2 and the body is propelled downward with an initial velocity of 1 m/sec2.

Work backwards

Velocity function + c, P(0,1) (why?)

Position function + c (from velocity function)

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SummaryThe Mean Value Theorem tells us that If y = f(x) is continuous at every point on the closed interval [a, b] and differentiable at

every point of its interior (a,b), then there is at least one point c in (a,b) at which(The derivative at point c = the slope of the chord.)

It’s corollaries go on to tell us that where f ‘ is positive, f is increasing, where f ‘ = 0, f is a constant function, and where f ‘ is negative, f is decreasing.

We can work backwards from a derivative function to the original function, a process called antidifferentiation. However, as the derivative of any constant = 0, we need to know a point of the original function to get its specific antiderivative. Without a point of the function all we can determine is a general formula for f(x) + C.

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Homework

Page 203Exercises 15-33 odds,

37, 43, 45