AP CALCULUS AB Chapter 4: Applications of Derivatives Section 4.2: Mean Value Theorem.

AP CALCULUS AB

Chapter 4:Applications of Derivatives

Section 4.2:Mean Value Theorem

What you’ll learn about Mean Value Theorem Physical Interpretation Increasing and Decreasing Functions Other Consequences

…and whyThe Mean Value Theorem is an important

theoretical tool to connect the average and instantaneous rates of change.

Mean Value Theorem for Derivatives

If ( ) is continuous at every point of the closed interval , and

differentiable at every point of its interior , , then there is at least

( ) - ( )one point in , at which '( ) .

-

y f x a b

a b

f b f ac a b f c

b a

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.

Example 1: Exploring the Mean Value TheoremShow 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

)()(

)('

Example 1 continuedGiven 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

)()(

)('

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.

Example Explore the Mean Value Theorem

2Show that the function ( ) satisfies the hypothesis of the Mean Value

Theorem on the interval 0,2 . Then find a solution to the equation

( ) - ( )'( ) on this interval.

-

f x x

c

f b f af c

b a

2The function ( ) is continuous on 0,2

and differentiable on 0,2 .

Since (0) 0, (2) 4, and '( ) 2

( ) - ( )'( )

-4 0

22 0

2 2

1.

f x x

f f f x x

f b f af c

b a

c

c

c

Example 2: Mean Value Theorem Ex:

.12

5 So

2

5

52

142

122

114

14

4244

1211 22

4,1on 2

2

2

2

fc

c

c

mc

m

f

fxxf

xxf

Rolle’s Theorem:If f is continuous on the closed intervaland differentiable on the open intervaland if then there is at least one number c insuch that

Rolle’s Theorem

ba, ba,

bfaf ba,

.0 cf

a b

(a, f(a))

(b, f(b)

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

Example 2: Further Exploration of the MVTExplain 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 xxf

1,1

1,3)(

2

3

xx

xxxf

3

1

)( xxf

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 c3) Evaluate f(x) at c, use that point and the slope

from step 1 to find the equation of the tangent line

21)( xxf

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

Physical Interpretation of the Mean Value Theorem

The 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

)()(

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

Increasing Function, Decreasing Function

1 2

1 2 1 2

1 2 1 2

Let be a function defined on an interval and let and

be any two points in .

1. on if ( ) ( ).

2. on if ( ) ( ).

f I x x

I

f I x x f x f x

f I x x f x f x

increases

decreases

Corollary: Increasing and Decreasing Functions

Let be continuous on , and differentiable on , .

1. If ' 0 at each point of , , then increases on , .

2. If ' 0 at each point of , , then decreases on , .

f a b a b

f a b f a b

f a b f a b

Example Determining Where Graphs Rise or Fall

3Where is the function ( ) 2 12 increasing and where is it decreasing?f x x x

2

2

2

2

The function is increasing when '( ) 0.

6 12 0

2

2 or 2

The function is decreasing when '( ) 0.

6 12 0

2

2 2

f x

x

x

x x

f x

x

x

x

Corollary: Functions with f’=0 are Constant If '( ) 0 at each point of an interval , then there is a

constant for which ( ) for all in .

f x I

C f x C x I

Section 4.2 – Mean Value Theorem To find the intervals on which a function is

increasing or decreasing1. Locate the critical numbers of f in Use

these numbers to determine the test intervals. (when or is undefined, and the endpoints of the interval).

2. Determine the sign of at a value in each of the test intervals.

3. Decide whether f is increasing or decreasing on each interval.

.,ba

0 xf

xf

f’(x) a + b - c

f(x) inc. dec.

Section 4.2 – Mean Value Theorem Ex:

1 ,3

1 :numbers Critical

01 013

0113

113

143

522

23

x

xx

xx

xx

xxxf

xxxxf

1/3 1f’(x)

f(x)

+ +

inc. dec. inc.

.1,3

1on decreasing is

.,13

1,on increasing is

xf

xf

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?

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

Corollary: Functions with the Same Derivative Differ by a Constant If '( ) '( ) at each point of an interval , then there is a constant

such that ( ) ( ) for all in .

f x g x I

C f x g x C x I

. then ,3 if So,

31 35

32

2323

Cxyxdx

dy

xxdx

dxx

dx

d

Example:

Antiderivative A function ( ) is an of a function ( ) if '( ) ( )

for all in the domain of . The process of finding an antiderivative

is .

F x f x F x f x

x f

antiderivative

antidifferentiation

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)

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).

Find each antiderivative – don’t forget C!

1) f’(x) = -sin x

2) f’(x) = 2x + 6

3) f’(x) = 2x2 + 4x - 3

Specific AntiderivativesFind the antiderivative ofthrough 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

Example Finding Velocity and Position

2

Find the velocity and position functions of a freely falling body for

the following set of conditions:

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

Assume that the body is released at time 0.

Velocity: We know that ( ) 9.8 and (0) 0, so

( ) 9.8

(0) 0

0. The velocity function is ( ) 9.8 .

Position: We know that ( ) 9.8 and (0) 0,

t

a t v

v t t C

v C

C v t t

v t t s

2

2

so

( ) 4.9

(0) 0

0. The position function is ( ) 4.9 .

s t t C

s C

C s t t

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)

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.