absolute extreme values are either maximum or minimum points on a curve. they are sometimes called...

Absolute extreme values are either maximum or minimum points on a curve.

They are sometimes called global extremes.

They are also sometimes called absolute extrema.(Extrema is the plural of the Latin extremum.)

4.1 Extreme Values of Functions


Definition Absolute Extreme ValuesLet f be a function with domain D. Then f (c) is the

a. absolute minimum value on D if and only if f(x) < f (c) for all x in D.b. absolute maximum value on D if and only if f(x) > f (c) for all x in D.

Extreme values can be in the interior or the end points of a function.

2y x

,D Absolute Minimum

No AbsoluteMaximum


2y x

0,2D Absolute Minimum

AbsoluteMaximum


2y x

0,2D No Minimum

AbsoluteMaximum


2y x

0,2D No Minimum

NoMaximum


Extreme Value Theorem:

If f is continuous over a closed interval, [a,b] then f has a maximum and minimum value over that interval.

Maximum & minimumat interior points

Maximum & minimumat endpoints

Maximum at interior point, minimum at endpoint


Local Extreme Values:

A local maximum is the maximum value within some open interval.

A local minimum is the minimum value within some open interval.


Absolute minimum(also local minimum)

Local maximum

Local minimum

Absolute maximum

(also local maximum)

Local minimum

Local extremes are also called relative extremes.


Local maximum

Local minimum

Notice that local extremes in the interior of the function

occur where is zero or is undefined.f f

Absolute maximum

(also local maximum)


Local Extreme Values:

If a function f has a local maximum value or a local minimum value at an interior point c of its domain, and if exists at c, then

0f c

f


Critical Point:

A point in the domain of a function f at which

or does not exist is a critical point of f .0f

f

Note:Maximum and minimum points in the interior of a function always occur at critical points, but critical points are not always maximum or minimum values.


EXAMPLE 3 FINDING ABSOLUTE EXTREMA

Find the absolute maximum and minimum values of on the interval . 2/3f x x 2,3

2/3f x x

1

32

3f x x

3

2

3f x

x

There are no values of x that will makethe first derivative equal to zero.

The first derivative is undefined at x=0,so (0,0) is a critical point.

Because the function is defined over aclosed interval, we also must check theendpoints.


0 0f

To determine if this critical point isactually a maximum or minimum, wetry points on either side, withoutpassing other critical points.

2/3f x x

1 1f 1 1f

Since 0<1, this must be at least a local minimum, and possibly a global minimum.

2,3D

At: 0x

At: 2x 2

32 2 1.5874f

At: 3x 2

33 3 2.08008f


0 0f

2/3f x x 2,3D

At: 2x 2

32 2 1.5874f

At: 3x

Absoluteminimum:

Absolutemaximum:

0,0

3,2.08

2

33 3 2.08008f



y = x2/3

Finding Maximums and Minimums Analytically:

1 Find the derivative of the function, and determine where the derivative is zero or undefined. These are the critical points.

2 Find the value of the function at each critical point.

3 Find values or slopes for points between the critical points to determine if the critical points are maximums or minimums.

4 For closed intervals, check the end points as well.



Find the absolute maximum and minimum of the function

]2,1[,2452)( 23 onxxxxf

4106)(' 2 xxxf

41060 2 xx

Find the critical numbers

2530 2 xx

)1)(23(0 xx 13

2 xx or



]2,1[,2452)( 23 onxxxxf

Check endpoints and critical numbers

The absolute maximum is 2 when x = -2The absolute minimum is -13 when x = -1

22

1127

26

3

2

131

xfx



]3,0[,1

3)(

2

on

x

xxf 2

2

)1(

)1)(3()2)(1()('

x

xxxxf

320 2 xx


)1)(3(0 xx 13 xx or

2

2

)1(

32)('

x

xxxf



]3,0[,1

3)(

2

on

x

xxf

33

21

30

xfx

Check endpoints and critical numbers

The absolute maximum is 3 when x = 0, 3The absolute minimum is 2 when x = 1



2,0,sinsin)( 2 onxxxf

xxxxf cossin2cos)('


xxx cossin2cos0

)sin21(cos0 xx

0cos x 0sin21 x

2

3,

2

x

6

5,

6

x


Find the absolute maximum andminimum of the function

2,0,sinsin)( 2 onxxxf

02

22

34

1

6

5

02

4

1

6

00

xfx

The absolute maximum is 1/4 when x = /6, 5/6The absolute minimum is –2 when x =3/2

Critical points are not always extremes!

3y x

0f (not an extreme)


1/3y x

is undefined.f

(not an extreme)


If f (x) is a differentiable function over [a,b], then at some point between a and b:

f b f af c

b a

Mean Value Theorem for Derivatives

4.2 Mean Value Theorem


f b f af c

b a


Differentiable implies that the function is also continuous.



f b f af c

b a


Differentiable implies that the function is also continuous.

The Mean Value Theorem only applies over a closed interval.



f b f af c

b a


The Mean Value Theorem says that at some point in the closed interval, the actual slope equals the average slope.


y

x0

A

B

a b

Slope of chord:

f b f a

b a

Slope of tangent:

f c

y f x

Tangent parallel to chord.

c


If f (x) is a differentiable function over [a,b], and if f(a) = f(b) = 0, then there is at least one point c between a and b such that f’(c)=0:

Rolle’s Theorem


(a,0) (b,0)


Show the function satisfies the hypothesis ofthe Mean Value Theorem

3,0oncos)(

xxf

The function is continuous on [0,/3] and differentiable on(0,/3). Since f(0) = 1 and f(/3) = 1/2, the Mean Value Theorem guarantees a point c in the interval (0,/3) for which

f b f af c

b a

csin

03/

12/1

c = .498


(0,1)

(/3,1/2)

at x = .498, the slopeof the tangent line isequal to the slope of the chord.


Definitions Increasing Functions, Decreasing FunctionsLet f be a function defined on an interval I and let x1 and x2

be any two points in I.

1. f increases on I if x1 < x2 f(x1) < f(x2).2. f decreases on I if x1 > x2 f(x1) > f(x2).

A function is increasing over an interval if the derivative is always positive.

A function is decreasing over an interval if the derivative is always negative.

A couple of somewhat obvious definitions:


Corollary Increasing Functions, Decreasing FunctionsLet 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].


Find where the function is increasing and decreasing and find the local extrema.

xxxxf 249)( 23

xxxxf 249)( 23

24183)(' 2 xxxf

)86(30 2 xx

)86(0 2 xx

)2)(4(0 xx

2 4

0 0 f’(x)+-+

),4()2,( inc

)4,2(dec

x = 2, local maximum x = 4, local minimum


(2,20) local max

(4,16) local min

y

x0

y f x

y g x

These two functions have the same slope at any value of x.

Functions with the same derivative differ by a constant.

C


Find the function whose derivative is and whose graph passes through

f x sin x

0,2

cos sind

x xdx

cos sind

x xdx

so:

cosf x x C

2 cos 0 C


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

cos sind

x xdx

cos sind

x xdx

so:

cosf x x C

2 cos 0 C 2 1 C

3 C

cos 3f x x Notice that we had to have initial values to determine the value of C.


The process of finding the original function from the derivative is so important that it has a name:

Antiderivative

A function is an antiderivative of a function

if for all x in the domain of f. The process

of finding an antiderivative is antidifferentiation.

F x f x

F x f x

You will hear much more about antiderivatives in the future.

This section is just an introduction.


Since acceleration is the derivative of velocity, velocity must be the antiderivative of acceleration.

Example 7b: Find the velocity and position equations for a downward acceleration of 9.8 m/sec2 and an initial velocity of 1 m/sec downward.

9.8a t

9.8 1v t t

1 9.8 0 C

1 C

9.8v t t C (We let down be positive.)


Since velocity is the derivative of position, position must be the antiderivative of velocity.

9.8a t

9.8 1v t t

1 9.8 0 C

1 C

9.8v t t C 29.8

2s t t t C

The power rule in reverse: Increase the exponent by one and multiply by the reciprocal of the new exponent.


9.8a t

9.8 1v t t

1 9.8 0 C

1 C

9.8v t t C 29.8

2s t t t C

24.9s t t t C The initial position is zero at time zero.

20 4.9 0 0 C

0 C

24.9s t t t


In the past, one of the important uses of derivatives was as an aid in curve sketching. We usually use a calculator of computer to draw complicated graphs, it is still important to understand the relationships between derivatives and graphs.

4.3 Connecting f’ and f’’ with the Graph of f

First Derivative Test for Local Extrema at a critical point c


1. If f ‘ changes sign from positive to negative at c, then f has a local maximum at c.

local max

f’>0 f’<0

2. If f ‘ changes sign from negative to positive at c, then f has a local minimum at c.

3. If f ‘ changes does not change sign at c, then f has no local extrema.

local min

f’<0 f’>0

no extreme

f’>0 f’>0

First derivative:

y is positive Curve is rising.

y is negative Curve is falling.

y is zero Possible local maximum or minimum.



Definition Concavity

The graph of a differentiable function y = f(x) is

a. concave up on an open interval I if y’ is increasing on I. (y’’>0)b. concave down on an open interval I if y’ is decreasing on I. (y’’<0)

concave down

concave up


Second Derivative Test for Local Extrema at a critical point c

1. If f’(c) = 0 and f’’(c) < 0, then f has a local maximum at x = c.2. If f’(c) = 0 and f’’(c) > 0, then f has a local minimum at x = c.

+ +

Second derivative:

y is positive Curve is concave up.

y is negative Curve is concave down.

y is zero Possible inflection point(where concavity changes).



Definition Point of Inflection

A point where the graph of a function has a tangent line andwhere the concavity changes is called a point of inflection.

inflection point

23 23 4 1 2y x x x x

23 6y x x 0ySet

20 3 6x x 20 2x x

0 2x x 0, 2x

First derivative test:

y0 2

0 0

21 3 1 6 1 3y negative

21 3 1 6 1 9y positive

23 3 3 6 3 9y positive

Possible extreme at .0, 2x


Sketch the graph

zeros at x = -1, x = 2

23 6y x x

0ySet

20 3 6x x

20 2x x

0 2x x

0, 2x

First derivative test:

y0 2

0 0

maximum at 0x

minimum at 2x



23 6y x x

0ySet

20 3 6x x

20 2x x

0 2x x

0, 2x


Or you could use the second derivative test:

maximum at 0x minimum at 2x

6 6y x

0 6 0 6 6y negativeconcave downlocal maximum

2 6 2 6 6y positiveconcave uplocal minimum


6 6y x

We then look for inflection points by setting the second derivative equal to zero.

0 6 6x

6 6x

1 x

Possible inflection point at .1x

y1

0

0 6 0 6 6y negative

2 6 2 6 6y positive

inflection point at 1x


Make a summary table: x y y y

1 0 9 12 rising, concave down

0 4 0 6 local max

1 2 3 0 falling, inflection point

2 0 0 6 local min

3 4 9 12 rising, concave up


A Classic Problem

You have 40 feet of fence to enclose a rectangular garden along the side of a barn. What is the maximum area that you can enclose?

4.4 Modeling and Optimization

x x

40 2x

40 2A x x

240 2A x x

40 4A x

0 40 4x

4 40x

10x

10 40 2 10A

10 20A

2200 ftA 40 2l x

w x 10 ftw

20 ftl


To find the maximum (or minimum) value of a function:


1. Understand the Problem.2. Develop a Mathematical Model.3. Graph the Function.4. Identify Critical Points and Endpoints.5. Solve the Mathematical Model.6. Interpret the Solution.

What dimensions for a one liter cylindrical can will use the least amount of material?

We can minimize the material by minimizing the area.

22 2A r rh

area ofends

lateralarea

We need another equation that relates r and h:

2V r h

31 L 1000 cm

21000 r h

2

1000h

r

22

10 02

02A r r

r

2 20002A r

r

2

20004A r

r


22 2A r rh area ofends

lateralarea

2V r h

31 L 1000 cm21000 r h

2

1000h

r

22

10 02

02A r r

r

2 20002A r

r

2

20004A r

r

2

20000 4 r

r

2

20004 r

r

32000 4 r

3500r

3500

r

5.42 cmr

2

1000

5.42h

10.83 cmh



Find the radius and height ofthe right-circular cylinder oflargest volume that can beinscribed in a right-circularcone with radius 6 in. andheight 10 in. h

r

10 in

6 in


h

r

10 in

6 in

The formula for the volume ofthe cylinder is hrV 2

To eliminate one variable, weneed to find a relationshipbetween r and h.

6

1010

r

h

rh3

510

6

h

10-h

r10


h

r

10 in

6 in

hrV 2

322

3

510

3

510 rrrrV

2520 rrdr

dV

)4(50 rr

4,0 rr


h

r

10 in

6 in

Check critical points and endpoints.

•r = 0, V = 0•r = 4 V = 160/3•r = 6 V = 0

The cylinder will have a maximum volume when r = 4 in. and h = 10/3 in.

Determine the point on the curve y = x2 that is closest to the point (18, 0).


22)18( yxd

42)18( xxd

Substitute for x

42 )32436( xxxd

)3624()32436(2

1 32

124

xxxxx

dx

ds



)3624()32436(2

1 32

124

xxxxx

dx

ds

0dx

dsset 36240 3 xx 1820 3 xx

2x 4y



1820 3 xx

2x 4y

)942)(2(0 2 xxx

2

- 0 +

If the end points could be the maximum or minimum, you have to check.

Notes:

If the function that you want to optimize has more than one variable, use substitution to rewrite the function.

If you are not sure that the extreme you’ve found is a maximum or a minimum, you have to check.


For any function f (x), the tangent is a close approximation of the function for some small distance from the tangent point.

y

x0 x a

f x f aWe call the equation of the tangent the linearization of the function.

4.5 Linearization and Newton’s Method

The linearization is the equation of the tangent line, and you can use the old formulas if you like.

Start with the point/slope equation:

1 1y y m x x 1x a 1y f a m f a

y f a f a x a y f a f a x a

L x f a f a x a linearization of f at a

f x L x is the standard linear approximation of f at a.


Find the linearization of f(x) = x4 + 2x at x = 2

L x f a f a x a


f’(x) = 4x3 + 2

L (x) = f(3) + f’(3)(x - 3)

L (x) = 87 + 110(x - 3)

L (x) = 110x - 243

Important linearizations for x near zero:

1k

x 1 kx

sin x

cos x

tan x

x

1

x

1

21

1 1 12

x x x

13 4 4 3

4 4

1 5 1 5

1 51 5 1

3 3

x x

x x

f x L x

This formula also leads to non-linear approximations:



Estimate using local linearization. 37

2

1

2

1)('

)(

xxf

xxf L x f a f a x a

)3637)(36(')36()37( ffL

)1(12

16)37( L

0833.6)37( L


Estimate sin 31º using local linearization.

xxf

xxf

cos)('

sin)(

L x f a f a x a

180

)30(')30()31(

ffL

1802

3

2

1)31(

L

360

3180)31(

L

Need tobe in radians

Differentials:

When we first started to talk about derivatives, we said

that becomes when the change in x and

change in y become very small.

y

x

dy

dx

dy can be considered a very small change in y.

dx can be considered a very small change in x.


Let y = f(x) be a differentiable function.

The differential dx is an independent

variable.

The differential dy is: dy = f ’(x)dx


Example: Consider a circle of radius 10. If the radius increases by 0.1, approximately how much will the area change?

2A r2 dA r dr

2 dA dr

rdx dx

very small change in Avery small change in r

2 10 0.1dA 2dA (approximate change in area)


Compare to actual change:

New area:

Old area:

210.1 102.01

210 100.00


01.2A

2dAAbsoluteerror

%2100

2

A

dA

%01.2100

01.2

A

A

percenterror


True Estimated

Absolute Change

Relative Change

Percent Change

)()( afdxaff dxafdf )('

)(af

f)(af

df

%100)(

xaf

df%100

)(x

af

f


Newton’s Method

0 x

y

y = f(x)

Rootsought

x1

First

(x1,f(x1))

x2

Second

x3

Third

(x2,f(x2))

(x3,f(x3))

)( 1212 xxmyy

))((')(0 121 xxxfxf ))((')(0 121 xxxfxf

)(')(')( 1121 xfxxfxxf

)('

)(

1

112 xf

xfxx

This is Newton’s Method of finding roots. It is an example of an algorithm (a specific set of computational steps.)

Newton’s Method: 1

nn n

n

f xx x

f x

This is a recursive algorithm because a set of steps are repeated with the previous answer put in the next repetition. Each repetition is called an iteration.


Newton’s Method

213

2f x x

Finding a root for:

We will use Newton’s Method to find the root between 2 and 3.


Newton’s Method 213

2f x x


xxf )('

Guess x1 = 2

)('

)(

1

112 xf

xfxx

5.22

122

x

Newton’s Method 213

2f x x


xxf )('

Guess x2 = 2.5

)('

)(

2

223 xf

xfxx

45.25.2

125.5.23 x

Find where crosses .3y x x 1y

31 x x 30 1x x

3 1f x x x

23 1f x x


nx nf xn nf x 1

nn n

n

f xx x

f x

0 1 1 2 11 1.5

2

1 1.5 .875 5.75.875

1.5 1.34782615.75

2 1.3478261 .1006822 4.4499055 1.3252004

31.3252004 1.3252004 1.0020584 1


There are some limitations to Newton’s Method:

Wrong root found

Looking for this root.

Bad guess.

Failure to converge


First, a review problem:

Consider a sphere of radius 10 cm.

If the radius changes 0.1 cm (a very small amount) how much does the volume change?

34

3V r 24dV r dr

24 10cm 0.1cmdV 340 cmdV

The volume would change by approximately 40 cm3 .

4.6 Related Rates

Now, suppose that the radius is changing at an instantaneous rate of 0.1 cm/sec.

34

3V r 24

dV drr

dt dt

2 cm4 10cm 0.1

sec

dV

dt

3cm40

sec

dV

dt

The sphere is growing at a rate of 40 cm3/sec .

Note: This is an exact answer, not an approximation like we got with the differential problems.

4.6 Related Rates

Water is draining from a cylindrical tank at 3 liters/second. How fast is the surface dropping?

L3

sec

dV

dt

3cm3000

sec

Finddh

dt2V r h

2dV dhr

dt dt

(r is a constant.)

32cm

3000sec

dhr

dt

3

2

cm3000

secdh

dt r

(We need a formula to relate V and h. )

4.6 Related Rates

Steps for Related Rates Problems:

1. Draw a picture (sketch).

2. Write down known information.

3. Write down what you are looking for.

4. Write an equation to relate the variables.

5. Differentiate both sides with respect to t.

6. Evaluate.

4.6 Related Rates

Hot Air Balloon Problem:

Given:4

rad

0.14min

d

dt

How fast is the balloon rising?

Finddh

dttan

500

h

2 1sec

500

d dh

dt dt

2

1sec 0.14

4 500

dh

dt

h

500ft

4.6 Related Rates

Hot Air Balloon Problem:

Given:4

rad

0.14min

d

dt

How fast is the balloon rising?

Finddh

dttan

500

h

2 1sec

500

d dh

dt dt

2

1sec 0.14

4 500

dh

dt

h

500ft

2

2 0.14 500dh

dt

1

12

4

sec 24

ft140

min

dh

dt

4.6 Related Rates

4x

3y

B

A

5z

Truck Problem:Truck A travels east at 40 mi/hr.Truck B travels north at 30 mi/hr.

How fast is the distance between the trucks changing 6 minutes later?

r t d 1

40 410

130 3

10

2 2 23 4 z 29 16 z 225 z 5 z

4.6 Related Rates

4x

3y

30dy

dt

40dx

dt

B

A

5z

Truck Problem:


r t d 1

40 410

130 3

10

2 2 23 4 z 29 16 z

2 2 2x y z

2 2 2dx dy dz

x y zdt dt dt

4 40 3 30 5dz

dt

Truck A travels east at 40 mi/hr.Truck B travels north at 30 mi/hr.

4.6 Related Rates

250 5dz

dt 50

dz

dt

miles50

hour

4.6 Related RatesTruck Problem:


Truck A travels east at 40 mi/hr.Truck B travels north at 30 mi/hr.

absolute extreme values are either maximum or minimum points on a curve. they are sometimes called...

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