• Calculus Date: 12/18/13 Obj: SWBAT apply first derivative testhttp://youtu.be/PBKnttVMbV4 first derivative test inc. dec.Today –ICover the second derivative test (easier than 1st)https://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/graphingdirectory/

n class: Start WS3-3A; complete for homeworkTomorrow Friday: With Christian complete anyremaining worksheets, make sureYou understand material from thetest.,etc.• Announcements:• Break Packet online on Friday• Merry Christmas if I don’t see you

"Do not judge me by my successes, judge me by how many times

I fell down and got back up again.“Nelson Mandela

Second DerivativeIf ( ) is a function of , then the function

( ) denotes the first derivative of ( ).Now, the derivative of ( ) is denoted by

( ) and called the second derivative ofthe function

y f x xy f x f x

y f xy f x

y

( ).f x2

2Notation: ( ) is also denoted by d ff xdx

Second Derivative TestThe first derivative test tells whether the function, f(x), is increasing, decreasing or constant.

The second derivative test tells whether the derivative, f ’(x), is increasing, decreasing or constant,or whether a function is concave up or concave down and whether a point is a local max or min.

In both cases is increasing. However, in the first case curves down and in the second case curves up.

ff f

Second Derivative

( ) is ( ) i

so,

s

( ) 0

f xf x

f x

( ) is ( ) i

so,

s

( ) 0

f xf x

f x

Second Derivative

Concave upward

Look at these two graphs. Each is concave upward, but one is decreasing and the other is increasing. We need to be able to determine concavity from the function and not just from the graph.

x

y

x

y

x x

y y

Concave downward

Look at these two graphs. Each is concave downward, but one is decreasing and the other is increasing. We need to be able to determine concavity from the function and not just from the graph.

x

y

x

y

x x

y y

ConcavityLet f be a differentiable function on (a, b).

1. f is concave upward on (a, b) if f ' is increasing on aa(a, b). That is f ''(x) 0 for each value of x in (a, b).

concave upward concave downward

2. f is concave downward on (a, b) if f ' is decreasing on (a, b). That is f ''(x) 0 for each value of x in (a, b).

Inflection PointA point on the graph of f at which f is continuous and concavity changes is called an inflection point.

To search for inflection points, find any point, c in the domain where f ''(x) 0 or f ''(x) is undefined. Some call these points “hypercritical points”.

If f '' changes sign from the left to the right of c, then (c, f (c)) is an inflection point of f.

The Second Derivative Test

left right

f (c) is an inflection point

No change Concavity remains the same

Determine the sign of the second derivative of f to the left and right of the hypercritical point to determine if the function is concave up or concave down or neither.

conclusion

f (c) is an inflection point

up down

down up

Example: Inflection Points

.16)( 23 xxxf2( ) 3 12f x x x

Find all inflection points of

( ) 6 12f x x Possible inflection points are solutions of a) ( ) 0 b) ( ) 6 12 0 no solutions 2

f x f x DNEx

x

2 - 0 +

Inflection point at x 2

f

f

The First and Second Derivative Test

.16)( 23 xxxf

2( ) 3 12 0f x x x

Graph and find all the relative extrema of

4,0x

0 4

+ 0 - 0 +

Relative max.

f (0) = 1

Relative min. f (4) = -31

f

ff(x) is concave down for all x in (-∞, 2) because f” < 0. There is an Inflection point at (2, -15) f(x) = 0 and f(x) becomes concave upward

- 2 +

( ) 6 12f x x = 0 x = 2

The First Derivative Test

-2 -1 1 2 3 4 5 6 7 8 9 10

-35

-30

-25

-20

-15

-10

-5

5

x

y

(4,f(4)=-31)

(0,f(0)=1)

Sketch of function based on estimates from the first derivative test.

-2 -1 1 2 3 4 5 6 7 8 9 10

-35

-30

-25

-20

-15

-10

-5

5

x

y

(4,f(4)=-31)

(0,f(0)=1)

The First and Second Derivative Tests

There is a local maximum at (0,1) because >0 for all x in (-∞, 0) and < 0 for all x in (0,2) .

There is a local minimum at (4,-31) because <0 for all x in (0,4) and > 0 for all x in (4, ∞).

f f

f f

f(x) is concave down for all x in (-∞, 2) because f” < 0. There is an inflection point at (2, -15) and f(x) =0 and where f(x) becomes concave upward

What else can we say aboutthe rate of change?

Let c be a critical point of f(x). If f”(c) exists, then

If f ''(c) > 0 then f(c) is a local minimum. If f ''(c) < 0 then f(c) is a local maximum. If f ''(c) = 0 inconclusive, use the first derivative

test to classify the extrema.

Also the Second Derivative Testcan be used to find local extrema

Let c be a critical point of f(x). If f”(c) exists, then If f ''(c) > 0 then f(c) is a local minimum. If f ''(c) < 0 then f(c) is a local maximum. If f ''(c) = 0 inconclusive, use the first derivative

test to classify the extrema.

Using the function below, find any relative extrema

Second Derivative Test

2( ) 3 12 0f x x x 4,0x

( ) 6 12f x x f(0)= -12 f(4) = 12

.16)( 23 xxxf

32

2

3136)(

xxxf Setting up the chart

interval Test points

Sign of f” f ‘ concave

(-∞, -1) -2 + inc upward

(-1,1) 0 - dec downward

(1,∞) 2 + inc upward

Graph 56)( 24 xxxf

1. Take the derivative f’(x)2. Find the critical points f’(x) = 0; f’(x) =

DNE3. Make sign chart

1. Label critical points – put 0 or DNE on graph2. Evaluate the derivative at points on either side of

extrema to determine the sign. Use Yvars for faster calculations. Mark the chart as + or – in these areas

3. Evaluate f(x) at the critical points to determine actual values of extrema.

4. Add arrows to show increasing or decreasing regions in f(x)

4. Write out all extrema and use the value of the derivative and the appropriate interval to justify your answer.

The Point of Diminishing Returns

( ) 120 6S t t

If the function represents the total sales of a particular object, t months after being introduced, find the point of diminishing returns.

2 3( ) 100 60S t t t

2( ) 120 3S t t t

S concave up on

S concave down on 0,20 20,

The point of diminishing returns is at 20 months (the rate at which units are sold starts to drop).

10 20 30 40

5000

10000

15000

20000

25000

30000

t

S(t)

S concave up on

S concave down on

0,20

20,

Inflection point

The Point of Diminishing Returns

-3 -2 -1 1 2 3 4 5 6

-1

1

2

3

4

5

6

7

8

x

y

Increasing/Decreasing/Constant

-3 -2 -1 1 2 3 4 5 6

-1

1

2

3

4

5

6

7

8

x

y

Increasing/Decreasing/Constant

.,on increasing is then

,, intervalan in of each valuefor 0 Ifbaf

baxxf

.,on decreasing is then

,, intervalan in of each valuefor 0 Ifbaf

baxxf

.,on constant is then

,, intervalan in of each valuefor 0 Ifbaf

baxxf

Increasing/Decreasing/Constant

Generic Example

-3 -2 -1 1 2 3 4 5 6

-1

1

2

3

4

5

6

7

8

x

y

c

( ) 0to the right of

f xc

A similar ObservationApplies at aLocal Max.

( ) 0to the left of

f xc

The First

Derivative Test

Another Example

3 3( ) 3 .f x x x

2

233

1( )3

xf xx x

Find all the relative extrema of

0, 3x

Stationary points: 1x

Singular points:

-1 0 1

+ ND + 0 - ND - 0 + ND +

Relative max. Relative min.

f

f

0, 3x

Stationary points: 1x Singular points:

3(1) 2f

3

3( 1) 2f

3

3( 1) 2f 3(1) 2f

Evaluate the derivative at points on either side of extrema to determine the sign. Use Yvars for faster calculations ND – derivative not defined

2

233

1( )3

xf xx x

-2 -1 1 2 3

-3

-2

-1

1

2

x

y

Local max. 3( 1) 2f

Local min. 3(1) 2f

Graph of 3 3( ) 3 . f x x x

-1 0 1

+ ND + 0 - ND - 0 + ND +f

f 3 33( 1) 2f 3(1) 2f

Example:Graph 23 23 4 1 2y x x x x

There are roots at and .1x 2x

23 6y x x

0ySet

20 3 6x x

20 2x x

0 2x x

0, 2x

First derivative test:

y

0 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 extrema at .0, 2x

We can use a chart to organize our thoughts.

y y=4 y=0

Example:Graph 23 23 4 1 2y x x x x

There are roots at and .1x 2x

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

Possible extreme at .0, 2x

y y=4 y=0

Example:Graph 23 23 4 1 2y x x x x

23 6y x x First derivative test:

y0 2

0 0

NOTE: On the AP Exam, it is not sufficient to simply draw the chart and write the answer. You must give a written explanation!

There is a local maximum at (0,4) because for all x in and for all x in (0,2) .

0y( ,0) 0y

There is a local minimum at (2,0) because for all x in(0,2) and for all x in .

0y(2, )0y

Example:Graph 23 23 4 1 2y x x x x

23 6y x x

First derivative test:

y0 2

0 0

There is a local maximum at (0,4) because for all x in and for all x in (0,2) .

0y( ,0) 0y

There is a local minimum at (2,0) because for all x in(0,2) and for all x in .

0y(2, )0y

Chapter 5Applications of the

Derivative

Sections 5.1, 5.2, 5.3, and 5.4

Applications of the Derivative

Maxima and Minima Applications of Maxima and Minima The Second Derivative - Analyzing Graphs

Absolute Extrema

Absolute Minimum

Let f be a function defined on a domain D

Absolute Maximum

The number f (c) is called the absolute maximum value of f in D

A function f has an absolute (global) maximum at x = c if f (x) f (c) for all x in the domain D of f.

Absolute Maximum

Absolute Extrema

c

( )f c

Absolute Minimum

Absolute ExtremaA function f has an absolute (global) minimum at x = c if f (c) f (x) for all x in the domain D of f.

The number f (c) is called the absolute minimum value of f in D

c

( )f c

-3 -2 -1 1 2 3 4 5 6

-1

1

2

3

4

5

6

7

8

x

y

Generic Example

-3 -2 -1 1 2 3 4 5 6

-1

1

2

3

4

5

6

7

8

x

y

Generic Example

-3 -2 -1 1 2 3 4 5 6

-1

1

2

3

4

5

6

7

8

x

y

Generic Example

Relative ExtremaA function f has a relative (local) maximum at x c if there exists an open interval (r, s) containing c such that f (x) f (c) for all r x s.

Relative Maxima

Relative ExtremaA function f has a relative (local) minimum at x c if there exists an open interval (r, s) containing c such that f (c) f (x) for all r x s.

Relative Minima

Generic Example

-3 -2 -1 1 2 3 4 5 6

-1

1

2

3

4

5

6

7

8

x

y

( ) 0f x

( )f x DNE

The corresponding values of x are called Critical Points of f

Critical Points of f

a. ( ) 0f c

A critical number of a function f is a number c in the domain of f such that

b. ( ) does not existf c(stationary point)

(singular point)

Candidates for Relative Extrema1.Stationary points: any x such that x is in

the domain of f and f ' (x) 0.

2.Singular points: any x such that x is in the domain of f and f ' (x) undefined

3. Remark: notice that not every critical number correspond to a local maximum or local minimum. We use “local extrema” to refer to either a max or a min.

Fermat’s Theorem

If a function f has a local maximum or minimum at c, then c is a critical number of f

Notice that the theorem does not say that at every critical number the function has a local maximum or local minimum

Generic Example

-3 -2 -1 1 2 3 4 5 6

-1

1

2

3

4

5

6

7

8

x

y

( )not a local extrema

f x DNE

Two critical points of f that donot correspond to local extrema

( ) 0not a local extrema

f x

Example

3 3( ) 3 .f x x x

2

233

1( )3

xf xx x

Find all the critical numbers of

0, 3x

Stationary points: 1x Singular points:

Graph of 3 3( ) 3 . f x x x

-2 -1 1 2 3

-3

-2

-1

1

2

x

y

Local max. 3( 1) 2f

Local min. 3(1) 2f

Extreme Value TheoremIf a function f is continuous on a closed interval [a, b], then f attains an absolute maximum and absolute minimum on [a, b]. Each extremum occurs at a critical number or at an endpoint.

a b a ba b

Attains max. and min.

Attains min. but no max.

No min. and no max.

Open Interval Not continuous

ExampleFind the absolute extrema of 3 2 1( ) 3 on ,3 .

2f x x x 2( ) 3 6 3 ( 2)f x x x x x

Critical values of f inside the interval (-1/2,3) are x = 0, 2

(0) 0(2) 4

1 72 8

3 0

ff

f

f

Absolute Max.

Absolute Min.Evaluate

Absolute Max.

ExampleFind the absolute extrema of 3 2 1( ) 3 on ,3 .

2f x x x

Critical values of f inside the interval (-1/2,3) are x = 0, 2

Absolute Min.

Absolute Max.

-2 -1 1 2 3 4 5 6

-5

ExampleFind the absolute extrema of 3 2 1( ) 3 on ,1 .

2f x x x 2( ) 3 6 3 ( 2)f x x x x x

Critical values of f inside the interval (-1/2,1) is x = 0 only

(0) 01 72 8

1 2

f

f

f

Absolute Min.

Absolute Max.

Evaluate

-2 -1 1 2 3 4 5 6

-5

ExampleFind the absolute extrema of 3 2 1( ) 3 on ,1 .

2f x x x 2( ) 3 6 3 ( 2)f x x x x x

Critical values of f inside the interval (-1/2,1) is x = 0 only

Absolute Min.

Absolute Max.

Start Herea. Reviewing Rolle’s and MVTb. Remember nDeriv and Yvarsc. Increasing, Decreasing,Constantd. First Derivative Test

Finding absolute extrema on [a , b] 0. Verify function is continuous on the interval. Determine the function’s domain.1. Find all critical numbers for f (x) in (a , b).2. Evaluate f (x) for all critical numbers in (a , b).3. Evaluate f (x) for the endpoints a and b of the

interval [a , b]. 4. The largest value found in steps 2 and 3 is the

absolute maximum for f on the interval [a , b], and the smallest value found is the absolute minimum for f on [a , b].

Rolle’s Theorem

• Given f(x) on closed interval [a, b]– Differentiable on open interval (a, b)

• If f(a) = f(b) … then– There exists at least one number

a < c < b such that f ’(c) = 0

f(a) = f(b)

a bc

The Mean Value Theorem (MVT)aka the ‘crooked’ Rolle’s Theorem

If f is continuous on [a, b] and differentiable on (a, b)There is at least one number c on (a, b) at which

ab

f(a)

f(b)

c

Conclusion:Slope of Secant Line

EqualsSlope of Tangent Line

abafbfcf

)()()('

We can “tilt” the picture of Rolle’s TheoremStipulating that f(a) ≠ f(b)

How is Rolle’s Connected to MVT?

The Mean Value Theorem (MVT)aka the ‘crooked’ Rolle’s Theorem

If f is continuous on [a, b] and differentiable on (a, b)There is at least one number c on (a, b) at which

ab

f(a)

f(b)

c

Conclusion:The average rate of change

equals the instantaneous rate of change

evaluated at a point

abafbfcf

)()()('

We can “tilt” the picture of Rolle’s TheoremStipulating that f(a) ≠ f(b)

Finding c• Given a function f(x) = 2x3 – x2

– Find all points on the interval [0, 2] where

– Rolle’s? • Strategy

– Find slope of line from f(0) to f(2)– Find f ‘(x)– Set f ‘(x) equal to slope … solve for x

( ) ( ) '( )f b f a f cb a

f(3) = 39 f(-2) = 64

f b f a 64 39 5b a 2 3

For how many value(s) of c is f ‘ (c ) = -5?

If , how many numbers on [-2, 3] satisfythe conclusion of the Mean Value Theorem.

2 2f x x 12 x 4

A. 0 B. 1 C. 2 D. 3 E. 4

CALCULATOR REQUIRED

X X X

Given the graph of f(x) below, use the graph of f to estimate thenumbers on [0, 3.5] which satisfy the conclusion of the Mean Value

Theorem.

Relative Extrema

3 2( ) 6 1f x x x

2( ) 3 12 0f x x x

Example: Find all the relative extrema of

4,0xStationary points:

Singular points: None

First Derivative Test

• What if they are positive on both sides of the point in question?

• This is called aninflection point

Domain Not a Closed IntervalExample: Find the absolute extrema of

1( ) on 3, .2

f xx

Notice that the interval is not closed. Look graphically:

Absolute Max.

(3, 1)

Optimization Problems1. Identify the unknown(s). Draw and label a diagram as

needed.

2. Identify the objective function. The quantity to be minimized or maximized.

3. Identify the constraints.

4. State the optimization problem.

5. Eliminate extra variables.

6. Find the absolute maximum (minimum) of the objective function.

Optimization - ExamplesAn open box is formed by cutting identical squares from each corner of a 4 in. by 4 in. sheet of paper. Find the dimensions of the box that will yield the maximum volume.

xx

x

4 – 2x

4 – 2xx

(4 2 )(4 2 ) ; in 0,2V lwh x x x x

2( ) 16 32 12V x x x 4(2 3 )(2 )x x

Critical points: 22, both in [0, 2]3

x

3

(2) 0(0) 0

2 4.74 in3

VV

V

The dimensions are 8/3 in. by 8/3 in. by 2/3 in. giving a maximum box volume of V 4.74 in3.

2 316 16 4V x x x x

An metal can with volume 60 in3 is to be constructed in the shape of a right circular cylinder. If the cost of the material for the side is $0.05/in.2 and the cost of the material for the top and bottom is $0.03/in.2 Find the dimensions of the can that will minimize the cost.

2 60V r h 2(0.03)(2) (0.05)2C r rh

top and bottom

sidecost

Optimization - Examples

2 60V r h

22

60(0.03)(2) (0.05)2r rr

2

60hr

So

2(0.03)(2) (0.05)2C r rh

2 60.06 rr

2

60.12C rr

2

60 gives 0.12C rr

36 2.52 in. which yields 3.02 in.

0.12r h

Sub. in for h

So with a radius ≈ 2.52 in. and height ≈ 3.02 in. the cost is minimized at ≈ $3.58.

Graph of cost function to verify absolute minimum:

2.5

-2 -1 1 2 3 4 5 6 7 8 9 10

-35

-30

-25

-20

-15

-10

-5

5

x

y

(4,f(4)=-31)

(0,f(0)=1)

2 3Given ( ) 130 15 8 find (2).s t t t s

2( ) 30 24s t t t

( ) 30 48s t t

then, (2) 30 48(2) 66s

Second Derivative - Example

Definition of Concavity and Figure 3.24

Sketch 4 graphs a)1 decreasing and concave upb)1 increasing and concave up, c)1 decreasing and concave

down,d)1 increasing and concave down

x

y

x

y

x

y

x

y

a b

c d

x x

x x

y y

y y