• Calculus Date: 12/17/13

Obj: SWBAT apply first derivative

testhttp://youtu.be/PBKnttVMbV4 first derivative test inc. dec.

Today – Cover the first derivative test

In class: Start WS3-3A; complete for homework

Tomorrow Cover the second derivative test (easier than 1st)

Friday: With Christian complete any

remaining worksheets, make sure

You understand material from the

test.,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

-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 If

baf

baxxf

.,on decreasing is then

,, intervalan in of each valuefor 0 If

baf

baxxf

.,on constant is then

,, intervalan in of each valuefor 0 If

baf

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

( ) 0

to the right of

f x

c

A similar ObservationApplies at aLocal Max.

( ) 0

to the left of

f x

c

The First

Derivative Test

The First Derivative Test

left right

f (c) is a relative maximum

f (c) is a relative minimum

No change No relative extremum

Determine the sign of the derivative of f to the left and right of the critical point.

conclusion

The First Derivative Test.16)( 23 xxxf

2( ) 3 12 0f x x x

Find all the relative extrema of

0)4(3 xx4,0x

0 4

+ 0 - 0 +

Relative max. f (0) = 1

Relative min. f (4) = -31

f

fEvaluate the derivative at points on either side of extrema to determine the sign.

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 Derivative Test

Here is the actual function.

Another Example

3 3( ) 3 .f x x x

2

233

1( )

3

xf x

x 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 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

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 2

1 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

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.

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 Extrema

A 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 Extrema

A 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 Extrema

1.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

( ) 0

not a local extrema

f x

Example

3 3( ) 3 .f x x x

2

233

1( )

3

xf x

x 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 .2

f 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 7

2 8

3 0

f

f

f

f

Absolute Max.

Absolute Min.Evaluate

Absolute Max.

ExampleFind the absolute extrema of 3 2 1

( ) 3 on ,3 .2

f 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 .2

f 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) 0

1 7

2 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 .2

f 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

ab

afbfcf

)()(

)('

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

ab

afbfcf

)()(

)('

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 af c

b a

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

f b f a 64 395

b 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, .

2f x

x

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: 2

2, both in [0, 2]3

x

3

(2) 0

(0) 0

24.74 in

3

V

V

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 r

r

2

60h

rSo

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

2 60.06 r

r

2

60.12C r

r

2

60 gives 0.12C r

r

36

2.52 in. which yields 3.02 in.0.12

r 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

Second Derivative

If ( ) is a function of , then the function

( ) denotes the first derivative of ( ).

Now, the derivative of ( ) is denoted by

( ) and called the second derivative of

the function

y f x x

y f x f x

y f x

y f x

y

( ).f x2

2Notation: ( ) is also denoted by

d ff x

dx

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

In both cases is increasing. However, in the first case

curves down and in the second case curves up.

f

f f

Second Derivative

( ) is

( ) i

so,

s

( ) 0

f x

f x

f x

( ) is

( ) i

so,

s

( ) 0

f x

f x

f x

Second Derivative

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.

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

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 DNE

x

x

2

- 0 +

Inflection point at x 2

f

f

-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 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