Chapter 4: Applications of Differentiation 4.1 Maximum and ...

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Chapter 4: Applications of Differentiation

In this chapter we will cover:

4.1 Maximum and minimum values. The critical points method for finding extrema.

4.3 How derivatives affect the shape of a graph. The first derivative test and the second derivative test.

4.4 Indeterminate forms and l’Hospital’s Rule

4.5 Summary of curve sketching

4.7 Optimization problems

4.9 Antiderivatives

4.10 Chapter Review

4.1. Maximum and minimum values. The critical points method for finding extrema.

Motivation:

Finding extrema (maximum and minimum) values of a function is an extremely important application in

mathematics (in particular, in our class it is an application of the derivative). For example, problems of maximizing

profit, minimizing cost, maximizing (or minimizing) areas or volumes and many other similar problems (see for

example, the problems in section 4.7) are such typical applications of the methods which we will learn in this

chapter. An entire branch of mathematics called optimization is dedicated to finding extrema of functions.

Goals:

define points of extrema (local and global) ;

develop the critical point method for finding global extrema of a function;

I. Definition of points of extrema. The extreme value theorem:

Questions: What are points of extrema for a function ?

What are maximum and minimum points for a function?

Do all functions have extrema (over an interval ) ?

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Definition 1:

A. Global (absolute) extrema:

Consider a function RDf : and Dc . Then )(cf is a:

global (or absolute) maximum value of )(xf on D if (1) Dxxfcf for )()( (in this case the point c

is called the x- value where )(xf achieves its global maximum) ;

global (or absolute) minimum value of )(xf on D if (2) Dxxfcf for )()( (in this case the point c

is called the x- value where )(xf achieves its global minimum) ;

Example 1:

For example, in Figure 1, we see that the global maximum of )(xf is 5 achieved for

3c and its global minimum is 2 achieved for 6c .

B. Local extrema: )(cf is a:

local maximum value of )(xf on D if such that 0 a (3) acacxxfcf , for )()( (that is, near

c, in this case c is called the x – value where )(xf achieves its local maximum ) ;

local minimum value of )(xf on D if such that 0 a (4) acacxxfcf , for )()( (that is, near c,

in this case c is called the x – value where )(xf achieves its local minimum ) ;

Example 2:

For example, in Figure 2, we see that )(xf has a global maximum at d : )(df ,

a global minimum at a: )(af , and local minima at c and e, and local maxima at b

and d.

Note that, in general, a global maximum is the greatest local maximum (except when

it is an endpoint) and a global minimum is the smallest local minimum (again,

except when it is an endpoint).

See also Figure 3 where a few local minima and maxima are displayed.

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Note again that endpoints (of the interval where the function is considered) can be global extrema, but not local

extrema, since they would not satisfy a condition of the type (3) (or (4)). Therefore, local extrema are always inside

the interval.

Figure 4:

Consider also Figure 4, where the function 234 18163)( xxxxf is graphed on

the interval 4,1 . For this example, identify:

local minima: global minimum:

local maxima: global maximum:

Now that we can identify from a graph points of extrema, let us try to answer the following essential question:

Do all functions defined over a certain interval have a local (or global) maximum and a local (or global) minimum

over that interval ?

Example 2:

a) Let 3)( :bygiven ,,,: xxff what are its local minima and maxima ?

b) Let 3)( :bygiven ,1,11,1: xxff what are its local minima and maxima ?

c) Let 3)( :bygiven ,1,11,1: xxff what are its local minima and maxima ?

d) Let 2)( :bygiven ,,0,: xxff what are its local minima and maxima ?

We see that some of these functions (such as the function given in c) have local extrema, while (most of) the others

do not. The following important theorem establishes which functions are guaranteed to have an extrema over their

corresponding interval.

Theorem 1 (Extreme value theorem):

Consider a continuous function Rbaf ,: . Then )(xf achieves a global maximum value )(cf and a global

minimum value )(df at some points badc ,, .

This theorem is given without proof, since the proof is quite advanced (based on Cantor’s completeness axiom).

Draw a few typical figures of a continuous function on a closed interval to convince yourself of its validity. Then

look again at the functions in Example 2 in the light of this theorem.

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Remark 1:

1. Note that we need two conditions to be verified in order for global extrema to exist: the function )(xf needs to be

continuous and the interval where the function is considered needs to be closed. Otherwise, we may well have that

the function does not have a (local or global) extremum on the interval ;

2. The functions for which we want to find extrema in this chapter almost always satisfy these two conditions (that is

they are continuous functions considered on a closed interval) and therefore global extrema are guaranteed to exist

for these functions.

Figure 5: A function which satisfies Figure 6: A function which does not satisfy the conditions of

the EVT the conditions of EVT: it has a global minimum but no

global maximum, this is in agreement with the theorem.

II. Fermat’s theorem . Finding points of global extrema for a function:

Therefore, so far we have learned how to identify the local and global extrema, and conditions under which such

extrema are guaranteed to exist. Next, we consider the problem of finding such extremum values for a given

function (expression) for which the graph is not given.

Note (see Figure 10 in the textbook) that for most functions local extrema occur at stationary points of the function

(that is points where 0)(' cf ) although this is not always the case. However, these points are good candidates for

extrema, as the following theorem states:

Theorem 2 (Fermat): If Rbaf ,: has a local maximum or a local minimum at bac , and if )(' cf exists,

then 0)(' cf .

So: If Rbaf ,: and bac , and )(cf is a local extrema and )(' cf exists, then 0)(' cf .

(If interested, see the proof of this theorem in the textbook).

Note that 0)(' cf does not in general imply that )(cf is a local extrema as Example 2 (a to c) showed. However,

stationary points of )(xf are valuable candidates for extrema of )(xf .

Example 3: Consider 1,01,1: f given by: xxf )( . In this case, the local extrema are again not points

where 0)(' cf (0 is the local minimum and )0('f does not exist).

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Remark 2:

However, for a function Rbaf ,: which is continuous on ba, , if ],[ bac is an extrema, then either:

bac , and )(' cf exists 0)(' cf (from Fermat’s theorem) ;

OR

bac , and )(' cf does not exist (as in example 3 above) ;

OR

c is either a or b (the endpoints of the interval ).

These 3 possibilities are the only 3 possibilities for an extrema of a function Rbaf ,: .

This is the basis of the following definition and of the Critical point method which follows.

Definition 2 (critical points) : A critical point (read a valuable candidate of an extrema) of a function Rbaf ,:

is a value bac , such that either 0)(' cf or )(' cf does not exist.

Note, that after we find the critical values of a given function, the only other candidates of an extrema for the

function are the endpoints of the interval.

Example 4: Consider xxxf 4)( 5/3 . Determine the critical points of )(xf .

The reasoning outlines in Remark 2 above gives us the procedure of finding the global extrema of any continuous

function defined on a closed interval: Rbaf ,: .

This procedure is oulined in the Critical Point Method below:

Example 5:

a) Find the absolute extrema of 13)( 23 xxxf for 42

1 x .

Critical Point Method for finding the global extrema of a continuous function Rbaf ,: :

Step 1: Find all critical values of )(xf (that is values c where either 0)(' cf or )(' cf does not exist). Find

)(cf for all of these critical values .

Step 2: Find )( and )( bfaf .

Step 3: The largest of the values calculated in Steps 1 and 2 is the global maximum of )(xf on ba, ;

the smallest of these values is the global minimum of )(xf on ba, .

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Do problems 47 and 48 from Exercise set 4.1 in the textbook.

Homework: problems 1,2,3,5,6,7,11,14,15,19,23,25,28,30,32,34,38,49,50,51,52,54,56,57 from problem set 4.1.

4.3 How derivatives affect the shape of a graph. The first derivative test and the second derivative test.

Goals:

prove the First Derivative Test and learn how to use it to find all local extrema of a function )(xf on a given

interval;

prove the Second Derivative Test and learn how to use it to find all local extrema of a function )(xf on a

given interval;

I. The first derivative test :

In order to develop this test, we pose the question: What does )(' xf say about )(xf ?

To answer this question, remember that in section 3.0.2 we introduced )(' xf as the slope of the tangent line to fG

at x .

The basis of the first derivative test is given by the following:

Theorem 1:

Consider )(xf differentiable on an interval I .

a) If 0)(' xf on I then )(xf is strictly increasing on I (that is : Ixxxfxfxx 212121 ,any for ) .

b) If 0)(' xf on I then )(xf is strictly decreasing on I (that is : Ixxxfxfxx 212121 ,any for ) .

Proof: Follows easily using the mean value theorem from )(xf . Use also your visual intuition to see why this

theorem must be true.

This theorem allows us to construct a table of monotonicity for a function )(xf , as long as we can determine the

sign of its derivative.

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This table of monotonicity will allow us to find easily all local and global extrema of a function )(xf on an interval

I.

Example 1:

Consider 51243)( 234 xxxxf on ]3,3[I . Find where )(xf is increasing and where it is decreasing on I,

then construct a monotonicity table and use this table to determine all points of extrema of )(xf on I .

The method shown in Example 1 suggests the following theorem:

Theorem 2 (The First Derivative Test):

Consider RDf : , )(xf continuous on D, and let Dc a critical number of )(xf . Then :

a) If )(' xf changes sign at c from positive to negative (as we move from left to right) then )(xf has a local

maximum at c .

b) If )(' xf changes sign at c from negative to positive (as we move from left to right) then )(xf has a local

minimum at c .

Proof:

Is immediate, if we construct a monotonicity table for )(xf around c.

Example 2:

Do problems 9 and 10 in the Exercise set 4.3 .

II. The second derivative test :

In order to develop this test, we pose the question: What does )('' xf say about )(xf ?

Figure 1: a concave up function (Figure 1.a) and a concave down function (Figure 1.b)

Definition 1:

a) If )(' xf is increasing on some interval I (so if 0)('' xf on I), then )(xf is called concave up on I;

b) If )(' xf is decreasing on some interval I (so if 0)('' xf on I), then )(xf is called concave down on I .

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A typical shape of a concave up function is (as shown in Figure 1a) . For this type of function, )(' xf is

increasing (so 0)('' xf ).

A typical shape of a concave down function is (as shown in Figure 1b) . For this type of function, )(' xf is

decreasing (so 0)('' xf ).

Look also at Figure 7 in the textbook and determine the regions where )(xf is concave up and where it is concave

down. Convince yourself that on these regions )(' xf has the expected monotonicity behavior (so )('' xf has the

expected sign).

Therefore, the sign of )('' xf provides the concavity of the function )(xf .

Definition 2:

A point fGxfxP on ))(,( is called an inflection point of a continuous function )(xf if )(xf changes concavity at P

(or, in other words, if )('' xf changes sign at P).

Therefore, a point where )(' xf changes sign is a point of local extremum for )(xf , and a point where )('' xf

changes sign is a point of inflection for )(xf .

Example 3:

Draw a monotonicity table and a concavity table (that is a table which shows the sign of )('' xf on fD ) for the

function 34 4)( xxxf .

Use this information to sketch the graph of )(xf .

Theorem 2 (Second Derivative Test):

Consider RDf : , )(xf continuous on D, and let Dc a stationary point of )(xf , that is let 0)(' cf .

a) If 0)('' xf , then )(xf has a local minimum at c;

b) If 0)('' xf , then )(xf has a local maximum at c.

Proof:

a) Use the definition of a concave up function (see Definition1a) with the fact that 0)(' cf to determine a

monotonicity table of )(xf near c, which shows that c is a point of local minimum for )(xf .

b) Similarly.

Example 4: Problem 19 from Exercise Set 4.3 .

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Comparison of the 3 methods for finding extrema:

So far we have developed the 3 main methods for finding the points of extrema of a function )(xf : the critical point

method (in section 4.1), the first derivative test (Theorem 2 in section 4.3) and the second derivative test (Theorem 3

in section 4.3).

A comparison of the advantages and disadvantages of these tests reveal that the first derivative test is usually the

best (preferred) test to use when determining the points of extrema of a function since:

For the Critical Point Method, only the global extrema are found. No information is known about the other

critical points (that is, if they are local extrema or no extrema at all) . Also, this method is quite tedious and

long.

The First Derivative Test determine all local extrema and the entire monotonicity behavior of )(xf on I. It is

the best since it provides a complete information about the monotonicity of the function )(xf if we can

determine the sign of )(' xf ;

The Second Derivative Test determines the local extrema but only for stationary points. It does not work for

the singular points of )(xf or for endpoints.

Therefore, when possible, it is preferred to use the First Derivative Test when determining the extrema of a function.

Example 5: Problem 12 from Exercise Set 4.3

Homework: Problems 1,4,5,7,8,10,11,13,16,20,21,33,36,39 and 45 from Exercise Set 4.3 .

4.4 Indeterminate forms and l’Hospital’s Rule

In this section we turn back to calculating limits of functions.

So far we know how to calculate limits of the form 23

12lim

2

2

1

xx

xx

x and

23

12lim

2

2

xx

xx

x but how do we calculate

limits like: 1

)ln(lim

1 x

x

x ,

1

)ln(lim

x

x

x or

x

e x

x lim ?

l’ Hospital’s Rule, given by the next theorem, provides a general and simple yet effective method to solve such

limits.

Theorem 1 (l’Hospital’s Rule):

Consider )( and )( xgxf differentiable and 0)(' xg near a point a (except possibly at a ) . Suppose that

0)(lim and 0)(lim

xgxfaxax

(such that

0

0

)(

)(lim

xg

xf

ax)

or that )(or )(lim and )(or )(lim

xgxfaxax

(such that

)(

)(lim

xg

xf

ax) .

Then

)('

)('lim

)(

)(lim

xg

xf

xg

xf

axax when

)('

)('lim

xg

xf

ax exists (or when it is or ).

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Proof:

Figure 1:

For the actual proof (in a simplified case), let us assume that 0)()( agaf , that )(' and )(' xgxf are continuous near

a , and that 0)(' xg near a .

In this case: )(

)(lim

)()(

)()(lim

)()(lim

)()(lim

)('

)('

xg

xf

agxg

afxf

ax

agxgax

afxf

ag

af

axax

ax

ax

.

For the more general proof (based on the mean value theorem) see Appendix F.

Q.E.D.

Example 1:

Using l’Hospital’s Rule (make sure that you check all the conditions before applying this rule, and especially the

condition of a case of study of the type

0

0 or

and that

)('

)('lim

xg

xf

ax exist) , calculate the following limits:

a) 23

12lim

2

2

1

xx

xx

x b)

1

)ln(lim

1 x

x

x c)

1

)ln(lim

x

x

x

d) x

e x

x lim e) 2

limx

ex

x f)

3

)ln(lim

x

x

x

g) x

x

x

)sin(lim

0 h) 30

)tan(lim

x

xx

x

i) )ln(lim

0xx

x

j)

)tan()cos(

1lim

2

xxx

11

k) x

xx

0lim (recall that there are 3 exponential cases of study : 1 and , 0 00 which are solved by using the

notation (1) )(

0)(lim xg

xxfL

and applying ln to the equality (1) .

l) Problems 15, 19 and 20 from the Exercise set 4.4 .

Homework: Problems 1,2,3,4,5,8,10,12,21,22,23,25,28,32,40,42,43,44,49,52,55,56,63,71,72,73 and 86 from the

Exercise Set 4.4

4.5 Summary of curve sketching

We can now use many of the tools we have studied so far in Calculus, especially the table of monotonicity, the table

of concavity and the asymptotes to develop a systematic and effective method for sketching the graph of any

function )(xf .

There are many important reasons for sketching the graph of a particular function by hand versus graphing it by

calculator:

even the best graphing devices have to be used intelligently and the graphs produced have to be fully

understood;

it is very important when graphing a given function to choose an appropriate viewing rectangle to avoid

getting a misleading graph (see for example the graphs of 3)( 2 xxf , 228)( xxf ,

xxxf 150)( 3 , xxf 150sin)( and other examples studied in section 1.4) . Notions of Calculus (and

algebra) are necessary to determine the domains and ranges of these functions (and therefore the appropriate

drawing rectangles);

using Calculus we discover the most important characteristics of a given function (such as asymptotes,

minima, maxima and points of inflection) which can be easily overlooked or at best very hard to find

accurately using only a graphical device. For example,

Figure 1 shows the graph of 218218)( 23 xxxxf . It appears quite reasonable, it is the graph of a

cubic, similar with the graph of 3)( xxf , apparently with no maximum or minimum value over this

interval. However, with the use of Calculus (the first derivative test), we find a maximum of )(xf at

75.0x and a minimum at 1x . Calculus also insures that there are no other points of extrema for this

function (if RD ).

Figure 1: The graph of 218218)( 23 xxxxf Figure 2: The graph of Figure 1 zoomed in

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Therefore, there should be a close interaction between Calculus and graphing devices, as each can be of

assistance to other: Calculus can provide the rigorous and accurate framework for finding the domain, range,

asymptotes and all points of extrema of a given function, while a calculator (or another graphing device) can

be used for carrying on necessary calculation or for checking our work).

Guideline (steps) for sketching the graph of a generic function )(xf :

1. Find the domain D of )(xf (the set of values x where )(xf is well defined ) by imposing

conditions for expressions in x ;

2. Find 0)( solvingby : xfOx

Find )0( gcalculatinby : fOy ;

3. (often optional) check for symmetries:

with the y axis : )(xf is symmetric with the y axis if )(xf is even , that is if

Dxxfxf ),()(

with origin: )(xf is symmetric with the origin if )(xf is odd, that is if Dxxfxf ),()(

check if )(xf is periodic :

)(xf is periodic if DxxfpxfRp ),(such that . The smallest such p is called the

fundamental period of )(xf . If )(xf is periodic with a fundamental period of T , then it needs to be

graphed only on an interval of length T , typically on T,0 .

4. Find the asymptotes:

a) H.A. : at )(lim Calculate : D)in (if xfx

at )(lim Calculate : D)in (if xfx

b) V.A. are lines ax (points excluded from the domain D) such that either )(or )(lim

xfax

or

)(or )(lim

xfax

(or both).

5. Draw the monotonicity and the concavity table of )(xf (that is, determine the signs of

)('' and )(' xfxf on D ) and use the information from 1-4 above to complete this table.

6. Sketch the graph of )(xf using the information from the table of monotonicity and concavity drawn in

5.

Example 1: Use the steps outlined above to sketch the graphs of:

a) 1

2)(

2

2

x

xxf b) 23)( 23 xxxf c)

1)(

2

x

xxf d) xexxf )(

Homework: Problems 1,3,9,10,11,13,16,22,24,30,33,37,47 and 52 from Exercise Set 4.5

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4.7 Optimization problems :

In this section we solve practical problems in which we use the methods studied in sections 4.1 and 4.3 to find the

extreme values of a practical quantity of interest.

There are many problems of this type, for example: a business person who wants to minimize costs and to

maximize profit, a traveler wants to minimize his/her traveling time, or practical problems which come from

physical or mathematical principles (such as the problem which generates the geodesic curves (the shortest curve

between two points on a given surface) , Fermat’s principle of light which states that light follows the path which

takes the least time. See also the “My dog knows calculus problem” (discussed here:

http://www.indiana.edu/~jkkteach/Q550/Pennings2003.pdf and mentioned in section 1.1) .

In this section we solve problems such as these, and also problems in which we are interested in maximizing areas,

volumes, and profits and in minimizing distances, times and costs.

The problems in this section are typically word problems, somewhat similar with the “related rates” problems,

studied in section 3.9, but with the notable difference that this time our main goal is to find the extreme values of a

practical quantity of interest (usually called the objective function), and not a derivative.

It is important to follow the following guidelines when solving the optimization problems in this section:

1. Understand the problem: read the problem carefully until it is clearly understood. Determine the

objective function and the variables that this function depends on.

2. Draw a diagram: In most problems it is useful to draw a diagram to determine the given and the required

quantities and the relation between these.

3. Introduce notation: Assign a symbol for the objective function (let us call it Q ) and for other unknown

quantities in the diagram (for example, call these ,...),,,, yxcba . Label the diagram with these symbols.

4. Express Q in terms of these symbols: ,...,,,, yxcba . Our final preliminary goal is to express Q in terms of

one unknown variable only.

5. Use the diagram (or equations/relations stated in the problem) to find relations between the unknown

independent variables such that Q depends on one independent (unknown) variable only. Set then

)(xfQ . Find the domain of this function D .

6. Use the methods studied in sections 4.1 and 4.3 (usually the First Derivative Test) to find the absolute

minimum or the absolute maximum (as required) of )(xfQ on D .

Examples :

1. A farmer has 2400 feet of fence and he wants to fence off a rectangular field. What are the dimensions of the

field which has the largest area ?

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2. A cylindrical can is to be made to hold )L 1( cm 1000 3 of oil, as in the Figure 3 below.

What are the dimensions (find hr and ) which will minimize the cost of the metal to manufacture the can ?

3. Find the point on the parabola xy 22 which is closest to the point )4,1( .

Hint: it is easier here to minimize 2)4,1(,, yxd instead of )4,1(,, yxd .

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4. See and solve the “My dog knows calculus” problem at

http://www.indiana.edu/~jkkteach/Q550/Pennings2003.pdf

Homework: Problems 1,2,4,5,7,8,10,12,13,15,18,19,21,24,26,29,30,34,38,48,52,69 and 79 from Exercise Set 4.7

4.9 Anti-derivatives:

In Chapter 3 we defined h

xfhxfxf

h

)()(lim)('

0

for a continuous function )(xf , when this limit exists, and

developed methods to calculate the derivative )(' xf quickly for many functions, using a table of derivatives.

Anti-derivatives:

Sometimes we are interested in the inverse problem: If ( )f x is seen as a rate of change (the derivative) of some

unknown function, what is that unknown function? In other words, the derivative of which function produces the

known (given) function )(xf ?

This type of problem is often met and very important in science. For example, a physicist often knows the force (the

acceleration) of a particle and wants to find the position. An engineer who can measure the variable rate at which

water is leaking from a tank wants to know the amount leaked over a certain time period.

Definition: Consider a continuous function :f D R . An anti-derivative of ( )f x is a function ( )F x such that

(2) '( ) ( )F x f x .

NOTE: ( )F x is also denoted by ( )f x dx .

Based on this definition and on the table of derivatives, we can find the anti-derivative of many functions.

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Example 1: Find xdx , 2x dx , dxxn , sin( )x dx .

NOTE: An anti-derivative is not unique, but unique up to an arbitrary constant. If ( )F x is an anti-derivative of

( )f x , then ( )F x C (where C is an arbitrary constant) represents the entire family of anti-derivatives of ( )f x (or

the general anti-derivative of ( )f x ).

Operations:

( ) ( ) ( ) ( )f x g x dx f x dx g x dx

( ) ( )f x dx f x dx

( ) ( ) ( ) ( )f x g x dx f x dx g x dx (true?)

Look at the table of anti-derivatives below , understand and memorize these formulas

Figure 1: Anti-derivative of basic functions

Example 2: Find all functions x

xxxxgxg

52)sin(4)('such that )( .

(In other words, calculate

dx

x

xxx

52)sin(4 ) .

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Example 3: Find all functions 2120)('such that )( xexFxF x and 2)0( F .

(In other words calculate the particular anti-derivative dxxexF x

2120)( such that 2)0( F .

Example 4: Find 4612)(''such that )( 2 xxxFxF and 1)1( ,4)0( FF .

Example 5: Do Examples 6 and 7 from section 4.9 in the textbook.

Homework: Problems 1,4,5,8,10,12,14,16,22,23,26,37,41 and 49 from Exercise Set 4.9