Solution manual for precalculus functions and graphs 12th edition by swokowski

Note: For exercises similar to numbers 67–74, pick a multiplier that will make all of the exponents of the terms in thedenominator a multiple of the index.. Instead, we will group the te

Chapter 1: Fundamental Concepts of Algebra

1.1 Exercises

1 (a) Since and have opposite signs, the product is negative.B C BC

(b) Since B  !# and C  !, the product B C# is positive

(c) Since B  ! B { is negative} and C  ! C { is positive}, the quotient is negative.B

CThus, B is the sum of two negatives, which is negative

C  B(d) Since C  ! and B  ! C  B  !,

2 (a) Since and have opposite signs, the quotient is negative B C B

C(b) Since B  ! and C  !# , the product BC# is negative

(c) Since B  C  ! and BC  !, B  C  ! (d) Since C  ! and C  B  ! C C  B  !,

BC

3 (a) Since ( is to the left of % on a coordinate line, (  %

(b) Using a calculator, we see that 1 Hence, 1 (c)

# ¸ "Þ&( #  "Þ& È##& œ "& Note:È##& Á „"&

4 (a) Since $ is to the right of ' on a coordinate line, $  '

(b) Using a calculator, we see that 1 Hence, 1 (c)

% ¸ !Þ(* %  !Þ) È#)* œ "( Note:È#)* Á „"(

5 (a) Since " , " (b) Since # , #

"" œ !Þ!* œ !Þ!*!*á ""  !Þ!* $ œ !Þ' œ !Þ''''á $  !Þ'''(c) Since ## and , ##

( œ $Þ"%#)&( 1¸ $Þ"%"&*$ (  1

6 (a) Since " , " (b) Since & , &

( œ ! "%#)&( (  !Þ"%$ ' œ !Þ)$ œ !Þ)$$$á '  !Þ)$$

(c) Since È# ¸ "Þ%"%, È#  "Þ%

7 (a) “ is negative” is equivalent to B B  ! We symbolize this by writing “ is negative B Í B  ! .”

(b) is nonnegative C Í C   ! (c) ; is less than or equal to 1 Í ; Ÿ 1(d) is between and % # Í #   % (e) > is not less than & Í >   &

(f) The negative of is not greater than D $ Í D Ÿ $ (g) The quotient of and is at most : ; ( Í : Ÿ ( (h) The reciprocal of is at least A * Í "   *

(i) The absolute value of is greater than B ( Í k kB  (

Note: An informal definition of absolute value that may be helpful is

ksomethingk œœ itself itself itself itself



if is positive or zero

if is negative

8 (a) is positive , Í ,  ! (b) is nonpositive = Í = Ÿ !

(c) A is greater than or equal to % Í A   %

(d) is between and - "& "$ Í "&  -  "$ (e) is not greater than : # Í : Ÿ #

(f) The negative of is not less than 7 # Í 7   #

(g) The quotient of and is at least < = "& Í <=   "& (h) The reciprocal of is at most 0 "% Í 0" Ÿ "%(i) The absolute value of is less than B % Í k kB  %

(b) &Î # œ &Î  # œ &Î#k k c d (c) k k k k"  * œ  "   * œ "  * œ "!c d

13 (a) Since % 1 is positive, k% 1kœ % 1

(b) Since 1 % is negative, k1 % œ k 1 % œ % 1

(c) Since ŠÈ#  "Þ&‹ is negative, ¹È#  "Þ& œ ¹ ŠÈ#  "Þ& œ "Þ& ‹ È#

14 (a) Since ŠÈ$  "Þ(‹ is positive, ¹È$  "Þ( œ¹ È$  "Þ(

(b) Since Š"Þ( È$‹ is negative, ¹"Þ( È$ œ  "Þ( ¹ Š È$ œ‹ È$  "Þ(

(c) ¸" "¸ ¸$ &¸ ¸ #¸ ˆ #‰ #

&$ œ "&"& œ "& œ  "& œ"&

15 (a) Eß F œ (  $ œ % œ %k k k k (b) Fß G œ &  ( œ "# œ "#k k k k

23 Eß F œ B  %k k Ê kB  % Ÿ $k 24 Eß F œ B  # œ B  #k k k k Ê kB  #   %k

Note: In Exercises 25–32, you may want to substitute a permissible value for the variable to first test if the expression

inside the absolute value symbol is positive or negative

25 Pick an arbitrary value for that is less than B $, say &

Since $  & œ # is negative, we conclude that if B  $, then $  B is negative

31 Since B  %  !# for every , B B  % œ B  %k # k #

32 Since B  "  !# for every , B B  " œ  B  " œ B  "k # k # #

(b) ,"Þ(* ‚ "!# ‚ *Þ)% ‚ "!$ œ " ('" $'! ¸ "Þ('" ‚ "!, '

45 Construct a right triangle with sides of lengths È# and The hypotenuse will have length " ÊŠÈ#‹# " œ# È$.Next construct a right triangle with sides of lengths È$ and È# The hypotenuse will have length

ÊŠÈ$‹#ŠÈ#‹# œÈ&

46 Use G œ # <1 with < œ " #, , and to obtain , , and "! #1 %1 #!1 units from the origin

47 The large rectangle has areaœwidth‚lengthœ + ,  - The sum of the areas of the two small rectangles is+,  +- Since the areas are the same, we have + ,  - œ +,  +-

49 (a) Since the decimal point is places to the right of the first nonzero digit, & %#( !!! œ %Þ#( ‚ "!, &

(b) Since the decimal point is places to the left of the first nonzero digit, ) !Þ!!! !!! !*$ œ *Þ$ ‚ "!)

(c) Since the decimal point is places to the right of the first nonzero digit, ) )"! !!! !!! œ )Þ" ‚ "!, , )

50 (a) )& #!! œ )Þ&# ‚ "!, % (b) !Þ!!! !!& % œ &Þ% ‚ "!'

(c) ,#% *!! !!! œ #Þ%* ‚ "!, (

51 (a) Moving the decimal point places to the right, we have & )Þ$ ‚ "! œ )$! !!!& ,

(b) Moving the decimal point places to the left, we have "# #Þ* ‚ "!"#œ !Þ!!! !!! !!! !!# *

(c) Moving the decimal point places to the right, we have ) &Þ'% ‚ "! œ &'% !!! !!!) , ,

55 It is helpful to write the units of any fraction, and then “cancel” those units to determine the units of the final

answer , miles seconds minutes hours days year mi

second minute hour day year

gramsmole atomsmole

59 frames seconds minutes hours frames

second minute hour

† † † %) œ %Þ"%(# ‚ "!

" "

'

60 calculations seconds minutes hours days calculations

second minute hour day

61 (a) " ft#œ "%% in , so the force on one square foot of a wall is # "%% in#‚ "Þ% lb inÎ #œ #!"Þ' lb

(b) The area of the wall is %! ‚ ) œ $#! ft , or # $#! ft#‚ "%% in ft#Î # œ %' !)!, in #

The total force is ,%' !)! in#‚ "Þ% lb inÎ #œ '% &"#, lb

Converting to tons, we have ,'% &"# lbÎ #!!! lb tonÎ œ $#Þ#&' tons

62 (a) We start with %!! adults, "&! yearlings, and #!! calves {totalœ (&!}

Number of Adultsœsurviving adultssurviving yearlings

so the number of adults is %!!

*!% of these ($'!) are part of the %!! adults this year The other adults represent%!

)!% of last year’s yearlings, so the number of yearlings is &!

13 A common mistake is to write B B œ B$ # ', and another is to write B# $ œ B&.

The following solution illustrates the proper use of the exponent rules

34 ˆ'B(Î&‰ˆ#B)Î&‰œ ' † #BÐ(Î&ÑÐ)Î&Ñœ "#B"&Î&œ "#B$

59 È& '% œÈ& $#È& # œÉ& # &È& # œ #È& # 60 È% &"# œÈ% #&'È%# œÈ% %%È% # œ %È% #

61 In the denominator, you would like to have È$ How do you get it? Multiply by È$ , or, equivalently, È$ Of

62 Ê"& œÈÈ"œ È" †ÈÈ& œ"&È&

& & &

Note: For exercises similar to numbers 67–74, pick a multiplier that will make all of the exponents of the terms in the

denominator a multiple of the index

67 The index is Choose the multiplier to be # È#C so that the denominator contains only terms with even exponents

77 Ë& Ë& Ë& Ë& ÈÈ È ÈÈ È È È

& & &

79 È$ È$ È$ 80 É$

$> @% # *> @ œ" % #(> @ œ $>@$ ' # #<  = $œ #<  =

81 ÈB C œ' % É B$ # C# #œÉ B$ #É C# #œ B C œ B Ck kk k k k$ # $ # since is always nonnegative.C#

82 ÈB C% "!œÉ B# # C& #œ B C œ B Ck kk k# & #k k&

B C  $) "#œ B# % C  $$ %œ B# C  $$ œ B# C  $$ B C  $# # C  $

B  # "#C œ% B  #$ %C œ% B  # $ C B  # # B  # C

85 +< #œ +#< Á +Ð< Ñ# since #< Á <# for all values of ; for example, let < < œ "

86 Squaring the right side gives us +  " #œ +  #+  "# Squaring the left side gives us +  "#

+  #+  " Á +  "# # for all values of ; for example, let + + œ "

87 +, BCœ + ,BC BC Á + ,B C for all values of and ; for example, let B C B œ " and C œ #

93 (a) È1  " ¸ #Þ!$&" (b) È$ "(Þ"  &"Î%¸ %Þ!("(

94 (a) #Þ'  "Þ$ #¸ !Þ&*"( (b) &È (¸ (!Þ')!(

95 $#!! "Þ!% ")!¸ #$# )#&Þ()$ ,

96 2 œ "%&% ft Ê œ "Þ#È2 œ "Þ#È"%&% ¸ %&Þ) mi

97 [ œ #$! kg Ê P œ !Þ%'È$[ œ !Þ%'È$ #$! ¸ #Þ)# m

98 P œ #& ft Ê [ œ ! !!"'P #Þ%$œ ! !!"' #& #Þ%$¸ $Þ** tons

99 , œ (& and A œ ")! Ê [ œ A œ ")! ¸ &#Þ'

,  $& (&  $&

È$ È$, œ "#! A œ #&! Ê [ œ A œ #&! ¸ &'Þ*

100 (a) 2 œ (# in and A œ "(& lb Ê W œ !Þ"!*" A!Þ%#& !Þ(#&2 œ !Þ"!*" "(& !Þ%#& (# !Þ(#&¸ #"Þ(' ft #

(b) 2 œ '' in Ê W œ !Þ"!*" A" !Þ%#& '' !Þ(#& A % increase in weight would be represented by "! "Þ"A andthus W œ !Þ"!*" "Þ"A# !Þ%#& '' !Þ(#& W ÎW œ "Þ"# " !Þ%#&¸ "Þ!%, which represents a % increase in % W

14 Using trial and error, we obtain (B  "!B  ) œ (B  % B  ##

15 The factors for B  %B  &# would have to be of the form ˆB  ‰ and ˆB  ‰

The factors of are and , but their sum is (& " & ' not )% Thus, B  %B  &# is irreducible

16 $B  %B  ## is irreducible

17 $'B  '!B  #& œ 'B  & 'B  & œ 'B  &# # 18 *B  #%B  "' œ $B  % $B  % œ $B  %# #

ference of two squares}

26 &B  "!B  #!B  %! œ & B  #B  %B  ) œ & B B  #  % B  #

œ & B  % B  # œ & B  # B  # B  # œ & B  # B  #

29 We might first try to factor B  %B  %  *C# # by grouping since it has more than terms, but this would prove to$

be unsuccessful Instead, we will group the terms containing and the constant term together, and then proceed asB

in Example 2(c)

B  %B  %  *C œ B  ## # # $C #œ B  #  $C B  #  $C

30 B  %C  'B  * œ B  'B  *  %C œ B  $  #C œ B  $  #C B  $  #C# # # # # #

31 C  #& C  & C  & C  &

C  "#&œ C  & C  &C  #& œC  &C  #&

34 &+  "#+  % #&+  #!+  % &+  # +  # + +  # +

+  "' ƒ +  #+ œ +  % +  # +  # † &+  # &+  # œ +  % &+  #

35 % "" % $=  " "" "#=  %  "" "#=  (

$=  " $=  " # œ $=  " #  $=  " # œ $=  " # œ $=  " #

36 % = % = &=  # %  &=  #= &=  #=  %

&=  # # &=  # œ &=  # #  &=  # # œ &=  # # œ &=  # #

48 %B  "# &B ( % &B ( % B  $  &B  ( B  $ "'B  *

B  'B  *# B  *# B  $ œ B  $B  *# B  $ œ B  *# œ B  *#

49 The lcd of the entire expression is Thus, we will multiply both the numerator and denominator by +, +,

, ++ ,

" "

+,

, ++, † +,

" "

+, † +,

,  + ,  + ,  +,  + ,  +

>  & >  & >  & >  "! >  #&

>  & œ >  &† >  & œ œ

" " +  +,  , +  +,  ,+  , œ +  , † +  +,  , œ +  ,

Note: Exercises 73–90 are worked using the factoring concept given as the third method of simplification in Example 7.

73 The smallest exponent that appears on the variable is B $

80 $B  # # %B  & %  %B  & Œ " $B  # $ œ $B  # %B  & ) $B  #  %B  &c d

81 $B  " 'ˆ ‰"# #B  & "Î# #  #B  & "Î# ' $B  " & $

œ $B  " #B  & $B  "  ") #B  & $B  " #B  &

B $B  "&  )B B &B  "& &B B  $

B  & œ B  & œ  B  &

The values for Y and Y do not agree Therefore, the two expressions are not equal." #

93 In the second figure, the dimensions of area I are B and B  C The area of I is B  C B, and the area of II is

B  C C E œ B  C

œ B  C B  B  C C

œ B  C B  C

The area {in the first figure}

{in the second figure}

{in the third figure}

G œ ''Þ&  "$Þ)A  &2  'Þ)C0 A œ &* 2 œ "'$ C œ #&

G œ ''Þ&  "$Þ) &*  & "'$  'Þ) #& œ "&#&Þ(0 caloriesFor the -year-old male, use&&

with , , and

G œ '&&  *Þ'A  "Þ*2  %Þ(C7 A œ (& 2 œ "() C œ &&

G œ '&&  *Þ' (&  "Þ* "()  %Þ( && œ "%&%Þ(7 calories

(b) As people age they require fewer calories The coefficients of and are positive because large peopleA 2require more calories

%B  )  B  # œ &B  ' Ê &B  ' œ &B  ' or e ! œ !f, indicating an identity The solution is ‘ „#e f.

8 ”#B  &# #B  &$ œ%B  #&"!B  &# •† #B  & #B  & Ê # #B  &  $ #B  & œ "!B  & Ê

%B  "!  'B  "& œ "!B  & Ê "!B  & œ "!B  & or e ! œ !f, indicating an identity

The solution is ‘  „˜ ™&

B œ %, which is not in the domain of the given expressions No solution

11 Divide both sides by a nonzero constant whenever possible In this case, divides evenly into both sides.&

(&B  $&B  "! œ !# {divide by } & Ê "&B  (B  # œ !# {factor} Ê

$B  # &B  " œ ! {zero factor theorem} Ê B œ # "

14 ”B  #$B B  #" œB  %#% •† B  # B  # Ê $B B  #  " B  # œ % Ê

$B  'B  B  # œ %# Ê $B  (B  # œ !# Ê $B  " B  # œ ! Ê

B œ " #

$ e is not in the domain of the given expressionsf

15 #&B œ *# Ê B œ# #&* Ê B œ „É#&* Ê B œ „$&

21 The expression is B  B  $!# The associated quadratic equation is B  B  $! œ !#

Using the quadratic formula, B œ , „ ,  %+-, to solve for with B + œ " , œ ", , and - œ $! gives us:

Now multiply the first factor by and the second factor by # ' #B  $ 'B  " œ !

So the final factored form of "#B  "'B  $# is #B  $ 'B  "

24 "&B  $%B  "' œ ! + œ "& , œ $% - œ "', , , so B œ $% „ ""&'  *'! œ $% „ %' œ# )

Thus, "&B  $%B  "' œ & B #  ˆ ‰#& ‘ † $ B  ˆ ‰)$ ‘œ &B  # $B  )

25 We must first isolate the absolute value term before proceeding.

k$B  #  $ œ (k Ê k$B  # œ %k Ê $B  # œ % or $B  # œ % Ê

$B œ ' or $B œ # Ê B œ # or B œ #

$

26 # &B  #  " œ &k k Ê # &B  # œ 'k k Ê k&B  # œ $k Ê

&B  # œ $ or &B  # œ $ Ê &B œ " or &B œ & Ê B œ " or B œ "

&

27 $ B  "  & œ ""k k Ê $ B  " œ 'k k Ê kB  " œ #k

Since the absolute value of an expression is nonnegative, kB  " œ #k has no solution

28 kB  $  ' œ 'k Ê kB  $ œ !k Since the absolute value of an expression can only

equal if the expression itself is , ! ! B  $ œ !k k Ê B  $ œ ! Ê B œ $

29 Since there are four terms, we first try factoring by grouping.

*B  ")B  %B  ) œ !$ # Ê *B B  #  % B  # œ !# Ê

*B  % B  # œ !# Ê B œ# % B œ # Ê B œ „# #

or ß

30 Notice that we can factor an out of each term, and then factor by grouping.B

%B  "!B œ 'B  "&B% $ # Ê %B  "!B  'B  "&B œ !% $ # Ê B %B  "!B  'B  "& œ !$ # Ê

B #B #B  &  $ #B  & œ !c # d Ê B #B  $ #B  & œ !# Ê

31 C$Î#œ &C Ê C$Î# &C œ ! Ê C Cˆ "Î# & œ !‰ Ê C œ ! or C"Î#œ &

C"Î#œ & Ê ˆC"Î#‰#œ &# Ê C œ #& The solutions are C œ ! and C œ #&

Note: The following guidelines may be helpful when solving radical equations.

Guidelines for Solving a Radical Equation

(1) Isolate the radical If we cannot get the radical isolated on one side of the equals sign because there is morethan one radical, then we will split up the radical terms as evenly as possible on each side of the equals sign.For example, if there are two radicals, we put one on each side; if there are three radicals, we put two onone side and one on the other

(2) Raise both sides to the same power as the root index Note: Remember here that

ˆ+  ,È8‰#œ +  #+,# È8  , 8#and that ˆ+  ,È8‰# is not+  , 8# #

(3) If your equation contains no radicals, proceed to part (4) If there are still radicals in the equation, go back

to part (1)

(4) Solve the resulting equation

(5) Check the answers found in part (4) in the original equation to determine the valid solutions

Note: You may check the solutions in any equivalent equation of the original equation, that is, an equation

which occurs prior to raising both sides to a power Also, extraneous real number solutions are introducedwhen raising both sides to an even power Hence, all solutions must be checked in this case Checkingsolutions when raising each side to an odd power is up to the individual professor