Solution Manual For College Algebra, 13th Edition by Margaret L. Lial, John Hornsby, David I. Schnei

CLASSROOM EXAMPLE SOLUTIONS BEVERLY FUSFIELD

COLLEGE ALGEBRA THIRTEENTH EDITION

Margaret L. Lial American River College

John Hornsby University of New Orleans

David I. Schneider University of Maryland

Callie J. Daniels St. Charles Community College

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ISBN-13: 978-0-13-597882-5 ISBN-10: 0-13-597882-3

Chapter R Review of Basic Concepts Section R.1

Fractions, Decimals, and Percents

Classroom Example 1 (page 2) (a)

12 3  4 3 4 3 3     1  20 5  4 5 4 5 5

(b)

8 1 8 1 1   1  48 6  8 6 6

(c)

90 5  18 5 18 5 5     1  162 9  18 9 18 9 9 Another strategy is to choose any common factor and work in stages.

90 10  9 5  2  9 5 5    1  162 18  9 9  2  9 9 9

Classroom Example 4 (page 4) 5 18 5  18 Multiply numerators.   9 25 9  25 Multiply denominators.

90 Multiply.  225 2  45 The greatest common factor  5  45 of 90 and 225 is 45.

2 2  45  52  1  52  5 45 5

Classroom Example 5 (page 4) 5

(a)

Classroom Example 2 (page 3) 37 To write 10 as a mixed number, divide the

numerator by the denominator.

7  Quotient Denominator  8 59  Numerator 56 3  Remainder

(b)

37 7 3 10 10

Classroom Example 3 (page 3) To write 3 54

as an improper fraction, multiply the

denominator of the fraction by the natural number, and then add the numerator to obtain the numerator of the improper fraction: 5  3  15 and 15  4  19 The denominator of the improper fraction is the same as the denominator in the mixed number, 5.

4 5  3  4 19 3   5 5 5

9 3 9 5 Multiply by 3 , the    10 5 10 3 reciprocal of 53 .

95 Multiply numerators.  10  3 Multiply denominators.

45  30 3  15 The greatest common factor  2  15 of 45 and 30 is 15.

3 1  or 1 2 2 2

3 1 11 10 3   4 3 4 3 11 3   4 10 11  3  4  10 33  40

Write the mixed numbers as improper fractions.

Multiply by 103 , the reciprocal of 103 .

Multiply.

Lowest terms

Classroom Example 6 (page 5) (a)

1 5 1 5   9 9 9 6  9 2  3

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Add numerators.

Keep same denominator.

Write in lowest terms.

1

2

Chapter R Review of Basic Concepts

(b)

7 2 7 3 2 2      30 45 30 3 45 2



(c)



25 90



5 18

3 1 3 2 1 5      10 4 10 2 4 5





(d)

21 4  90 90

6 5  20 20 1 20

3 1 27 3  3 1  8 2 8 2



27 3 4   8 2 4



27 12  8 8



15 8

30  3  90 and 45  2  90, so the LCD  90.

Write equivalent fractions with the common denominator.

Add numerators Keep the same denominator.

Write in lowest terms.

10  2  20 and 4  5  20, so the LCD  20.

Write equivalent fractions with the common denominator.

Subtract numerators Keep the same denominator.

42.830 71.000  3.074 116.904

(b)

Classroom Example 9 (page 8) (a)

30.2 has one decimal place, and 0.052 has three decimal places, so the product has four decimal places.

30.2  0.052 604 1510 1.5704

(b)

0.06 has two decimal places, and 0.12 has two decimal places, so the product has four decimal places.

0.06  0.12 12 6 0.0072

(c)

0.37 5.476

Write each mixed number as an improper fraction.

The LCD is 8.

Write equivalent fractions with the common denominator.

Subtract. Write as a mixed number.

Move each decimal point two places to the right.

14.8 37 547.6 37 177 148 29 6 29 6 0

Classroom Example 7 (page 7) 15 100

(a)

0.15 

(b)

0.009 

(c)

2.5  2

9 1000

(d)

5 25  10 10

Classroom Example 8 (page 7) (a)

32.50  21.72 10.78

42.83  71  3.074 Place the digits of the decimal numbers in columns by place value. Attach zeros as placeholders.

3.1 3.76 Move each decimal point one place to the right.

1.21 31 37.60 31 66 62 40 31 9 (continued on next page)

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Section R.1 Fractions, Decimals, and Percents

(continued) We carried out the division to three decimal places so that we could round the answer to one decimal place.

3.76  3.1  1.2 Classroom Example 10 (page 9) (a)

19.5  1000 Move the decimal point three places to the right.

19.5  1000  19, 500

(b)

960.1  10 Move the decimal point one place to the left.

960.1  10  96.01

Classroom Example 11 (page 9) (a)

3 50 Divide 3 by 50. Add a decimal point and as many 0s as necessary to 3.

0.06 50 3.00 3 00 0 3  0.06 50

(b)

(c)

Write 0.3 as a percent.

Move the decimal two places right.

0.3 = 30%

(d)

Write 0.165 as a percent.

Move the decimal two places right.

0.165 = 16.5%

Classroom Example 13 (page 11) (a)

(b)

(c)

Write 65% as a fraction.

Drop the % symbol and divide by 100. Then write the fraction in lowest terms.

65 13  5 13 65%    100 20  5 20 135 27  5 27 7   , or 1 100 20  5 20 20 A number greater than 1 is more than 100%.

135% 

1.5%  

1.5 100 1.5 10  100 10

Drop the % symbol.

Divide by 100.

Multiply by 1 in the form 10 10

to eliminate the decimal

in the numerator.

15 Multiply.  1000 35 3 Write in lowest terms.   200  5 200

Classroom Example 14 (page 11)

1 11

(a)

0.0909 11 1.0000 99 10 0 100 99 1

3 Write 50 as a percent.

3 3   100% 50 50 3 100 %   50 1 3  50  2 %  50  1 3 2 %  1  6%

This is a repeating decimal.

1  0.09  0.091 11

(b)

Classroom Example 12 (page 10) (a)

Write 85% as a decimal.

Move the decimal two places left.

85% = 0.85

(b)

Write 110% as a decimal.

Move the decimal two places left.

110% = 1.10 or 1.1

Multiply by 1 in the form 100%.

Multiply and factor.

Divide out the common factor.

Simplify.

Write 13 as a percent.

1 1 Multiply by 1 in   100% the form 100%.

3 3 1 100   % 3 1 1  100  % Multiply and factor.

3 1 100 Divide out the  % common factor.

3 1  33 %, or 33.3% Simplify.

3

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3

4

Chapter R Review of Basic Concepts

Section R.2

Classroom Example 7 (page 20)

Sets

Classroom Example 1 (page 16) (a)

{5, 6, 7, …, 10} is finite; 8 is an element.

(b)

1, , , ,  is infinite; 8 is not an 1 2

1 3

1 4

element. (c)

{x|x is a fraction between 9 and 10} is infinite; 8 is not an element.

(d)

{x|x is a natural number between 7 and 9} is finite; 8 is an element.



(a)

(b)

(c)

Classroom Example 2 (page 16) (a)

{9, 10, 11}

(b)

{7}



S  12, 7.25,  3,  56 , 0, 0.45, 9, 21 , 999 3

(d)

The natural numbers in set S are

 9 (or 3),

.

21 (or 7), 999 3

The whole numbers in set S are

0, 9 (or 3),

.

21 (or 7), 999 3

The integers in set S are

12, 0, 9 (or 3),

.

21 (or 7), 999 3

The rational numbers in set S are

12,  7.25,  , 0, 0.45, 9 (or 3), 5 6

.

21 (or 7), 999 3

Classroom Example 3 (page 17)

 

(a)

B  D; false

(e)

The irrational number in set S is  3 .

(b)

D  B; true

(f)

All elements of S are real numbers.

(c)

D  A; false

Section R.3

(d)

U  A; true

Classroom Example 1 (page 25)

Classroom Example 4 (page 18)

Real Numbers

(a)

6.85  6.85

(a)

A contains the elements of U that are not in A: {1, 3, 5, 7, 9}

(b)

 50  50

(b)

B  contains the elements of U that are not in B: {1, 2, 4, 5, 7, 8}

(c)



(c)

U  

(d)

If y  2, y 

(d)

  U

2 2 2   3 3 3 2  2

Classroom Example 2 (page 27)

Classroom Example 5 (page 18) (a)

{15, 20, 25, 30  {12,18, 24, 30}  {30}

(b)

{3, 6, 9,12,15,18}  {6,12,18, 24}  {6,12,18}

(c)

{6, 7, 8}  {678}  

(a)

Add the absolute values. The numbers have the same sign, both negative, so the sum will be negative. 4   9    4  9     4  9  13

(b)

The numbers have different signs. Subtract the lesser absolute value from the greater. Because the sum has the same sign as the number with the greater absolute value, and 4.6  3.8 , the

Classroom Example 6 (page 19) (a)

{1, 3, 5, 7, 9}  {3, 6, 9,12}  {1, 3, 5, 6, 7, 9,12}

(b)

{9,10,11,12}  {10,12,14,16}  {9,10,11,12,14,16}

(c)

{2, 6, 10, 14, }  {4, 8, 12, 16, }  {2, 4, 6, 8, 10, }

sum is negative. 4.6  3.8    4.6  3.8     4.6  3.8  0.8

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Section R.3 Real Numbers

(c)

(d)

To subtract 11 from –5, change to addition and replace 11 by its additive inverse (opposite), –11. Then apply the addition rules. 5  11  4   11    4  11     4  11  15

(opposite), 23 . Then apply the addition rules.

3  2 3 2      4  3 4 3 9 8 The LCD is 3  4  12.   3 3 9 8 2 4 12 12 4  3  12 ; 3  4  12 17  12

Classroom Example 3 (page 28)

(b)

(c)

The numbers have different signs, so the product is negative.

7  2  14 The numbers have the same sign, so the product is positive. 0.9  15  13.5 0  3  0 because the product of zero and any

(e) (f) (g)

The numbers have the same sign, so the product is positive.

3  5  15     4  6  24 5 Write in lowest terms.  15  8533  85 8 24 

is a complex fraction.

11 16  83 3 16   11 8 11 16

a  a  b1 ; Multiply by 16 b 11

,

the reciprocal of the divisor.

Multiply numerators.

Multiply denominators.

Write in lowest terms.

48 6 8 6  11  11 88 8

Classroom Example 4 (page 29) (a)

103  10  10  10  1000 The base is 10, and the exponent is 3.

(b)

(3) 4  (3)(3)(3)(3)  81 The base is −3, and the exponent is 4.

(c)

34   (3  3  3  3)  81 The base is 3, and the exponent is 4.

(d)

2  52  2  5  5  50 The base is 5, and the exponent is 2.

(e)

(2  5) 2  102  10  10  100 The base is 10, and the exponent is 2.

Classroom Example 5 (page 30) (a)

3  9  25  4  3  9  32  4  27  8  19

(b)

(30  5)  3  15  7  25  3  15  7  75  15  7  5  7  12

number is zero. (d)

 83

48  88 6  11

To subtract  23 from 34 , change to addition and replace  23 by its additive inverse

(a)

(h)

5

(c)

24  11 16  11 5 1    9  3 2 96 15 3

(d)

7 2  (9) 49  9 40 5    6(3)  1(2) 18  2 16 2

Classroom Example 6 (page 31)

For parts (a)–(d), x = –4, y = 3, and z = –6.

10 is undefined.

0

(a)

0  0 because 0  0.5  0. 0.5

The numbers have the same sign, so the quotient is positive. 10 2 5

(b)

6 x 2  5 y  3z  6(4) 2  5(3)  3(6)  6(16)  15  18  96  15  18  129 4 y  3( x  1) 2 4(3)  3(4  1) 2  z9 6  9 12  3(5) 2 12  3(25)   3 3 12  75 63    21 3 3

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6

Chapter R Review of Basic Concepts

(c)

x y 4 3   4 3  4 3  1  1  0  0 z x 6 4 3  2 1   2 2 2 2

Classroom Example 7 (page 33) (a)

commutative property associative property

(12  2 x)  18  (2 x  12)  18  2 x  (12  18)  2 x  30

(b)



(b)

 

 24 s

3  4 1

7   7

(b)

2 y   2   y   2

(c)

5 43  43  43(5) 415    ,z  0  2 5 z 2(5) z10 z  z2

(d)

2 3a 3 3 a3  3a 3    4   2 2  bc  bc 4 b2 c4

3 5

3(5)

5 3 4

5 4

3 4

5(4)

y 3(4)  220 y12

   

   

2



    2

9a 3(2) 9a 6  b 2 c 4(2) b 2 c8

2

Classroom Example 3 (page 42)

Classroom Example 8 (page 33)

(a)

80  1

(b)

80  1

(a)

8(m  2n)  8m  16n

(c)

(8)0  1

(d)

(8)0  1

(b)

(3r  5s )  3r  5s

(e)

(c)

22t  55  11(2t  5)

3b   1, b  0

(d)

3 5 p  12 q  28 4 6





8 0

Classroom Example 4 (page 43)

 85 p  83 q  21

1 1  103 1000

(a)

103 

(b)

51  

3m  5n  3(13)  5(9)  39  45  6  6

(c)

4   9

2m  3 n 2(13)  3 9 26  3 9   mn 13  (9) 4 26  3(9) 26  27 1    4 4 4

(d)

mn 4 

Classroom Example 11 (page 35)

(e)

(mn) 4 

d  P, Q   14   8  22  22

Classroom Example 5 (page 44)

Classroom Example 9 (page 34) Pd  P  120  146  120  26  26

Classroom Example 10 (page 34) (a) (b)

The distance between −8 and 14 is 22.

Section R.4

Integer and Rational Exponents

Classroom Example 1 (page 40) (a)

m6  m8  m6  8  m14

 90r 8

 715

5

 associative property

  commutative property associative 4  s  54    9  property commutative   54    94  s property

4

(a)

4 (35t )  74  (35) t 7

54 s  94   s (54)   94

3

Classroom Example 2 (page 41)

 20t

(c)

5r 6r  3r   (5)(6)(3) r

1 5

2



1 4   9

2



1 81  16 16 81

m n4 1 1  (mn)4 m 4 n 4

(a)

158  158  3  155 153

(b)

y4  y 4  ( 9)  y13 y 9

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Section R.5 Polynomials



  6  61  361

(c)

35r 6 7 r 6  ( 4) 7r10   4 5 5 25r

(e)

2162 3  2161 3

(d)

34a8b11 2a8 12b11 5 2a 4b6 2b6    4 3 3 51a12b5 3a

(f)

1003 2  1001 2  is not a real number.

(a)



  5 x 2 x   5 x 2 x 

1 4 3

5x 2 x

3

1( 3) 4( 3)

3

3 12

40  5 x3 12 (8)  9 x

30r 4 s 9 2 4  ( 6) 9  3 2 10 12 2r10  r  r s  12 s 3 45r 6 s 3 3 3s

(c)

4b  4b   4 b 4  b 4 b 4 b  1 3

2 3( 2)

1 3 4

2

2

3

(a)

181 2  187 2 181 2  7 2 184   3  18 183 183 18

(b)

1003 2  163 4  1001 2



 16 

1 4 3

3

 103  23 

(b)

3 2

2

Classroom Example 9 (page 47)

Classroom Example 6 (page 44) 3

3

1( 3)

42  3 b 6b3 45 b 3  4 12 44 b 12 4 b  45  4 b 3  ( 12)  49 b9 b9  9 4

4 z 3 4  5 z 2 5  20 z 3 4  2 5  20 z 23 20

(d)

 5m 4 3   m 4   23   5  n   8n 

2

13



Classroom Example 7 (page 46)

Section R.5



5 2 m8 3 m 4 3  n 4 3 81 3 n5 3



25m8 3  4 3 25m 4  2n 4 3  5 3 2n3

Polynomials

Classroom Example 1 (page 52) (a)

x5  5 x3  8 is a polynomial. The first term, x5 , has the greatest degree, so this is a polynomial of degree 5. It is a trinomial because it has three terms.

(b)

x 2  7 xy 3 is a polynomial. The first term has degree 2, while the second term has degree 4, so the polynomial has degree 4. There are two terms, so it is a binomial.

(a)

491 2  7

(b)

1441 2  12

(c)

12

(144)

(d)

641 6  2

(e)

(64)1 6 is not a real number

(f)

641 6  2

(g)

(125)1 3  5

Classroom Example 2 (page 52)

(h)

641 3  4

(a)

 12

(c)



  3  27



  5  125

3

(b)

253 2  251 2

(c)

45 2   41 2

(d)

64

3

3

(b)

   2  32

23



 641 3

5



2

17 x  10 x  x  9 x  10 x  5x 3

2

3

2

(c)

  4  16 2

6m  11m  21  m  6m  35 4

2

4

2

 (6  1)m 4  (11  (6))m 2  (21  35)  7m 4  5m 2  14

3

5

4 is not a polynomial because there is y a variable in the denominator.  y8 

 (17  9) x3  (10  10) x 2  (1  5) x  8 x3  4 x

Classroom Example 8 (page 46) 813 4  811 4

1000  125 8

(c)

1( 4) 3( 4)

(a)

7

10r s  5r s   25r s  15r s  3 6

6 3

3 6

 (10  25)r 3 s 6  (5  15)r 6 s 3  35r 3 s 6  10r 6 s 3

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6 3

8

Chapter R Review of Basic Concepts



 

6 z 2  5 z  3  4 3z 2  2 z  9

(d)



 



 6 z 2  30 z  18  12 z 2  8 z  36 2

(b)

( a  b) 4  ( a  b) 2 ( a  b) 2  (a 2  2ab  b 2 )(a 2  2ab  b 2 )  a 4  2a3b  a 2b 2  2a3b  4a 2b 2  2ab3  a 2b 2  2ab3  b 4  a 4  4a3b  6a 2b 2  4ab3  b 4

(c)

( s  4t )3  ( s  4t )( s  4t ) 2  ( s  4t )( s 2  8st  16t 2 )  s 3  8s 2t  16 st 2  4s 2t  32 st 2  64t 3  s 3  12 s 2t  48st 2  64t 3



 (6  12) z  (30  8) z  (18  36)  6 z 2  22 z  18

Classroom Example 3 (page 53) 2

3t  2t  7 4t  5  15t 2  10t  35 12t  8t 2  28t 3

 5(3t 2  2t  7)  4t (3t 2  2t  7)

Classroom Example 7 (page 56)

12t 3  23t 2  38t  35

36 x5  24 x 4  12 x3 36 x5 24 x 4 12 x3    6x4 6 x4 6 x4 6x4

Classroom Example 4 (page 54) (a)

(b)

(c)

(7 y  3)(4 y  5)  (7 y )(4 y )  (7 y )(5)  3(4 y )  3(5)  28 y 2  35 y  12 y  15  28 y 2  23 y  15

 6x  4  2x







 x3 4 x 2  10 x  10 x  25  x3 4 x 2  25 5

 4 x  25 x

3

 6x  4 



35m5 n 4  49mn3  12mn 35m5 n 4 49mn3 12mn    7m2 n2 7m2 n2 7m2 n2 7m2 n2 2 2 Divide each term by 7 m n .

7n 12  5m 3 n 2   m 7mn

Classroom Example 9 (page 56) 4n 2  n  1

(a)

(7 m  10)(7m  10)  (7 m) 2  102  49m 2  100

(b)

(4r 2  9)(4r 2  9)  (4r 2 ) 2  92  16r 4  81

(c)

(5 x 2  8 y 4 )(5 x 2  8 y 4 )  (5 x 2 )2  (8 y 4 ) 2  25 x 4  64 y8

(d)

(8 z  3) 2  (8 z ) 2  2(8 z )(3)  32  64 z 2  48 z  9 (5 z  12q3 ) 2  (5 z ) 2  2(5 z )(12q 3 )  (12q 3 ) 2  25 z 2  120 zq 3  144q 6

3n  2 12n3  11n 2  5n  8 12n3  8n 2 3n 2  5n 3n 2  2n 3n  8 3n  2  10 12n3  11n 2  5n  8 10  4n 2  n  1  3n  2 3n  2

Classroom Example 6 (page 55) (a)

2 x

Classroom Example 8 (page 56)

Classroom Example 5 (page 54)

(e)

1

1 3 4 12 x3  12 x  2 x  2x 6 6 x4

(6 p  11)(6 p  11)  (6 p)(6 p )  6 p (11)  11(6 p)  11(11)  36 p 2  66 p  66 p  121  36 p 2  121 x3 (2 x  5)(2 x  5)  x3  2 x (2 x)  2 x(5)  (5)(2 x)  (5)(5) 

4

Divide each term by 6 x .

(4 x  3)  7 y (4 x  3)  7 y 

 (4 x  3) 2  (7 y ) 2  (4 x) 2  2(4 x)(3)  (3) 2  (7 y ) 2  16 x 2  24 x  9  49 y 2

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Section R.6 Factoring Polynomials

Classroom Example 10 (page 57)

(b)

The positive factors of 12 are 1 and 12, 2 and 6, and 3 and 4. The factors of −3 are −1 and 3, and 1 and −3. Try different combinations of these factors until the correct one is found.

12t 2  5t  3  (4t  3)(3t  1)

(c)

The positive factors of 3 are 1 and 3. Because the middle term is negative, we consider only the negative factors of 16: −1 and −16, −2 and −8, and −4 and −4. Trying the different combinations of these factors, we find that none give the correct product. Therefore, 3x 2  15 x  16 cannot be factored with integer coefficients and is prime.

(d)

First, factor out the GCF, 3: 24 x 2  42 x  15  3(8 x 2  14 x  5)

8x2  4 x 2  0 x  2 8 x 4  0 x3  12 x 2  7 x  18 8 x 4  0 x3  16 x 2  4 x 2  7 x  18 4 x 2  0 x  8 7 x  10 8 x 2  12 x 2  7 x  18 7 x  10  8x2  4  2 2 x 2 x 2

Section R.6

Factoring Polynomials

Classroom Example 1 (page 61) 2

4

2

2

(a)

6a  18a  6a (1  3a )

(b)

14 x3 y 2  28 x 2 y 3  21x 2 y 2  7 x 2 y 2 (2 x  4 y  3)

(c)

24( x  2)3  16( x  2) 2  6( x  2)  2( x  2) 12( x  2) 2  8( x  2)  3  2( x  2) 12( x 2  4 x  4)  8 x  16  3

 2( x  2) 12 x 2  48 x  48  8 x  16  3  2( x  2)(12 x 2  56 x  67)

Classroom Example 2 (page 62) (a)

(b)

(c)

r 2 s  3r 2  5s  15  (r 2 s  3r 2 )  (5s  15)  r 2 ( s  3)  5( s  3)  (r 2  5)( s  3)

4m 2  n 2  4n  m 2 n  (4m 2  m 2 n)  (4n  n 2 )  m 2 (4  n)  n(4  n)  (m 2  n)(4  n) 3

2

3

Now factor 8 x 2  14 x  5 . The positive factors of 8 are 1 and 8, and 2 and 4. The positive factors of 5 are 1 and 5. Try different combinations of these factors until the correct one is found.

24 x 2  42 x  15  3(4 x  5)(2 x  1) Classroom Example 4 (page 64) (a)

49 x 2  28 xy  4 y 2  (7 x  2 y ) 2

(b)

81a 2b 2  90ab  25  (9ab  5)2

Classroom Example 5 (page 65) (a)

64r 2  49  (8r  7)(8r  7)

(b)

169u 6  144v 4  (13u 3  12v 2 )(13u 3  12v 2 )

(c)

(2c  3d ) 2  16 f 2   2c  3d   4 f   2c  3d   4 f    2c  3d  4 f  2c  3d  4 f 

2

9 y  15 y  6 y  10  (9 y  15 y )  (6 y  10)  3 y 2 (3 y  5)  2(3 y  5)  (3 y 2  2)(3 y  5)

(d)

The positive factors of 5 are 5 and 1. The factors of −12 are −1 and 12, −2 and 6, −3 and 4, −4 and 3, −6 and 2, and −12 and 1. Try different combinations of these factors until the correct one is found.

5 z 2  4 z  12  (5 z  6)( z  2)

x 2  18 x  81  25 y 2

  x  9  25 y 2   x  9  5 y   x  9  5 y  2

Classroom Example 3 (page 63) (a)

9

  x  9  5 y  x  9  5 y 

(e)

4 x 2  y 2  10 y  25  4 x 2  ( y 2  10 y  25)  4 x 2  ( y  5) 2   2 x  ( y  5)  2 x  ( y  5)   (2 x  y  5)(2 x  y  5)

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10

Chapter R Review of Basic Concepts

Classroom Example 6 (page 66)

(c)

(a)

t 3  1000  (t  10)(t 2  10t  100)

(b)

r 3  8s 3  (r  2 s )(r 2  2rs  4 s 2 )

(c)

  x  3

2 5

  x  3

2 5

125u 9  216v12  (5u 3  6v 4 )(25u 6  30u 3v 4  36v8 )

Section R.7

(b)

Let u  3 x  1 . Then 8(3 x  1) 2  10(3x  1)  25  8u 2  10u  25  (4u  5)(2u  5)   4(3 x  1)  5 2(3 x  1)  5  (12 x  4  5)(6 x  2  5)  (12 x  1)(6 x  7) Let u  3 x  1 . Then (3x  1)3  27  u 3  27  (u  3)(u 2  3u  9)  (3 x  1  3) (3x  1) 2  3(3x  1)  9  (3 x  2) (9 x 2  6 x  1)  (9 x  3)  9

The domain of

(a)

(b)

Let u  m 2 . Then 15m 4  m 2  6  15u 2  u  6  (3u  2)(5u  3)  (3m 2  2)(5m 2  3)

(a) (b)

 5 5  3 y 2  5 

28 y 5  21y 2  7 y 5 4 y



 7 y 5 y  3 y 3



(c)

Simplify the exponents.

Check by multiplying the factors. (b)

  6n 3n  2

18n 4 3  12n1 3  6n1 3 3n 4 3 1 3  2n1 3 1 3 13

33

12 x 2  30 x 6 x(2 x  5) 6x   2 (2 x  5)(2 x  5) 2 x  5 4 x  25 x 2  8 x  16 ( x  4) 2 ( x  4) 2    2 x(4  x) 2 x( x  4) 8x  2 x2 x   4  4 x x4 or   2x 2x 2x

Classroom Example 3 (page 74)

The least exponent of the variable x in 28 y 5  21 y 2 is y 5 . Because 7 is the greatest common numerical factor, factor out 7 y 5 .

Factor.

 x  7 x  1 consists of all real  x  3 x  1

Classroom Example 2 (page 73)

Classroom Example 8 (page 67) (a)

Rational Expressions

numbers except those that make the denominator equal 0.  x  3 x  1  0  x  3  0  x  3 or x 1  0  x  1 The domain is  x | x  3, 1 .

 (3 x  2)(9 x 2  15 x  13)

(c)

1   x  3 2  x 

Classroom Example 1 (page 72)

Classroom Example 7 (page 67) (a)

 x  3 2 5   x  33 5 2 5    x  3  x  3 2 5   2 5   x  33 5  2 5  



6 z 6 28 8 z 4   7 9z2 3 4n 2  3n  10 2n  1 (n  2)(4n  5) 2n  1    2n 2  3n  2 n  4 (n  2)(2n  1) n  4 4n  5  n4 5 z 2  16 z  3 30 z 2  6 z  3 z 2  z  12 2 z  8z 2 2 5 z  16 z  3 2 z 3  8 z 2  2  z  z  12 30 z 2  6 z (5 z  1)( z  3) 2 z 2 ( z  4) z    ( z  4)( z  3) 6 z (5 z  1) 3

 6n1 3 3n  2

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Section R.8 Radical Expressions

(d)

x 2  1 xy  2 y  3x  6  x3  1 xy  3 x  y  3 y ( x  2)  3( x  2) ( x  1)( x  1)   2 ( x  1)( x  x  1) x( y  3)  ( y  3) ( x  1)( x  1) ( y  3)( x  2)   ( x  1)( x 2  x  1) ( x  1)( y  3) x2  2 x  x 1

Classroom Example 4 (page 76)

(b)

(c)

3 2 9 4z 9  4z     10 z 4 15 z 2 30 z 4 30 z 4 30 z 4

2

7 2m 7 2m    m5 5m m5 m5 7  2m 2m  7  or 5m m5 4 6  x 2  2 x  15 x 2  25 4 6   ( x  3)( x  5) ( x  5)( x  5) Factor the denominators.

The LCD is ( x  3)( x  5)( x  5).

4 x  20  (6 x  18) ( x  3)( x  5)( x  5)

Apply the distributive property in the numerator.



2 x  2 ( x  3)( x  5)( x  5)

Classroom Example 5 (page 77) (a)

(b)

x 6  12 x

   3x  4  x 6   6 x  1 x 2 3  42

2

2

2

x

1 x2



1  z 11 z ( z  1)( z  1) z 11  z 11 z 1  1  z 11 z ( z  1)( z  1) 1z  z 11 z



x y

x y

xy ( x  y ) xy ( x  y )  2  ( y  x)( y  x) y  x2 xy  ( y  x)

Radical Expressions

Classroom Example 1 (page 83) 27  271 3  3

(a)

3

(b)

 4 10, 000  10, 0001 4  10

(c)

3

216  (216)1 3  6

(d)

4

81 is not a real number.

(e)

3

125  125    512  512 

(f)

 5 243   (243)1 5  (3)  3





z ( z  1)  z ( z  1)  ( z  1)( z  1)  z ( z  1) 2 z z2  z  z2  z  2  2 2 z  1  z  z 2z  z  1



1251 3 5  5121 3 8

Classroom Example 2 (page 84)

  2 8 3

(a)

163 4  4 16

(b)

642 3   3 64   42  16

3

(c)

1213 2  

(d)

y7 8  8 y7

(e)

7z4 5  7 5 z4

(f)

12q1 4  12 4 q

(g)

5 x  2 y 1 6  6 5 x  2 y

2

Combine like terms in the numerator.

3  42

x y

13

4( x  5) 6( x  3)   ( x  3)( x  5)( x  5) ( x  3)( x  5)( x  5) 

x y

1  1y x 2 y 2  xy x 1  y 1 x xy  1  2 2  y x y 2  x2  12 x 2  y 2 x2 y 2  2 2 x2 y 2 2

Section R.8 2

(a)

Classroom Example 6 (page 78)

 121  11  1331 3

3

Classroom Example 3 (page 84) (a)

7

n3  n3 7

(b)

4

10 x  10 x 

(c)

15 3 r

(d)

2 5 3x 2

14

   15r 4

43

   2 3 x  2  3 x

Copyright © 2021 Pearson Education, Inc.

8

5

8 16

8 5 16 5

11

12

(e)

Chapter R Review of Basic Concepts 3



r 2  s4  r 2  s4



Classroom Example 7 (page 87)

13

Classroom Example 4 (page 84) z2  z

(a) (b)

3

(a)

10

(b)

9

(c)

6 3

z3  z3 3  1

z   z 3 2

(c)

z6 

(d)

81r 8 s10  9r 4 s 5  9r 4 s 5

(e)

4 ( 3) 4  3  3

(f)

5

14 5 pq  11 5 pq  (14  11) 5 pq  3 5 pq

(b)

75ab3  b 12ab  3  25ab 2b  b 4  3ab  5b 3ab  2b 3ab  5b  2b  3ab  3b 3ab

(c)

Classroom Example 5 (page 85)

(b)

5

n3  5 n 2  5 n3  n 2  5 n5  n

(c)

3

4  3 21x 2 y  3 84 x 2 y





 3xy 2  2 x 2 y 3 3 x 2 y

Classroom Example 9 (page 88) (a)

3 2

 5  4 7   3 2  5   3 2 4 7 

Distributive property

 3 10  3  4  2  7 a  b  ab Commutative and associative properties

6 6 a a a   12 6 12 b b2 b

(e)

6

(f)

5 4

(g)

6

(h)

 3xy 2 3 3 x 2 y  2 x 2 y 3 3x 2 y

11 11 11   169 13 169

 3 10  12 14

17  20 17 5 

12

(b)

5

Multiply.

 11  17  11  17    11   17  2

2

 11  17  6

5  9 cannot be found using the product rule.

3

3 81x 5 y 7  3 24 x8 y 4

 3 27  3 x3 x 2 y 6 y  3 8  3 x 6 x 2 y 3 y

5  45  5  45  225  15

(d)

42  18 42  42 18  41 9  9 4

(a)

2

x  10 x  25  ( x  5)  x  5

(a)

19

Classroom Example 8 (page 87)

m10  m10 5  m 2

2

(h)

a3b18   a 3b18   a3 9b18 9  a1 3b 2  b2 3 a

3

(3 x  4) 2  3x  4

(g)

25  25 10  21 2  2

4

(c)

Classroom Example 6 (page 86) (a)

288  2  144  2  144  12 2

(b)

8 3 125  8  5  40

(c)

3

128a 6b8c10  3 2  64a 6b6b 2 c9 c





 3 64a 6b6 c9 2b 2 c  4a 2b 2 c3 3 2b 2 c

5  32 3  2   5  2 16 3  2   5  4 2 3  2   15  5 2  12 2  4 2 2  15  5 2  12 2  8 77 2

Classroom Example 10 (page 89)



(a)

2 2 7 2 7   7 7 7 7

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Section R.8 Radical Expressions

(b)

3

3 3 4 34 433 12 3 12  3    9 3 9 3 32  3 3 3 33

Classroom Example 11 (page 90) (a)

3

a 5b

3

2 5

a b

 

(b)

4

3

a5  3 b

3

2

3

3

2

a  b a b

b 3 b 3 b2

5

3

a3

3

b

4

a b

2

 

3

b 3 b3

 

a 3

b b a 3 b2 a 3 b2  bb b2

4 4 4 6 3 6 3 6 43 4      4 4 8 4 16 x8 x16 x2 x x x



x2 4 6  4 3 x4

Classroom Example 12 (page 90)







      2 3  5  2 3  5  3  5    

2 3 5 2 3 5 2   2 3 5 3 5 3 5 32  5 95

4

2

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13

Chapter 1 Equations and Inequalities Section 1.1

Linear Equations

Classroom Example 1 (page 107) 4 3x  5  3  8 x  7  12 x  20  3  8 x  7 12 x  20  4  8 x 24  4 x 6 x Solution set: {6}

P  $2580, t  129  34 yr, and r  0.06

3x  6 1 2 33  x x 10 2 5 5 33   3x  6 1  2 10   x   10  x    10 5 2  5 3 x  6  5 x  4 x  66 2 x  6  4 x  66 6 x  60 x  10 Solution set: {−10}

Classroom Example 3 (page 108) (a)

6 x  9  4 x  13 2 x  22  x  11 conditional equations; {11}

(b)

10  14 x  7  2 x  5 10  14 x  14 x  35 10  35 contradiction; 

(c)

3  2 x  1  5 x  3  x 6 x  3  5 x  3  x x  3  3  x identity; {all real numbers}

Classroom Example 4 (page 109) d  rt 

d t r 2

S  kr  kr 

(b)



S  k r 2  r S k r  r 2

11y  8  2  4 y  5w  6 z 11y  8  8 y  10 w  6 z 3 y  10 w  6 z  8 10 w  6 z  8 y 3

Classroom Example 5 (page 110)

Classroom Example 2 (page 107)

(a)

(c)





I  Prt  2580 0.06 34  $116.10

Section 1.2 Applications and Modeling with Linear Equations Classroom Example 1 (page 113) Let x = the length of the original rectangle. Then, x − 2 = the width of the original rectangle. x + 3 = the length of the new rectangle, so (x − 2) + 3 = x + 1 = the width of the new rectangle.

The perimeter of the new rectangle is 2  x  3  2  x  1  2 x  6  2 x  2  4 x  8 . The perimeter of the new rectangle is 4 in. less than 8 times the width of the original rectangle, so we have 4 x  8  8  x  2  4 . Solving, we have 4 x  8  8  x  2  4 4 x  8  8 x  16  4 4 x  8  8 x  20 28  4 x 7x Thus, the length of the original rectangle is 7 in. and the width of the original rectangle is 5 in.

Classroom Example 2 (page 114) Let x = the distance to her grandmother’s house. r t d Going

40

x 40

x

Returning

48

x 48

x

The driving time returning was 1 hour less than the driving time going, so we have x x 1  .

40 48 (continued on next page)

14

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Section 1.3 Complex Numbers

(continued)

15

Classroom Example 5 (page 99)

Solving, we have x x 1  40 48  x   x  40  48   1  40  48    40   48  48 x  1920  40 x 1920  8 x  240  x The distance from Krissa’s home to her grandmother’s home is 240 mi.

P  1.06 F  7.18 P = 70, so we have 70  1.06 F  7.18 62.82  1.06 F  59.26  F The flow rate must be approximately 59.26 L per second to remove 70% of the contaminants.

Classroom Example 6 (page 117) (a)

The year 2019 is 9 years after the year 2010.

Let x = 9 and find the value of y. y  416 x  8042 y  416 9  8042  11786 The per capita health care expenditures in the year 2019 were $11,786.

(b)

Let y = 12,000 and find the value of x y  416 x  8042 12000  416 x  8042 3958  416 x x  9.5 Per capita health care expenditures are projected to reach $12,000 about 9.5 years after 2010, or in 2019.

Classroom Example 3 (page 115) Let x = the amount of 25% antifreeze solution (in liters).

Gallons of Gallons of Strength Pure Solution Antifreeze 25%

x

0.25x

10%

5

0.10  5  0.5

15%

x5

0.15  x  5

The number of gallons of pure antifreeze in the 25% solution plus the number of gallons of pure antifreeze in the 10% solution must equal the number of gallons of pure antifreeze in the 15% solution.

0.25 x  0.5  0.15  x  5 0.25 x  0.5  0.15 x  0.75 0.1x  0.25  x  2.5 L 2.5 liters of the 25% solution should be added.

Section 1.3

Complex Numbers

Classroom Example 1 (page 124) (a)

81  i 81  9i

(b)

55  i 55

Classroom Example 4 (page 16)

(c)

98  i 98  i 49  2  7i 2

Let x = amount invested at 2.4%.

Then 28,000 − x = amount invested at 3.1%.

Amount in Interest Interest Account Rate x 2.4% 0.024x 28,000 − x 3.1% 0.031(28,000 − x) 784 The amount of interest from the 2.4% account plus the amount of interest from the 3.1% account must equal the total amount of interest.

0.024 x  0.031 28, 000  x   784 0.024 x  868  0.031x  784 0.007 x  84 x  12, 000 Owen invested $12,000 at 2.4% and $28,000 − $12,000 = $16,000 at 3.1%.

Classroom Example 2 (page 125)

 21  21

(a)

21  21  i 21  i 21  i 2

(b)

5  30  i 5  i 30  i 2 150  i 2 25  6  5 6

(c)

42 i 42   3 i 3

(d)

63 i 63 63  i i 3 21 21 21

2

42  14 3

Classroom Example 3 (page 125) 15  75 15  i 75 15  i 25  3   5 5 5 5 3  i 3 15  5i 3    3i 3 5 5



Copyright © 2021 Pearson Education, Inc.



16

Chapter 1 Equations and Inequalities

Classroom Example 4 (page 126)

(b)

x 2  144  x  12i

(a)

4  5i   5  8i   1  3i

(c)

(b)

10  7i   5  3i   15  10i

( x  8) 2  24  x  8   24  x  8  2 6  x  8  2 6

Classroom Example 3 (page 134)

Classroom Example 5 (page 1279) (a)

x 2  10 x  20  0 x 2  10 x  20

5  3i 2  7i   10  5  7i   3i  2  3i  7i 



x 2  10 x  12  10

 10  35i  6i  21i 2  10  29i  21  31  29i

(b)

x  5   45  3 5 x  5  3 5

9  8i 9  8i   92  8i 2 difference of two squares

Classroom Example 4 (page 135) 4x2  6 x  5  0 x 2  64 x  54  0

x 2  64 x   54



x 2  64 x  12  64

Classroom Example 6 (page 128)

(b)

15 15 i 15i 15i      15i, or 0  15i i i i i 2 1

(b)

i 14  i 16  i 2  i16

8

2

4 4

1

 11  i 2  i 2  1

Section 1.4

 x  34    1611 2

11   411 i 16 x   34  411 i

x  34  i

2 b  b 2  4ac 6  6  4(1)(3)  2a 2(1) 6  36  12 6  48   2 2 6  4 3   3  2 3 2

x

   i   i    i 1

2

x2  6 x  3  x2  6 x  3  0 a  1, b  6, c  3

  i 1 i  i 8

2

Classroom Example 5 (page 135)

Classroom Example 7 (page 129) i 33  i 32  i  i 4

   54   12  64 

9   54  169 x 2  32 x  16

5  5i 5  5i 3  i 15  5i  15i  5i 2    3i 3i 3i 9  i2 15  20i  5 10  20i    1  2i 9   1 10

(a)

2

 x  52  45

4  5i 2  42  2 45i   5i 2

 81  64i 2  81  64  1  81  64  145, or 145  0i

(a)

2

x 2  10 x  25  20  25

 16  40i  25i 2  16  40i  25  9  40i

(c)

  20   12 10

Quadratic Equations

2

Classroom Example 1 (page 132)

Classroom Example 6 (page 136)

10 x 2  x  2  0 (2 x  1)(5 x  2)  0

4 x 2  3x  5  4 x 2  3x  5  0 a  4, b  3, c  5

2 x  1  0 or 5 x  2  0 2 x  1 x   12

or or



5x  2 x  52

Solution set:  12 , 52



Classroom Example 2 (page 133) (a)

2 b  b 2  4ac (3)  (3)  4(4)(5)  2a 2(4) 3  9  80 3  71 3  i 71    8 8 8 3 71 i   8 8

x

x 2  29  x   29

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Section 1.5 Applications and Modeling with Quadratic Equations

Classroom Example 7 (page 136)



x3  125   x  5 x 2  5 x  25

(c)



difference of cubes (section R.4)

x  5  0 or x  5 or

x 2  5 x  25  0 x

5  52  4(1)(25) 2(1)

5  25  100 2 5  75 x 2 5  5 3i x 2 5 5 3 x  i 2 2  5 5 3  i Solution set: 5,   2 2  

17

2 x2  6 x  7  2 x2  6 x  7  0 a  2, b  6, c  7 b 2  4ac  (6) 2  4(2)(7)  92 Therefore, there are two distinct irrational solutions.

Section 1.5 Applications and Modeling with Quadratic Equations Classroom Example 1 (page 142)

x

Let x = the width of the rectangle. Then 2x = the length of the rectangle. The box is formed by cutting 10 cm + 10 cm from both the length and the width.

Thus the width of the bottom of the box is x − 20, the length of the bottom of the box is 2x − 20, and the height of the box is 10 cm.

Classroom Example 8 (page 137) (a)

(b)

1 3V V   r 2 h  3V   r 2 h   r2  h 3 h 3V 3V r r   h h h  3V  h r h 2my 2  ny  3 p  2my 2  ny  3 p  0 Use the quadratic formula with a = 2m, b = −n, and c = −3p to solve for y: y 

 ( n)  ( n) 2  4(2m)(3 p) 2(2m) n  n 2  24mp 4m

Classroom Example 2 (page 143)

Classroom Example 9 (page 138) (a)

(b)

4 x 2  12 x  9  0 a  4, b  12, c  9 b 2  4ac  (12) 2  4(4)(9)  0 Therefore, there is one distinct rational solution.

2

2

V  lwh 7500  (2 x  20)( x  20)(10) 7500  20 x 2  600 x  4000 0  20 x 2  600 x  3500 0  x 2  30 x  175 0  ( x  5)( x  35) x5 0 or x  35  0 x  5 or x  35 Reject the negative solution because length cannot be negative. The dimensions of the box are 35 cm by 70 cm.

3x  x  5  3 x  x  5  0 a  3, b  1, c  5 b 2  4ac  12  4(3)(5)  59 Therefore, there are two distinct nonreal complex solutions.

Let x = the length of the shorter leg.

Then 2x − 10 = the length of the longer leg, and (2x − 10) + 20 = 2x + 10 = the length of the hypotenuse. Using the Pythagorean theorem, we have x 2   2 x  10   2 x  10 2

2

x 2  4 x 2  40 x  100  4 x 2  40 x  100 5 x 2  40 x  100  4 x 2  40 x  100 x 2  80 x  0 x  x  80  0 x  0 or x  80  0  x  80 . Since length cannot be 0, we reject that solution. The shorter leg is 80 m, the longer leg is 2(80) −10 = 150 m, and the hypotenuse is 2(80) + 10 = 170 m.

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18

Chapter 1 Equations and Inequalities

Classroom Example 3 (page 144) (a)

Section 1.6 Other Types of Equations and Applications

s  4.9t 2  73.5t 100  4.9t 2  73.5t  4.9t 2  73.5t  100  0 Solve using the quadratic formula, with a = 4.9, b = −73.5, and c = 100.

Classroom Example 1 (page 154) (a)

b  b 2  4ac 2a (73.5)  (73.5) 2  4(4.9)(100)  2(4.9) 73.5  5402.25  1960  9.8 73.5  3442.25   1.51 or 13.49 9.8 The projectile will be 100 m above the ground after 1.51 sec and after 13.49 sec. t

(b)

When the projectile returns to the ground, the height s will be 0 ft, so let s = 0 and solve for t: 0  4.9t 2  73.5t  t (4.9t  73.5)  0  t  0 or 4.9t  73.5  0  t  15 The first solution, 0, represents the time at which the projectile was on the ground before being launched. The projectile will return to the ground 15 sec after being launched.

2 x 2  3x  2 x  3  10 x  2 x 2  2 x 2 x2  9 x  3  2 x2  2 x 7x  3 x  73

The restriction x  1 does not affect the result. Therefore, the solution set is

(a)

S  1.43x 2  7.64 x  42.8 For 2012, x = 6.

2

Let S = 40, then solve for x: 40  1.43 x 2  7.64 x  42.8 0  1.43 x 2  7.64 x  2.8

7.64  74.3856 7.64  8.62  2.86 2.86 x  5.7 or x  0.34 The negative result does not make sense in the context of the problem, so disregard it.

According to the model, sales reached 40 million about 5.7 years after 2006, or in 2011. 

7 x5 1   2 x  3 x x  3x The least common denominator is x  x  3  x 2  3x , which is equal to 0 if x = 3

x 2  5 x  x  3  7 x2  4 x  4  0 ( x  2)( x  2)  0  x  2 The restrictions x  0 or x  3 do not affect the result. Therefore, the solution set is 2 .

This is $2 million greater than the actual figure of $35.2 million.

b  b 2  4ac x 2a 7.64  7.64 2  4  1.432.8 x 2  1.43

3 7

or x = 0. Therefore, 0 and 3 cannot possibly be solutions of this equation. x5 1  7  x  x  3    x  x  3  2  x  3 x   x  3x  x  x  5   x  3  7

S  1.43 6  7.64 6  42.8  37.2 million.

(b)

 .

Classroom Example 2 (page 155)

Classroom Example 4 (page 127) (a)

2 x  3 5x  x x 1 2 The least common denominator is 2(x + 1), which equals 0 when x = −1. Therefore, −1 cannot be a solution of the equation.  2x  3 5x   2  x  1    2  x  1 x  2 x 1  x  12 x  3  2 5 x   2 x 2  2 x

(b)

x 5 5 x5 x5 The least common denominator is x − 5, which equals 0 when x = 5. Therefore, 5 cannot be a solution of the equation. x   5   5    x  5   x  5    x  5  x5 x  5  x  5  5 6 x  25  5  6 x  30  x  5 The only possible solution is 5. However, the variable is restricted to real numbers except 5.

Therefore, the solution set is .

Copyright © 2021 Pearson Education, Inc.

Section 1.6 Other Types of Equations and Applications

(b)

x 5 50   2 x  5 x  5 x  25 The least common denominator is  x  5 x  5  x 2  25 , which is equal to 0 if

2  2  0 4  0 Thus, −2 is not a solution.

x  5 .

6  4(6)  12  0 6  24  12  0 6  36  0 66 0 00 Thus, 6 is a solution.

Solution set: {6}

19

Check x = 6: ?

 x  5 x  5 

x 5    x 5 x 5

 50    x  5 x  5  2  x  25  x  x  5  5  x  5  50

x 2  5 x  5 x  25  50 x 2  25  x  5 Because of the restriction x  5, the solution set is .

Classroom Example 5 (page 159) 3x  1  x  4  1 3x  1  x  4  1

Classroom Example 3 (page 156)

 3x  1   x  4  1 3x  1   x  4   2 x  4  1 2

Let x = the amount of time (in hours) it takes Lisa and Keith to rake the leaves.

Part of the r t Job Accomplished Lisa

1 5

Keith

1 4

x

1 x 5

x

1 x 4

Lisa and Keith must accomplish 1 job (raking the leaves), so we must solve the following equation.

1 1 x  x 1 5 4 1  1 20  x  x   20  1 5 4  4 x  5 x  20 20 9 x  20  x   2 92 9 It takes Lisa and Keith 2 92 hr working together to rake the leaves.

Classroom Example 4 (page 158) x  4 x  12  0 x  4 x  12 x 2  4 x  12 x 2  4 x  12  0 ( x  6)( x  2)  0  x  6 or x  2 Check x = −2: ?

2  4(2)  12  0

2

2

3x  1  x  4  2 x  4  1 2x  4  2 x  4

 2 x  4 2   2 x  4 

2

4 x 2  16 x  16  4 x  16 4 x 2  20 x  0 x 2  5 x  0  x( x  5)  0  x  0 or x  5 Check x = 0: ?

3(0)  1  (0)  4  1 1 2  1 1  1 Thus, x = 0 is not a solution.

Check x = 5: ?

3(5)  1  (5)  4  1 4 3 111 Thus, x = 5 is a solution. The solution set is {5}.

Classroom Example 6 (page 160) 3

5 x 2  12 x  6  3 x  0 3

5 x 2  12 x  6  3 x

 5x  12x  6    x  3

2

3

3

3

5 x 2  12 x  6  x 5 x 2  13x  6  0 5 x  3 x  2  0  x  53 or x  2

(continued on next page)

2  8  12  0 2  4  0

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20

Chapter 1 Equations and Inequalities

(continued)

Check x = 259:

Check x  53 :

259  31 2  6 259  31 4  8  0



?

256  6 4 256  8  0 16  6(4)  8  0 00 Thus, x = 259 is a solution.

Check x = 19:

   6  3 53 ? 0

2 35 3  12 53 5

3 9  36  6  3 3  0 5 5 5 3 3 3 3 000 5 5

?

19  31 2  6 19  31 4  8  0

Thus, x  53 is a solution.

16  6 4 16  8  0 4  6(2)  8  0 00 Thus, x = 19 is a solution.

Solution set: {19, 259}.

Check x = 2:

 

?

  6  3 2 0

3 5 2 2  12 2

20  24  6  3 2  0 3 232 000 Thus, x = 2 is a solution.

3

Solution set:

  3 ,2 5

15u 2  4u  3  0

5u  33u  1  0  u 

x3 2  216

 x   216 x   216   6  36 32 23

23

2

3

 x  72 5  4 52  x  7 2 5   45 2   x7 

2

5 15   3

 4

5

Classroom Example 8 (page 162)

 x  31 2  6  x  31 4  8  0 14 Let u   x  3 , so

5  4  3

1 ?

3

3 3  9  12 3 15    4    3  15    5 5  25  5 27 12  333 5 5 5 Thus, x  is a solution.

3 Check x = −3: 2

15  3

2

14 12 u   x  3    x  3 .   Substituting, we have u 2  6u  8  0 u  4u  2  0  u  4 or u  2 Now solve for x, by replacing u with 2

2

2

x  7  25  32  x  25 or x  7  25  32  x  39 Solution set: {−39, 25}

(a)

3 1 or u   5 3

Now solve for x, by replacing u with x 1 : 3 5 1 x 1   x  ; x 1    x  3 5 3 3 5 Check x  : 3

Solution set: {36} (b)

15 x 2  4 x 1  3  15 x 2  4 x 1  3  0

Let u  x 1 . Substituting, we have

Classroom Example 7 (page 161) (a)

(b)

 x  31 4 : 14 4   x  3  44  x  3  259  x 14 2   x  3  24  x  3  19  x

2

1

 4  3

?

3

 1  1 1 4 15     4     3  15     3  3  3 9 3 5 4  333 3 3 Thus, x  3 is a solution.  5 Solution set: 3,   3

Copyright © 2021 Pearson Education, Inc.

Section 1.7 Inequalities

Classroom Example 9 (page 163)

Classroom Example 5 (page 171)

18 x 4  29 x 2  3  0 Let u  x 2 ; then u 2  x 4 . With this substitution, the equation becomes 18u 2  29u  3  0 . Solving, we have 18u 2  29u  3  0 2u  39u  1  0  u  32 or u  19

x 2  2 x  15  0 Step 1: Find the values of x that satisfy x 2  2 x  15  0 .

Now solve for x: x 2  32  x   x 2  19  x  

3   26 2

21

x 2  2 x  15  0   x  5 x  3  0  x  5 or x  3 Step 2: The two numbers divide a number line into three regions.

or

1   13 . Checking each proposed 9

solution, we find that all are solutions.



Solution set:  26 ,  13

Section 1.7



Step 3: Choose a test value to see if it satisfies the inequality x 2  2 x  15  0 .

Inequalities

Classroom Example 1 (page 168)

Interval

2 x  7  5 2 x  12 x6 Solution set:  x | x  6

Test Value

A:  , 3

−5

Classroom Example 2 (page 170)

B:  3, 5

?

52  2 5  15  0

20  0

False

3  4x  2x  8 5  6 x  56  x  x   56

C: 5,  



Solution set: ,  56 

?

0

02  2 0  15  0 15  0 True

6

62  2 6  15  0 90 False

?

Solution set: [−3, 5]

Classroom Example 3 (page 170) R  45 x and C  30 x  5250 Set R ≥ C, and solve for x: 45 x  30 x  5250 15 x  5250 x  350 The break-even point is at x = 350. This product will at least break even only if the number of units produced and sold is in the interval 350,   .

Classroom Example 4 (page 171)

Is x 2  2 x  15  0 True or False?

Classroom Example 6 (page 172) 3x 2  11x  4  0 Step 1: Find the values of x that satisfy 3x 2  11x  4  0 .

3x 2  11x  4  0  3x  1 x  4  0  x   13 or x  4

Step 2: The two numbers divide a number line into three regions.

1  6x  8  4 9  6 x  12 3 x2 2

Solution set:  32 , 2 (continued on next page)

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22

Chapter 1 Equations and Inequalities

(continued)

Step 3: Choose a test value to see if it satisfies the inequality 3x 2  11x  4  0 .

Interval



A: ,  13

B:



 13 , 4

Test Value

Is 3x 2  11x  4  0 True or False?

−1

3  1  111  4  0 10  0 True

0

3 0  110  4  0 4  0 False

5

3 5  115  4  0 16  0 True

?

2



?

2



?

2

C: 4,  





Solution set: ,  13   4,  

The object will be greater than 128 ft above ground level between 1 and 8 seconds after it is launched.

Classroom Example 8 (page 174) 6 4 x3 Step 1: Rewrite the inequality so that 0 is on one side and there is a single fraction on the other side.

6 6 4 40 x3 x3 4  x  3 6  0 x3 x3 6  4 x  12 18  4 x 0 0 x3 x3 Step 2: Determine the values that will cause either the numerator or denominator to equal 0.

18  4 x  0  x  92 or x  3  0  x  3

The values 92 and 3 divide the number line into three regions. Use an open circle on 3 because it makes the denominator equal 0.

Classroom Example 7 (page 173) s  16t 2  144t 16t 2  144t  128 Step 1: Find the values of t that satisfy 16t 2  144t  128 16t 2  144t  128  0  16t 2  144t  128  16 t  8t  1  0  t  8 or t  1

Step 2: The two numbers divide a number line into three regions. (Note that time t ≥ 0.)

Step 3: Choose a test value to see if it satisfies the inequality 16t 2  144t  128 .

Test Value

Is 16t 2  144t  128 True or False?

A:  0, 1

0.5

16 0.5  144 0.5  128 68  128 False

B: 1,8

2

16  2  144  2  128 224  128 True

C: 8,  

10

16 10  144 10  128 160  128 False

Interval

?

2

Step 3: Choose a test value to see if it satisfies the 6 4 inequality, x3 Test Value

A:  , 3

0

?

6 4 03



4

?

6 4 43

5

?

6 4 53

B: 3, 92  C:  92 , 



Solution set:



3, 92 

?

2

2

Is x 6 3  4

Interval

?

Solution set: (1, 8)

Copyright © 2021 Pearson Education, Inc.

True or False? 2  4 False 6  4 True 3  4 False

Section 1.8 Absolute Value Equations and Inequalities

Classroom Example 9 (page 175) 3x  1 4 2x  3 Step 1: Rewrite the inequality so that 0 is on one side and there is a single fraction on the other side.

3x  1 3x  1 4 40 2x  3 2x  3 3x  1 4  2 x  3  0 2x  3 2x  3 3 x  1  8 x  12 0 2x  3 5 x  13 0 2x  3 Step 2: Determine the values that will cause either the numerator or denominator to equal 0. 5 x  13  0  x  13 or 2 x  3  0  x  32 . The 5

Section 1.8 Absolute Value Equations and Inequalities Classroom Example 1 (page 181) (a)

9  4x  7

9  4 x  7 or 9  4 x  7 Case 1 4 x  2 or 4 x  16 Subtract 9. x  12

Solution set: (b)

or

 , 4

x4

1 2

3x  2  x  5 3x  2  x  5 or 3x  2    x  5 Special case a  b

2 x  7

or 3x  2   x  5

Solve each linear equation.

x   72

values 32 and 13 divide the number line into three 5 regions.

4x  3

or



Solution set:  72 , 43



x  34

Classroom Example 2 (page 182) (a)

Test Value



A: , 32 B:

C:



3 13 , 2 5



0



2

 135 , 

Solution set:

5



, 32



Solution set: 1, 4

Is 23xx 13  4 True or False? ?

3(0) 1 4 2(0)  3  13  4 True ?

3(2) 1 4 2(2)  3

7  4 False

?

3(5) 1 4 2(5)  3 16  4 True 7 13 , 5



4 x  6  10

This inequality fits Case 2. 10  4 x  6  10 4  4 x  16 1  x  4

Step 3: Choose a test value to see if it satisfies the 3x  1 inequality, 4 2x  3 Interval

Divide by  4.

(b)

4 x  6  10

This inequality fits Case 3.

4 x  6  10 or 4 x  6  10 4 x  4 or 4 x  16 x  1 or x4 Solution set:  , 1   4,   Classroom Example 3 (page 183) 5  8 x  6  14 5  8x  8 5  8 x  8

or 5  8 x  8

8 x  13 or x  13 or 8



8 x  3 x   83

Solution set: ,  83    13 , 8

Copyright © 2021 Pearson Education, Inc.



23

24

Chapter 1 Equations and Inequalities

Classroom Example 4 (page 183) (a)

Classroom Example 5 (page 184)

7 x  28  0

(a)

“m is no more than 9 units from 3” means that m is 9 units or less from 3. Thus the distance between m and 3 is less than or equal to 9, or m  3  9.

(b)

“t is within 0.02 unit of 5.8” means that t is less than 0.02 unit from 5.8. Thus the distance between t and 5.8 is less than 0.02, or t  5.8  0.02.

The absolute value of a number will be 0 if that number is 0. Therefore 7 x  28  0 is equivalent to 7 x  28  0, which has solution set {−4}. (b)

6 x  9  2

The absolute value of a number is always nonnegative, so the inequality 6 x  9  2 is always true. The solution set is  ,   . (c)

2  5 x  5

There is no number whose absolute value is less than any negative number. The solution set of 2  5 x  5 is .

Classroom Example 6 (page 184)

We want y to be within 0.001 unit of 6, so we have y  6  0.001 or 5 x  2  6  0.001.

5 x  8  0.001 0.001  5 x  8  0.001 7.999  5 x  8.001 1.5998  x  1.6002 Values of x in the interval 1.5998, 1.6002 will

satisfy the condition.

Copyright © 2021 Pearson Education, Inc.

Chapter 2 Graphs and Functions Section 2.1

Rectangular Coordinates and Graphs

Classroom Example 5 (page 206) (a)

The coordinates of M are  7  (2) 5  13   9  ,     , 4 2 2   2 

(b)

Let (x, y) be the coordinates of Q. Use the midpoint formula to find the coordinates:  8  x 20  y  ,     4, 4 2 2 

Classroom Example 1 (page 202) (a)

(transportation, $9576)

(b)

(health care, $4928)

Classroom Example 2 (page 204) d ( P, Q ) 

2  32  8  5

2

8 x  4 8 x 8 x  0 2 20  y  4  20  y  8  y  12 2 The coordinates of Q are (0, 12).

 25  169  194

Classroom Example 3 (page 204)

5  02  1  2

d ( R, S ) 

2

Classroom Example 6 (page 206)

 25  9  34 d ( R, T )   4  0  3  2 2

The year 2014 lies halfway between 2011 and 2017, so we must find the coordinates of the midpoint of the segment that has endpoints (2011, 167.7) and (2017, 233.7)  2011  2017 167.7  233.7  M  ,   2 2

2

 16  25  41

4  52  3  12

d (S , T ) 

 81  4  85

The longest side has length

  2014, 228.8 The estimate of $228.8 billion is $16.4 billion more than the actual amount.

85

 34    41   85  2

2 ?

2

34  41  85 The triangle formed by the three points is not a right triangle.

Classroom Example 4 (page 205)

Classroom Example 7 (page 207)

Choose any real number for x, substitute the value in the equation and then solve for y. Note that additional answers are possible. x

y = −2x + 5

The distance between P (2, 5) and Q(0, 3) is

−1

y  2(1)  5  7

2  0  5  3 

44  8  2 2

0

y  2(0)  5  5

The distance between Q (0, 3) and R (8, 5) is

3

y  2(3)  5  1

2

2

(a)

8  0  5  3  64  64 2

2

 128  8 2 The distance between P (2, 5) and R (8, 5) is

Three ordered pairs that are solutions are (−1, 7), (0, 5), and (3, −1). Other answers are possible.

2

2  82  5  5  100  100  200  10 2

Because 2 2  8 2  10 2 , the points are collinear.

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25

26

Chapter 2 Graphs and Functions

(b)

y

x  3 y1

Find a fourth point by letting x = −2 and solving for y:

−9

x  3 9  1  3 8  2

2  3 y  1   2  y  1  9  y

−2

x  3 2  1  3 1  1

−1

x  3 1  1  3 0  0

The points to be plotted are (0, −1), (1, 0), (2, 7), and (−2, −9). Note that (−1, −2) is also on the graph.

0

x  3 0 1  3 1  1

7

x  3 7 1  3 8  2

3

Ordered pairs that are solutions are (−2, −9), (−1, −2), (0, −1), (1, 0) and (2, 7).

Other answers are possible. (c)

x

y   x2  1

−2

y  (2)2  1  3

−1

y  (1)2  1  0

0

y  02  1  1

1

y  12  1  0

2

y  22  1  3

(c)

Let y = 0 to find the x-intercept, and then let x = 0 to find the y-intercept: 0   x 2  1  1   x 2  1  x 2  1  x

 

y   02  1  1

Find a third point by letting x = 2 and solving

 

for y: y   22  1  3 . Find a fourth point

Ordered pairs that are solutions are (−2, −3), (−1, 0), (0, 1), (1, 0) and (2, −3).

Other answers are possible.

by letting x = −2 and solving for y:

Classroom Example 8 (page 208)

The points to be plotted are (−1, 0), (1, 0), (0, 1), (2, −3), and (−2, −3)

(a)

y    2  1  3 2

Let y = 0 to find the x-intercept, and then let x = 0 to find the y-intercept: 0  2 x  5  x  52 y  2(0)  5  y  5 Find a third point on the graph by letting x = −1 and solving for y: y  2  1  5  7 .

The three points are

 52 , 0 , (0, 5), and (−1, 7).

Note that (3, −1) is also on the graph.

Section 2.2

Circles

Classroom Example 1 (page 213) (a)

(h, k) = (1, −2) and r = 3

 x  h2   y  k 2  r 2 2

 x  12   y  2  32 (b)

Let y = 0 to find the x-intercept, and then let x = 0 to find the y-intercept: x  3 0 1  3 1  1 0  3 y  1  0  y  1  y  1 Find a third point by letting x = 2 and solving for y: 2  3 y  1  23  y  1  7  y .

 x  12   y  22  9 (b)

(h, k) = (0, 0) and r = 2

 x  h2   y  k 2  r 2  x  02   y  02  22

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x2  y2  4

Section 2.2 Circles

Factor and then divide both sides by 2:

Classroom Example 2 (page 214)

2

 x  1   y  2  9 2

(a)

27

2

This is a circle with center (1, −2) and radius 3.

2

1 3   2  x    2  y    50   2 2 2

2

1  3   x     y    25 2 2

Because c = 25 and 25 > 0, the graph is a circle. The  1 3 center is   ,  and the radius is 5.  2 2 Classroom Example 5 (page 216)

x2  y 2  4 This is a circle with center (0, 0) and radius 2.

(b)

Complete the square twice, once for x and once for y: x 2  6 x  y 2  2 y  12  0

 x  6 x  9   y  2 y  1  12  9  1 2

2

 x  32   y  12  2 Because c = −2 and −2 < 0, the graph is nonexistent.

Classroom Example 6 (page 217)

Determine the equation for each circle and then substitute x = −3 and y = 4.

Station A: Classroom Example 3 (page 215)

Complete the square twice, once for x and once for y: x 2  4 x  y 2  8 y  44  0

 x  4 x  4   y  8 y  16  44  4  16 2

2

Because c = 64 and 64 > 0, the graph is a circle. The center is (−2, 4) and the radius is 8.

Classroom Example 4 (page 216) 2 x 2  2 y 2  2 x  6 y  45 Group the terms, factor out 2, and then complete the square:

2 x  2 x  2 y  6 y   45 2  x  x   2  y  3 y   45 2

2

16  16

Station B: 2

 x  22   y  42  64

2

 x  12   y  42  42 3  12  4  42  42 42  42

2

1 9   2  x2  x    2  y 2  3 y     4 4 1 9  45  2    2   4 4

 x   6   y  0  52  x  62  y 2  25 2

3  62  42  25 32  42  25 9  16  25 25  25

Station C: 2

 x  52   y  2  102  x  52   y  22  100 3  52  4  22  100 82  62  100

64  36  100 100  100 Because (−3, 4) satisfies all three equations, we can conclude that the epicenter is (−3, 4).

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28

Chapter 2 Graphs and Functions

Section 2.3

Functions

(b)

Classroom Example 1 (page 222)





M   4, 0 ,  3,1 , 3,1

M is a function because each distinct x value has exactly one y value.





N   2, 3 , 3, 2 ,  4, 5 , 5, 4

N is a function because each distinct x value has exactly one y value.





P   4, 3 , 0, 6 ,  2,8 ,  4, 3

Function (c)

P is not a function because there are two y-values for x = −4.

Classroom Example 2 (page 223) (a)

Domain: {−4, −1, 1, 3} Range: {−2, 0, 2, 5} The relation is a function.

Not a function

(b)

Domain: {1, 2, 3} Range: {4, 5, 6, 7} The relation is not a function because 2 maps to 5 and 6.

(c)

Domain: {−3, 0, 3, 5} Range: {5} The relation is a function.

(d)

Classroom Example 3 (page 224)

Function

(a)

Domain: {−2, 4}; range: {0, 3}

Classroom Example 5 (page 226)

(b)

Domain: (, ) ; range: (, )

(a)

(c)

Domain: [−5, 5]; range: [−3, 3]

(d)

Domain: (, ) ; range: (, 4]

Classroom Example 4 (page 225) (a)

y  2 x  5 represents a function because y is always found by multiplying x by 2 and subtracting 5. Each value of x corresponds to just one value of y. x can be any real number, so the domain is all real numbers or (, ) .

Because y is twice x, less 5, y also may be any real number, and so the range is also all real numbers, (, ) .

Not a function

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Section 2.3 Functions

(b)

For any choice of x in the domain of y  x 2  3 , there is exactly one corresponding value for y, so the equation defines a function.

The function is defined for all values of x, so the domain is (, ) . The square of any number is always positive, so the range is [3, ) .

(c)

For any choice of x in the domain of x  y ,

(e)

3 , we see that y can be found by x2 dividing x + 2 into 3. This process produces one value of y for each value of x in the domain, so this is a function. The domain includes all real numbers except those that make the denominator equal to zero, namely x = −2. Therefore, the domain is (, 2)  (2, ) . Values of y can be negative or positive, but never zero. Therefore the range is (, 0)  (0, ) .

For y 

there are two possible values for y. Thus, the equation does not define a function. The domain is [0, ) while the range is (, ) .

Classroom Example 6 (page 228)

(d)

By definition, y is a function of x if every value of x leads to exactly one value of y.

Substituting a particular value of x, say 1, into y   x corresponds to many values of y. The ordered pairs (0, 2) (1, 1) (1, 0) (3, −1) and so on, all satisfy the inequality. This does not represent a function. Any number can be used for x or for y, so the domain and range of this relation are both all real numbers, (, ) .

29

(a)

f (3)  (3) 2  6(3)  4  13

(b)

f ( r )   r 2  6r  4

(c)

g (r  2)  3(r  2)  1  3r  7

Classroom Example 7 (page 228) (a)

f (1)  2(1) 2  9  7

(b)

f (1)  6

(c)

f (1)  5

(d)

f (1)  0

Classroom Example 8 (page 229) (a)

f ( x)  x 2  2 x  3 f (5)  (5) 2  2(5)  3  12 f (t )  t 2  2t  3

(b)

2x  3y  6  y 

2 x2 3

2 x2 3 2 16 f (5)  (5)  2   3 3 2 f (t )  t  2 3 f ( x) 

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30

Chapter 2 Graphs and Functions

Classroom Example 9 (page 231)

Classroom Example 2 (page 239) f ( x)  2 is a constant function. Its graph is a horizontal line with a y-intercept of 2.

The function is increasing on (,  1), decreasing on (–1, 1) and constant on (1, ).

Classroom Example 10 (page 231)

The example refers to the following figure.

Domain: (, ), range: {2} Classroom Example 3 (page 239)

x = 5 is a vertical line intersecting the x-axis at (5, 0).

Domain:{5}, range: (, ) Classroom Example 4 (page 240) (a)

The height of the kite is changing most rapidly from 0 to 5 minutes.

(b)

The height is constant from 15 minutes to 20 minutes.

(c)

After 15 minutes, the height of the kite is 150 feet.

Section 2.4

Linear Functions

Classroom Example 1 (page 238)

3x  4 y  0 Use the intercepts.

3 0  4 y  0  4 y  0  y  0 : y -intercept

3x  4 0  0  3 x  0  x  0 : x-intercept

The graph has just one intercept. Choose an additional value, say 4, for x.

3  4  4 y  0  12  4 y  0 4 y  12  y  3 Graph the line through (0, 0) and (4, −3).

f ( x)  32 x  6

Use the intercepts to graph the function. f (0)  32 0  6  6 : y-intercept 0  32 x  6  6  32 x  x  4 : x-intercept

Domain: (, ), range: (, )

Domain: (, ), range: (, )

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Section 2.5 Equations of Lines and Linear Models

31

Classroom Example 5 (page 241) (a)

m

4  (6) 10 5   2  2 4 2

(b)

m

88 0  0 5  (3) 8

(c)

10  10 20   the slope is 4  (4) 0 undefined. m

Classroom Example 6 (page 242) 2x − 5y = 10 Solve the equation for y.

2 x  5 y  10 5 y  2 x  10 2 y x2 5 2 The slope is , the coefficient of x.

5

Classroom Example 9 (page 244) The average rate of change per year is 1329.4  738.4 591   59.1.

2000  1990 10 Federal health care spending increased by an average of approximately $59.1 billion each year from 1990 to 2000.

Classroom Example 10 (page 245)

Classroom Example 7 (page 243) First locate the point (−2, −3). Because the slope is 4 , a change of 3 units horizontally (3 units to the 3 right) produces a change of 4 units vertically (4 units up). This gives a second point, (1, 1), which can be used to complete the graph.

(a)

C ( x)  120 x  2400

(b)

R ( x)  150 x

(c)

P ( x)  R ( x)  C ( x)  150 x  (120 x  2400)  30 x  2400

(d)

P ( x )  0  30 x  2400  0  x  80 At least 81 items must be sold to make a profit.

Section 2.5 Equations of Lines and Linear Models Classroom Example 1 (page 253) y  (5)  2( x  3) y  5  2 x  6 y  2 x  1

Classroom Example 8 (page 244) The average rate of change per year is 118.3  141.1 22.8   2.85 2017  2009 8 The graph confirms that the line through the ordered pairs falls from left to right, and therefore has negative slope. Thus, the amount spent by the federal government on R&D decreased by an average of $2.85 billion each year from 2009 to 2017.

Classroom Example 2 (page 253) 3  (1) 4  4  5 9 Now use either point for ( x1 , y1 ) :

First find the slope: m  y  3   94 [ x  (4)] 9( y  3)  4( x  4) 9 y  27  4 x  16 9 y  4 x  11 4 x  9 y  11

Classroom Example 3 (page 254) Write the equation in slope-intercept form: 3x  4 y  12  4 y  3 x  12  y  34 x  3 The slope is 34 , and the y-intercept is (0, −3).

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32

Chapter 2 Graphs and Functions

(b)

Classroom Example 4 (page 255) 42 1  2  2 2 1 Now, substitute  2 for m and the coordinates of one

First find the slope: m 

The line perpendicular to the equation has slope  23 . An equation of the line through (2, −4) that is perpendicular to 3 x  2 y  5 is y  (4)   23 ( x  2)  y  4   23 x  43 

of the points (say, (2, 2)) for x and y into the slopeintercept form y = mx + b, then solve for b: 2   12  2  b  3  b . The equation is

Classroom Example 7 (page 259)

y   12 x  3 .

(a)

y   23 x  83 or 2x +3y = −8.

8804  8312  164 4 1 Now use either point for ( x1 , y1 ) :

First find the slope: m  y  8312  164( x  1) y  8312  164 x  164 y  164 x  8148

(b) Classroom Example 5 (page 255)

The example refers to the following figure:

The year 2019 is represented by x = 6. y  164(6)  8148  9132 According to the model, average tuition and fees for 4-year colleges in 2019 will be about $9132.

Classroom Example 8 (page 261)

(a)

The line rises 5 units each time the x-value increases by 2 units. So the slope is 52 . The

Write the equation as an equivalent equation with 0 on one side. 3x  2(5  x)  2 x  38  3x  2(5  x)  2 x  38  0 Now graph the equation to find the x-intercept.

y-intercept is (0, 5), and the x-intercept is (−2, 0). (b)

An equation defining f is f ( x)  52 x  5 .

Classroom Example 6 (page 257) (a)

Rewrite the equation 3 x  2 y  5 in slopeintercept form to find the slope: 3x  2 y  5  y  32 x  52 The slope is 32 .

The line parallel to the equation also has slope 3 . An equation of the line through (2, −4) that 2

The solution set is {−4}.

Section 2.6

Classroom Example 1 (page 268) (a)

The function is continuous over (, 0)  (0, )

(b)

The function is continuous over its entire domain (, ).

is parallel to 3x  2 y  5 is

y  (4)  32 ( x  2)  y  4  32 x  3  y  32 x  7 or 3x − 2y = 14.

Graphs of Basic Functions

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Section 2.7 Graphing Techniques

33

Classroom Example 2 (page 271) (a)

(b)

Graph each interval of the domain separately.

If x < 1, the graph of f(x) = 2x + 4 has an endpoint at (1, 6), which is not included as part of the graph. To find another point on this part of the graph, choose x = 0, so y = 4. Draw the ray starting at (1, 6) and extending through (0, 4). Graph the function for x ≥ 1, f(x) = 4 − x similarly. This part of the graph has an endpoint at (1, 3), which is included as part of the graph. Find another point, say (4, 0), and draw the ray starting at (1, 3) which extends through (4, 0).

Graph each interval of the domain separately.

If x ≤ 0, the graph of f(x) = −x − 2 has an endpoint at (0, −2), which is included as part of the graph. To find another point on this part of the graph, choose x = −2, so y = 0. Draw the ray starting at (0, −2) and extending through (−2, 0). Graph the function for x > 0, f ( x)  x 2  2 similarly. This part of the graph has an endpoint at (0, −2), which is not included as part of the graph. Find another point, say (2, 2), and draw the curve starting at (0, −2) which extends through (2, 2). Note that the two endpoints coincide, so (0, −2) is included as part of the graph.

Classroom Example 4 (page 273) For x in the interval (0, 2], y = 20. For x in (2, 3], y = 20 + 2 = 22. For x in (3, 4], y = 22 + 2 = 24. For x in (4, 5], y = 24 + 2 = 26. For x in (5, 6], y = 26 + 2 = 28.

Section 2.7

Classroom Example 1 (page 279) Use this table of values for parts (a)−(c) g ( x)  2 x 2

h( x )  12 x 2

k ( x)  12 x

−2

8

2

1

−1

2

1 2

1 4

0

0

0

0

1

2

1 2

1 4

2

8

2

1

g ( x)  2 x

Classroom Example 3 (page 273) Create a table of sample ordered pairs: −6

−3

 32

0

3 2

3

6

y   13 x  2

−4

−3

−3

−2

−2

−1

0

 

x

(a)

x

Graphing Techniques

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2

2

34

(b)

Chapter 2 Graphs and Functions

Classroom Example 3 (page 282)

h( x)  12 x 2

(a)

x y

Replace x with –x to obtain  x  y . The result is not the same as the original equation, so the graph is not symmetric with respect to the y-axis. Replace y with –y to obtain x   y  x  y The

(c)

1  k ( x)   x  2 

result is the same as the original equation, so the graph is symmetric with respect to the xaxis. The graph is symmetric with respect to the x-axis only.

2

(b)

y  x 3

Replace x with –x to obtain y   x  3  y  x  3 . The result is the same as the

original equation, so the graph is symmetric with respect to the y-axis. Replace y with –y to obtain  y  x  3 . The result is not the same

Classroom Example 2 (page 281)

as the original equation, so the graph is not symmetric with respect to the x-axis.

Therefore, the graph is symmetric with respect to the y-axis only.

Use this table of values for parts (a) and (b) x

g ( x)   x

h( x )   x

−2

−2

2

−1

−1

1

0

0

0

1

−1

1

2

−2

2

(a)

g ( x)   x

(c)

Replace x with –x to obtain 2( x)  y  6  2 x  y  6 . The result is not the same as the original equation, so the graph is not symmetric with respect to the y-axis. Replace y with –y to obtain 2 x  ( y )  6  2 x  y  6 . The result is not the same as the original equation, so the graph is not symmetric with respect to the x-axis.

Therefore, the graph is not symmetric with respect to either axis. (d)

(b)

h( x )   x

2x  y  6

x 2  y 2  25 Replace x with –x to obtain ( x) 2  y 2  25  x 2  y 2  25 . The result is the same as the original equation, so the graph is symmetric with respect to the y-axis.

Replace y with –y to obtain x 2  ( y ) 2  25  x 2  y 2  25 . The result is the same as the original equation, so the graph is symmetric with respect to the x-axis.

Therefore, the graph is symmetric with respect to both axes. Note that the graph is a circle of radius 5, centered at the origin.

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Section 2.7 Graphing Techniques

35

Classroom Example 4 (page 284) (a)

y  2 x3 Replace x with –x and y with –y to obtain ( y )  2( x)3   y  2 x3  y  2 x3 . The result is the same as the original equation, so the graph is symmetric with respect to the origin.

(b)

y  2 x 2 Replace x with –x and y with –y to obtain ( y )  2( x) 2   y  2 x 2  y  2 x 2 .

The result is not the same as the original equation, so the graph is not symmetric with respect to the origin.

Classroom Example 5 (page 285) (a)

g ( x)  x5  2 x3  3x Replace x with –x to obtain g ( x)  ( x)5  2( x)3  3( x)   x5  2 x3  3x   ( x5  2 x3  3x)   g ( x)  g(x) is an odd function.

(b)

h( x )  2 x 2  3 Replace x with –x to obtain h( x)  2( x) 2  3  2 x 2  3  h( x)  h(x) is an even function.

(c)

k ( x)  x 2  6 x  9 Replace x with –x to obtain k ( x)  ( x) 2  6( x)  9  x 2  6 x  9  k ( x) and   k ( x)  k(x) is neither even nor odd.

Classroom Example 6 (page 286)

Compare a table of values for g ( x)  x 2 with 2

f ( x)  x  2 . The graph of f(x) is the same as the graph of g(x) translated 2 units up.

Classroom Example 7 (page 287)

Compare a table of values for g ( x)  x 2 with f ( x)  ( x  4) 2 . The graph of f(x) is the same as the graph of g(x) translated 4 units left.

x

g ( x)  x 2

f ( x)  ( x  4) 2

−7

49

9

−6

36

4

−5

25

1

−4

16

0

−3

9

1

−2

4

4

−1

1

9

Classroom Example 8 (page 288) (a)

f ( x)  ( x  1) 2  4

This is the graph of g ( x)  x 2 , translated one unit to the right, reflected across the x-axis, and then translated four units up. x

g ( x)  x 2

f ( x)   ( x  1) 2  4

−2

4

−5

−1

1

0

x

g ( x)  x 2

f ( x)  x 2  2

0

0

3

−2

4

6

1

1

4

−1

1

3

2

4

3

0

0

2

3

9

0

1

1

3

4

16

−5

2

4

6 (continued on next page)

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36

Chapter 2 Graphs and Functions

(continued)

Classroom Example 9 (page 289) (b)

The graphs in the exercises are based on the following graph.

f ( x )  2 x  3

This is the graph of g ( x)  x , translated three units to the left, reflected across the x-axis, and then stretched vertically by a factor of two.

(c)

x

g ( x)  x

f ( x)  2 x  3

−6

6

−6

−5

5

−4

−4

4

−2

−3

3

0

−2

2

−2

−1

1

−4

0

0

−6

(a)

g ( x)  f ( x)  2 This is the graph of f(x) translated two units down.

(b)

h( x)  f ( x  2) This is the graph of f(x) translated two units right.

h( x)  12 x  2  3

This is the graph of g ( x)  x , translated two units to the left, shrunk vertically by a factor of 2, and then translated 3 units down. x

g ( x)  x

h( x)  12 x  2  3

−2

undefined

−3

−1

undefined

−2.5

0

0

2

2  1.4

6

6  2.4

7

7  2.6

1 2

2  3  2.3

−2 1 2

8  3  1.6

−1.5

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Section 2.8 Function Operations and Composition

(c)

k ( x)  f ( x  1)  2 This is the graph of f(x) translated one unit left, and then translated two units up.

(c)

(d) (e)

37

( fg )( x)  ( x 2  3 x)(4 x  5)  4 x3  5 x 2  12 x 2  15 x  4 x3  7 x 2  15 x f  x 2  3x  g  ( x)  4 x  5

The domains of f and g are both (, ) . So, the domains of f + g, f − g, and fg are the intersection of the domains of f and g, f

(d)

(, ) . The domain of g includes those

F ( x)  f ( x) This is the graph of f(x) reflected across the y-axis.

real numbers in the intersection of the domains of f and g for which g ( x)  4 x  5  0  f

x   54 . So the domain of g is

,  54    54 ,  .

Classroom Example 3 (page 299) (a)

Section 2.8

Function Operations and Composition

Classroom Example 1 (page 298)

From the figure, we have f(1) = 3 and g(1) = 1, so (f + g)(1) = 3 + 1 = 4. f(0) = 4 and g(0) = 0, so (f − g)(0) = 4 − 0 = 4. f(−1) = 3 and g(−1) = 1, so (f g)(−1) = (3)(1) = 3

For parts (a)−(d), f ( x)  3x  4 and g ( x)  2 x 2  1 (a)

f (0)  3(0)  4  4 and g (0)  2(0) 2  1  1 , so

( f  g )(0)  4  1  5

(b)

f(−2) = 0 and g(−2) = 2, so

f (4)  3(4)  4  8 and g (4)  2(4)2  1  31 ,

so ( f  g )(4)  8  31  23 (c)

(b)

f (2)  3(2)  4  10 and g (2)  2(2) 2  1  7 , so ( fg )(2)  (10)(7)  70

(d)

2

f (3)  3(3)  4  5 and g (3)  2(3)  1  17 , f  5 so   (3)  g 17  

Classroom Example 2 (page 298)

For parts (a)−(e), f ( x)  x 2  3x and g ( x)  4 x  5 2

2

(a)

( f  g )( x)  ( x  3 x)  (4 x  5)  x  x  5

(b)

( f  g )( x)  ( x 2  3x)  (4 x  5)  x 2  7 x  5

  (2)   0 . f g

0 2

x

f(x)

g(x)

−2

−5

0

−1

−3

2

0

−1

4

1

1

6

From the table, we have f(1) = 1 and g(1) = 6, so (f + g)(1) = 1 + 6 = 7. f(0) = −1 and g(0) = 4, so (f − g)(1) = −1 − 4 = −5. f(−1) = −3 and g(−1) = 2, so (f g)(−1) = (−3)(2) = −6 f(−2) = −5 and g(−2) = 0, so

  (2)  f g

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f 5  g 0

is undefined.

38

Chapter 2 Graphs and Functions

(c)

f ( x)  3x  4, g ( x)   x

From the formulas, we have f (1)  3(1)  4  7 and g (1)   1  1 , so (f + g)(1) = 7 + (−1) = 6. f (0)  3(0)  4  4 and g (0)   0  0 , so (f − g)(1) = 4 − 0 = 4. f (1)  3(1)  4  1 and g (1)   1  1 ,

The screens show how a graphing calculator evaluates the expressions in this classroom example.

so (fg)(−1) =(1)(−1) = −1. f (2)  3(2)  4  2 and f  2  1. g (2)   2  2 , so   (2)  2 g

Classroom Example 4 (page 301)

Classroom Example 6 (page 303)

For parts (a) and (b), f ( x)  x  1 and g ( x)  2 x  5

Step 1: Find f ( x  h) :

(a)

f ( x  h)  3( x  h) 2  2( x  h)  4  3( x 2  2 xh  h 2 )  2 x  2h  4  3 x 2  6 xh  3h 2  2 x  2h  4 Step 2: Find f ( x  h)  f ( x) :

f ( x  h)  f ( x )  (3x 2  6 xh  3h 2  2 x  2h  4)  (3x 2  2 x  4)  6 xh  3h 2  2h Step 3: Find the difference quotient: f ( x  h)  f ( x) 6 xh  3h 2  2h   6 x  3h  2 h h

The domain and range of g are both (, ) .

However, the domain of f is [1, ) . Therefore, g ( x ) must be greater than or equal to 1: 2 x  5  1  x  2 . So, the domain of f  g is [2, ) . (b)

(a)

(b)

2 x

2 First find g(2): g (2)   1 . Now find 2 ( f  g )(2)  f ( g (2))  f (1)  1  4  5

First find f(5): f (5)  5  4  9  3 . Now find ( g  f )(5)  g ( f (5))  g (3)  23

( g  f )( x )  g ( f ( x))  2 x  1  5

The domain of f is [1, ) , while the range of f is [0, ) . The domain of g is (, ) .

Therefore, the domain of ( g  f ) is restricted to that portion of the domain of g that intersects with the domain of f, that is [1, ) .

Classroom Example 5 (page 303)

For parts (a) and (b), f ( x)  x  4 and g ( x) 

( f  g )( x)  f ( g ( x))  (2 x  5)  1  2 x  4

Classroom Example 7 (page 304)

For parts (a) and (b), f ( x)  (a)

5 2 and g ( x)  x4 x

5 5x  2 4 2 4x x     The domain and range of g are both all real numbers except 0. The domain of f is all real numbers except −4. Therefore, the expression for g ( x ) cannot equal −4. So, ( f  g )( x)  f ( g ( x)) 

2 1  4  x   . So, the domain of f  g x 2 is the set of all real numbers except for  12 ,

and 0. This is written

,  12    12 , 0  0, 

Copyright © 2021 Pearson Education, Inc

Section 2.8 Function Operations and Composition

(b)

2 2x  8  5 ( x  4) 5 The domain of f is all real numbers except −4, while the range of f is all real numbers except 0. The domain and range of g are both all real numbers except 0, which is not in the range of f. So, the domain of g  f is the set of all real numbers except for −4. This is written , 4  4,   ( g  f )( x)  g ( f ( x)) 

Classroom Example 8 (page 305) f ( x)  2 x  5 and g ( x)  3 x 2  x

( g  f )( x)  g (2 x  5)  3(2 x  5) 2  (2 x  5)  3(4 x 2  20 x  25)  2 x  5  12 x 2  58 x  70 ( f  g )( x)  f (3 x 2  x)  2(3 x 2  x)  5  6x2  2 x  5 In general, 12 x 2  58 x  70  6 x 2  2 x  5, so ( g  f )( x)  ( f  g )( x) .

Classroom Example 9 (page 306) ( f  g )( x)  4(3 x  2) 2  5(3 x  2)  8 Answers may vary. Sample answer: f ( x)  4 x 2  5 x  8 and g ( x)  3 x  2 .

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39

Chapter 3 Polynomial and Rational Functions Section 3.1

Quadratic Functions and Models

Classroom Example 1 (page 325) (a)

x

f ( x)  x 2  4 x  4

−5

1

−4

−4

−3

−7

−2

−8

−1

−7

0

−4

1

1

(c)

The graph of F  x   2  x  3  5 is the 2

graph of y  2 x 2 translated 3 units to the left and 5 units up.

domain:  ,   ; range:  , 5 Classroom Example 2 (page 326) f  x  x2  2 x  5

Rewrite by completing the square on x.





f  x  x2  2 x  5  x2  2 x  1  1  5





 x 2  2 x  1  1  5   x  1  6

domain:  ,   ; range:  8,   (b)

The graph of y  2 x 2 is a narrower version of the graph of y  x 2 . The graph of y  2 x 2 is a reflection of the graph of y  2 x 2 across the x-axis.

This equation is of the form y  ( x  h) 2  k , with h  1 and k  6. The graph opens upward and has the same shape as y  x 2 . The vertex is (−1, −6). It is a translation of the graph y  x 2 , 1 unit to the left and 6 units down. The axis is the vertical line x  1. The domain is (–∞, ∞). Since the smallest value of y is –6 and the graph opens upward, the range is [–6, ∞). The function is decreasing on (–∞, −1). The function is increasing on (–1, ∞).

domain:  ,   ; range: , 0

40

2

Copyright © 2021 Pearson Education, Inc

Section 3.1 Quadratic Functions and Models

Classroom Example 3 (page 327)

Classroom Example 5 (page 330)

f  x  2x2  x  6

(a)

Rewrite by completing the square on x.



41

Use the projectile height function with v0  112 and s0  75 : s t   16t 2  112t  75



f  x   2 x 2  x  6  2 x 2  12 x  6

   2 x  x     6  2 x    1 1  2 x 2  12 x  16  16 6

h   14 and k   49 . The graph opens upward and 8

Algebraic Solution: The ball reaches its maximum height at the vertex. Using the vertex formula, we have a = −16, b = 112, and c = 75. The axis of the b 112   3.5 . parabola is x  h   2a 2  16

has the same shape as y  x 2 . The vertex is

f 3.5  16 3.52  112 3.5  75  271 .

2

1 2

1 16

1 2

1 8

4

(b) 49 8

This equation is of the form y  ( x  h) 2  k , with





2

 14 ,  49 . It is a translation of the graph y  x , 8

unit to the left and

49 8

 

1 4

Thus, the ball reaches its maximum height of 271 feet at 3.5 seconds.

Graphing Calculator Solution: Graph y1  16 x 2  112 x  75 in the window [0, 8] × [–100, 350].

units down. The axis is the

vertical line x   14 . The domain is (–∞, ∞). The smallest value of y is  49 and the graph opens 8



,  . Evaluate f 0 to upward, so the range is   49 8

find the y-intercept. f 0  2  02  0  6  6 Thus, the y-intercept is (0, –6).

Find the x-intercepts by solving f  x   0.

2 x2  x  6  0

2 x  3 x  2  0 x

(c)

2 or x  2 3

The x-intercepts are 2, 0 and

 32 , 0.

Algebraic solution: Solve the quadratic inequality 16t 2  112t  75  200  16t 2  112t  125  0 Solve the corresponding equation 16t 2  112t  125  0 using the quadratic formula with a = −16, b = 112, and c = −125. 112  1122  4(16)(125) 2(16) 112  12, 544  8000 112  4544   32 32 112  4544 t  1.39 or 32 112  4544  5.61 t 32 The values 1.39 and 5.61 divide the number line into three intervals:(– ∞, 1.39), (1.39, 5.61), and (5.61, ∞). Use a test point in each interval to determine where the inequality is satisfied. (continued on next page) t

Classroom Example 4 (page 329) f  x   3x 2  12 x  8

a = −3, b = 12, and c = −8. The axis of the parabola is b 12   2 . The the vertical line x  h   2a 2  3

 

vertex is 2, f  2 . f  2  3 22  12  2  8  4.

Thus, the vertex is (2, 4).

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42

Chapter 3 Polynomial and Rational Functions

(continued)

Graphing Calculator Solution: Is 16t 2  112t  125  0 True or False?

Interval

Test Value

,1.39

0

16  02  112  0  125  0 125  0 False

1.39, 5.61

2

16  22  112  2  125  0 35  0 True

5.61,  

?

?

?

16  102  112  10  125  0 605  0 False Thus, the height of the ball will be greater than 200 ft between 1.39 seconds and 5.61 seconds.

Graphing Calculator Solution: Graph y1  16 x 2  112 x  75 and y2  200 in the window [0, 8] × [–100, 350].

(d)

10

Algebraic Solution: Use the quadratic formula to find the positive solution of 16t 2  112t  75  0 . 112  1122  (4)(16)(75) t 2(16) 112  17, 344   0.62 or 7.62 32 reject The ball will hit the ground after about 7.62 seconds.

Classroom Example 6 (page 3311) (a)

Using the data for the years 2010−2016 and the quadratic regression function on the graphing calculator, we have y  2.098 x 2  38.92 x  832.1 .

(b)

2018 corresponds to x = 18, the model predicts that the number of visits will be f 18  2.098 x 2  38.92 x  832.1  811 million This can be verified graphically.

[10, 20] by [650, 850]

Section 3.2

Synthetic Division

Classroom Example 1 (page 342)

x − 3 is in the form x − k, so k = 3.

3 4  15 11  10 12  9 6 4 3 2 4 4 4 x3  15 x 2  11x  10  4 x 2  3x  2  x3 x3

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Section 3.3 Zeros of Polynomial Functions

Classroom Example 2 (page 343)

(b)

f ( x)  3x 4  15 x3  50 x  25; k  4 4 3

15  12 3 3

0  50 25 12 48  8 12  2 17

6 11 12 5  20  4  8  12 0  20 1 2 3 0 5  40 Since f (4)  0 , x  4 is not a factor of f ( x) .

Classroom Example 3 (page 344)

To determine if k = 2 is a zero of f ( x)  2 x3  3x 2  18, divide synthetically.

2 2  3 0  18 4 2 4 2 1 2  14 The remainder is not zero, so 2 is not a zero of f(x).

(b)

f ( x)  x5  6 x 4  11x3  12 x 2  5 x  20 By the factor theorem, x  4 will be a factor of f ( x) if and only if f (4)  0 . Use synthetic division and the remainder theorem.

4 1

f (4)  17

(a)

To determine if k = −3 is a zero of f ( x)  x 4  4 x3  14 x 2  36 x  45 , divide synthetically.  4  14 36 45 3 21  21  45 1 7 7 15 0 The remainder is zero, so −3 is a zero of f(x).

Classroom Example 2 (page 348) f ( x)  6 x3  37 x 2  32 x  15 Because 5 is a zero of f, x  5 is a factor. Divide f ( x) by x  5 :

56 6

 37 30 7

To determine is k = 1 + 3i is a zero of f ( x)  x 4  x3  6 x 2  14 x  20 , divide synthetically. 1 6 14  20 1  3i  9  3i  12  6i 20 1 3i  3  3i 2  6i 0 The remainder is 0, so 1 + 3i is a zero of f(x).

Zeros of Polynomial Functions

Classroom Example 1 (page 347) (a)

4

2

f ( x)  3 x  48 x  8 x  32 By the factor theorem, x  4 will be a factor of f ( x) if and only if f (4)  0 . Use synthetic division and the remainder theorem.

4 3

0  48 8 32  12 48 0  32 3  12 0 8 0 Since f (4)  0 , x  4 is a factor of f ( x) .





  x  5 2 x  33 x  1

Classroom Example 3 (page 349) f  x   8 x 4  26 x3  27 x 2  11x  4

(a)

1  3i 1

Section 3.3

32 15  35  15 3 0

Thus f  x    x  5 6 x 2  7 x  3

3 1

(c)

43

p must be a factor of a0  4 and q must be a factor of a4  8 . Thus p can be 1, 2, or  4 , and q can be 1, 2, 4, or  8 . The possible p

rational zeros, q , are

1,  12 ,  14 ,  18 , 2, or  4 . (b)

By trial and error, we find that −1 is a zero: 1 8 8

 26 8  34

 27 34 7



11 7 4

4 4 0

Thus, f  x    x  1 8 x3  34 x 2  7 x  4



Now, use the quotient polynomial and synthetic division to find that 4 is a zero: 48 8

 34 32 2

7 8 1

4 4 0





Thus f  x    x  1 x  4 8 x 2  2 x  1 .

8 x  2 x  1   2 x  1 4 x  1 .

2

(continued on next page)

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44

Chapter 3 Polynomial and Rational Functions

(continued)

Classroom Example 6 (page 353)

Setting 2x −1 = 0 and 4x + 1 = 0 yields the zeros 12 and  14 . Thus, the rational zeros of f ( x ) are  14 , 12 , 1, and 4 and

f  x    x  1 x  4 2 x  1 4 x  1 .

The three zeros give  x  3 ,  x  2 , and

 x  5 as factors of f ( x) . Since f is to be of

degree 3, these are the only possible zeros by the number of zeros theorem. Therefore, f ( x ) has the form f  x   a  x  3 x  2 x  5 for some real number a. To find a, use the fact that f  1  6 : f  1  6  a  1  3 1  2 1  5  6  12a  a   12

Thus f  x  (b)





Since f  x  has only real coefficients, and 2  i is a zero, then 2  i is also a zero. Use synthetic division to divide f  x  by x   2  i  : 2i 1

Classroom Example 4 (page 351) (a)

f  x   x 4  x3  17 x 2  55 x  50

 x  5

  12 x  3 x  2   12 x3  192 x  15

1

1 2i 1 i

 17 1  3i  16  3i

55  35  10i 20  10i

Now use synthetic division to divide the quotient polynomial x3  1  i  x 2   16  3i  x   20  10i 

by x   2  i  : 2i 1 1

1 i 2i 3

 16  3i 6  3i  10

20  10i  20  10i 0



So, f  x    x   2  i   x   2  i  x 2  3x  10

Classroom Example 7 (page 354)

the form f  x   a  x  4 x  4 x  4 for

f  x   2 x 4  3 x3  5 x 2  4 x  1

f  2  24  a  2  4 2  4 2  4  24  8a  3  a Thus f  x   3  x  4 x  4 x  4

 3 x3  36 x 2  144 x  192

Classroom Example 5 (page 353)

By the conjugate zeros theorem, 3  i must also be a zero, so the polynomial has at least three zeros, −4, 3  i , and 3  i . For the polynomial to be of least degree, these must be the only zeros. The factor theorem gives us f  x    x   4  x  3  i   x  3  i    x  4  x  3  i   x  3  i 



  x   2  i   x   2  i   x  5 x  2 Thus, the zeros of f  x  are 2  i , 2  i , −5, and 2.

Because 4 is a zero of multiplicity 3, f ( x ) has some real number a. To find a, use the fact that f  2  24 :

 50 50 0

f  x   2 x 4  3 x3  5 x 2  4 x  1 has 4 variations in  1

2

3

4

sign. Thus, f has 4 or 4 −2 = 2 or 2 − 2 = 0 positive real zeros.

For negative zeros, consider the variations in signs for f   x  f   x   2   x   3   x   5   x   4   x   1 .

4

3

2

 2 x 4  3 x3  5 x 2  4 x  1 There are no changes in signs, so there are no negative real zeros.

Positive

Negative

Nonreal Complex

4

0

0

2

0

2

0

0

4

  x  4  x  3  i 2    2   x  4 x  6 x  9  1 2

    x  4  x  6 x  10 2

 x3  2 x 2  14 x  40

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Section 3.4 Polynomial Functions: Graphs, Applications, and Models

Section 3.4

Polynomial Functions: Graphs, Applications, and Models

Classroom Example 2 (page 363) (a)

The graph of f  x   x3  1 is the same as the graph of f  x   x3 , but translated 1 unit up. It

f  x   x4  2 x2  x  8

Since a = −1 < 0 and f has even degree, the end behavior is shaped

Classroom Example 1 (page 360) (a)

45

. (b)

includes the points (−1, 0), (0, 1), and (1, 2).

g  x   x3  3x 2  x  5

Since a = 1 > 0 and g has odd degree, the end behavior is shaped

(c)

h  x    x5  x 4  3x 2  2

Since a = −1 < 0 and h has odd degree, the end behavior is shaped Increasing on ,   . (b)

The graph of f  x    x  2 is the same as 5

the graph of f  x   x5 translated 2 units right.

(d)

k  x   x6  x3  1

Since a = 1 > 0 and f has even degree, the end behavior is shaped

It includes the points (1, −1), (2, 0), and (3, 1).

Classroom Example 3 (page 364) f  x   2 x3  3 x 2  11x  6

Step 1: p must be a factor of a0  6 and q must be a factor of a3  2. Thus, p can be ±1, ±2, ±3, ±6 and q p

can be ±1 or ±2. The possible rational zeros, q , are

Increasing on ,   . (c)

±1, ±2, ±3, ±6,  12 or  32 . The remainder theorem

The graph of f  x    12  x  3  5 is the 4

graph of f  x   x 4 translated 3 units left,

reflected across the x-axis, stretched vertically by a factor of 12 , and then translated 5 units up. It includes the points (−2, 4.5), (−3, 5), and (−4, 4.5).

shows that 2 is a zero.

2 2 3  11  6 4 14 6 2 7 3 0 The new quotient polynomial is 2 x 2  7 x  3.

2x2  7 x  3  0 (2 x  1)( x  3)  0 2 x  1  0  x   12

or x  3  0  x  3

The rational zeros are – 3,  12 , and 2. Thus f ( x)   x  3 2 x  1 x  2 .

Step 2: f 0  6, so plot 0, 6 . (continued on next page) Increasing on ,  3

Decreasing on  3,  

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46

Chapter 3 Polynomial and Rational Functions

(continued) Step 3: The x-intercepts divide the x-axis into four intervals: Graph Above Test Value of Sign of f or Below Interval Point f (x) (x) x-Axis (– ∞, –3)

3,  12   12 , 2

–5

–126

Negative Below

–2

12

Positive

1

–12

Negative Below

Above

(2, ∞) 3 42 Positive Above Plot the x-intercepts, y-intercept and test points with a smooth curve to get the graph.

Classroom Example 5 (page 367)

Algebraic Solution: f  x    x3  2 x 2  4 x  5

Use synthetic division to find f 1 and f  2 : 1 1

2 4 5 1 3 1 1 1  3 2

2 1

2 4 5 0 8 2 1 0  4  3

Since f 1  2 and f  2  3 , by the intermediate value theorem, there must be a real zero between 1 and 2.

Graphing Calculator Solution:

Classroom Example 4 (page 366) f  x    x  2  x  1 3

2

Step 1: Since the function is given in factored form, the zeros can be determined by inspection. They are −2 and 1. Plot these as x-intercepts of the graph.

Step 2: f 0  0  2 0  1  8 , so plot (0, 8) as 3

2

Classroom Example 6 (page 368) f  x   x4  5 x 2  3x  7

(a)

the y-intercept.

Step 3: The dominating term of f(x) is x5 , so the end behavior of the graph is

Since f  x  has real coefficients and the leading coefficient, 1, is positive, use the boundedness theorem. Divide f  x  synthetically by x − 1:

.

Because 1 is a zero of multiplicity 2, the graph will touch the axis at 1, and then turn and change direction. Because −2 is a zero of multiplicity 3, the graph will cross and is tangent to the x-axis at this zero. This causes a change is concavity at x = −2.

The x-intercepts divide the x-axis into three intervals: Graph Above Test Value of Sign of f or Below Interval Point f (x) (x) x-Axis (– ∞, –2)

–3

–16

Negative Below

(−2, 1)

–1

4

Positive

Above

(1, ∞) 1.5 10.7 Positive Above Plot the x-intercepts, y-intercept and test points with a smooth curve to get the graph.

0 5 3 7 1 1 6 9 1 1 6 9 2 Because all the numbers in the bottom row of the synthetic division are nonnegative, then f  x  has no zero greater than 1.

11

(b)

2 1

0 5 3 7  2 4  18 30 1  2 9  15 23 Because −2 < 0 and the numbers in the bottom row of the synthetic division alternate in sign, then f  x  has no zero less than −2.

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Section 3.4 Polynomial Functions: Graphs, Applications, and Models

47

Classroom Example 7 (page 369) f  x    x3  8 x 2  4 x  10

The greatest degree term is x3 , so the graph will have end behavior similar to the graph of f  x    x3 . There are at most three real zeros. The

y-intercept is (0, 10) because f 0  10. To

y  5.232 x 2  222.0 x  6966

approximate the zeros, graph y1   x3  8 x 2  4 x  10 in the window [–10, 4] by [–90, 20].

[−1, 11] by [6800, 8800]

(b)

y  2.065 x3  36.20 x 2  335.2 x  6916

[−1, 11] by [6800, 8800]

Classroom Example 8 (page 370)

(a)

Year 2006

x 0

Enrollment (in thousands) 6955

2008

2

7332

2010

4

7924

2012

6

8093

2014

8

8257

2016

10

8742

(c)

y  1.255 x 4  23.04 x3  118.0 x 2  48.27 x  6950

Using x = 0 to represent 2006, x = 2 to represent 2008, etc., use the regression feature of a calculator to determine the quadratic function that best fits the data.

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(continued on next page)

48

Chapter 3 Polynomial and Rational Functions





x3  2 x 2  5 x  10  x3  2 x 2  5 x  10

(continued)

 x  x  2  5  x  2 2



  x  2 x 2  5



We can set this equal to 0 and solve to find the remaining zeros.

 x  2   x 2  5  0

[−1, 11] by [6800, 8800]

(d)

quadratic: R 2  0.9731361218 cubic: R 2  0.981602943 quartic: R 2  0.9974956996 As we can see from the graphs, each regression is a close fit. However, the quartic value is closest to 1, so the quartic function is the best fit.

Summary Exercises on Polynomial Functions, Zeros, and Graphs Classroom Example 1 (page 378) (a)

f  x   x4   x3  3 x 2  5 x  10 has three 1

2

3

variations in sign, so f has 3 or 3 – 2 = 1 positive real zeros. f   x   x 4  x3  3 x 2  5 x  10 has one 

x  2  0 x2  5  0 x2 x 2  5 x   5   i 5 The rational zeros are –1 and 2.

The real complex zeros are i 5, i 5.

Section 3.5

Rational Functions: Graphs, Applications, and Models

Classroom Example 1 (page 382) 4 x To obtain the graph of f  x   4x , stretch the graph of f ( x) 

f ( x)  1x vertically by a factor of 4. Just as with the

graph of f ( x)  1x , y  0 is the horizontal asymptote and x  0 is the vertical asymptote.

1

variation in sign, so f has 1 negative real zero.

The degree of the function is 4, so the remaining zeros are nonreal complex.

(b)

Positive

Negative

Nonreal Complex

3

1

0

1

1

2

p must be a factor of a0  10 and q must be a factor of a4  1. Thus, q can be ±1, and p can be ±1, ±2, ±5, or ±10. The possible p

rational zeros, q , are ±1, ±2, ±5, or ±10. (c), (d) The remainder theorem shows that –1 is a zero. 1 1

1 3  5  10 1 2 5 10 1 2 5  10 0 The new quotient polynomial is x3  2 x 2  5 x  10. This polynomial can be factored.

Domain: , 0  0,   Range: , 0  0,  

The function is never increasing. It is decreasing on , 0 and 0,   .

Classroom Example 2 (page 383) 1 x3 To obtain the graph of f ( x)   x 1 3 , shift the graph f ( x)  

of f ( x)  1x 3 units to the right, then reflect the graph across the x-axis. Just as with the graph of f ( x)  1x , y  0 is the horizontal asymptote, but this graph has x  3 as the vertical asymptote. (continued on next page)

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Section 3.5 Rational Functions: Graphs, Applications, and Models

49

x 2  x  6  0   x  2 x  3  0  x  2, 3

(continued)

Thus, the vertical asymptotes are x = −2 and x = 3. Since the numerator has lower degree than the denominator, the horizontal asymptote is y = 0. (b)

Domain:  , 3  3,  

Range: , 0  0,  

(c)

The function is increasing on , 3 and 3,   . It never decreases.

Classroom Example 3 (page 384) f ( x) 

2

 x  22

4

To obtain the graph of f ( x) 

2

 x  22

 4 , shift the

graph of f ( x)  12 , 2 units to the right, stretch the x

graph vertically by a factor of 2, and then shift the graph 4 units down. Unlike the graph of f ( x)  12 , 2

 x  2 2

x2  x x 1 To find the vertical asymptote, set the denominator equal to zero: x  1  0  x  1 Thus, the vertical asymptote is x = 1. Since the degree of the numerator is exactly one more than the denominator, there is an oblique asymptote. Divide the numerator by the denominator and, disregarding the remainder, set the rest of the quotient equal to y to obtain the equation of the asymptote: x2 x  1 x2  x  0 f  x 

x2  x 2x  0 2x  2 2 Thus, the oblique asymptote is y = x + 2.

x

the vertical asymptote of f ( x) 

4x  2 x 1 To find the vertical asymptote, set the denominator equal to zero: x  1  0  x  1 Thus, the vertical asymptote is x = −1. Since the numerator and the denominator have the same degree, the horizontal asymptote is y = 4. f  x 

 4 will be

x  2 and the horizontal asymptote is y = −4.

Classroom Example 5 (page 388) x2 x2  x  6 Step 1: Find the vertical asymptotes by setting the denominator equal to 0: x 2  x  6  0   x  3 x  2  0  x  3 or x  2 Step 2: Find the horizontal asymptote. Since the numerator has lower degree than the denominator the horizontal asymptote is y = 0. There is no oblique asymptote.

Step 3: f (0)  20  2  13 , so the y-intercept is f  x 

Domain: , 2  2,   Range:  4,  

The function increases on  , 2 . It decreases on

2,   .

 

Classroom Example 4 (page 386) (a)

x2 x  x6 To find the vertical asymptotes, set the denominator equal to zero: f  x 

2

0 06

0, 13 .

Step 4:

x2  0  x  2  0  x  2, so the x2  x  6

x-intercept is (2, 0).

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(continued on next page)

50

Chapter 3 Polynomial and Rational Functions

(continued)

Step 6: Plot a point in each of the intervals

Step 5: Determine whether the graph will intersect the horizontal asymptote: f ( x)  2 x  2  0  x  2  0  x  2 . The graph

, 1 , 1, 12  , and  12 ,   .

x  x6

intersects its horizontal asymptote at (2, 0).

Step 6: Plot a point in each of the intervals , 2 , 2, 2 , 2, 3 and 3,  .

Interval

Test Value Point of f (x)

Sign of f (x)

Test Value Point of f (x)

Interval

Sign of f (x)

Graph Above or Below x-Axis

(– ∞, –1) –3

7

Positive

Above

Graph Above or Below x-Axis

1, 12 

0

−2

Negative

Below

 12 , 

1

1

Positive

Above

Step 7: Sketch the graph.

(– ∞, –2) –3

 56

Negative

Below

(−2, 2)

0

1 3

Positive

Above

(2, 3)

5 2

 92

Negative

Below

3,  

4

1 3

Positive

Above

Step 7: Sketch the graph Classroom Example 7 (page 389) 2 x 2  3x  4 x2  6 x  9 Step 1: Find the vertical asymptotes by setting the denominator equal to 0: f  x 

x 2  6 x  9  0   x  3  0  x  3 2

Classroom Example 6 (page 388) 4x  2 x 1 Step 1: Find the vertical asymptote by setting the denominator equal to 0: x  1  0  x  1 Step 2: Find the horizontal asymptote. Since the numerator and the denominator have the same degree, the horizontal asymptote is y = 4. There is no oblique asymptote. f  x 

Step 3: f (0) 

4  0  2  2, 0 1

so the y-intercept is

(0, –2).

Step 4: 4xx12  0  4 x  2  0  x  12 , so the

x-intercept is

 12 , 0.

Step 5: Determine whether the graph will intersect the horizontal asymptote: f ( x)  4xx12  4  4 x  2  4 x  4  2  4 . Thus,

Step 2: Find the horizontal asymptote. Since the numerator and the denominator have the same degree, the horizontal asymptote is y = 2. There is no oblique asymptote.

Step 3: f (0) 

2  0   3 0   4

0,  94  .

Step 4:

2

02  6  0  9

  94 , so the y-intercept is

2 x 2  3x  4  0  2 x 2  3x  4  0  2 x  6x  9

3  32  4(2)(4) 3  41   2(2) 4 x  2.35 or x  0.85 The x-intercepts are (−2.35, 0) and (0.85, 0). (continued on next page) x

the graph does not intersect its horizontal asymptote.

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Section 3.5 Rational Functions: Graphs, Applications, and Models

(continued)

Step 5: Determine whether the graph will intersect the horizontal asymptote: 2 x 2  3x  4 f ( x)  2 2 x  6x  9 2 x 2  3x  4  2 x 2  12 x  18  22  9 x   22 x 9 Thus, the graph intersects its horizontal asymptote at





 22 ,2 .

9

Step 6: Plot a point in each of the intervals , 3 , 3, 2.35 , 2.35, 0.85 and 0.85,   .

Graph Above or Below x-Axis

Interval

Test Value Sign of Point of f (x) f (x)

(– ∞, –3)

–4

16

Positive Above

3, 2.35

−2.5

4

Positive Above

2.35, 0.85

0

 94

Negative Below

.85,  

2

2 5

Positive Above

x2 x 1 x  x 2

x2  x 2x 2x  2 2 The oblique asymptote is y  x  2 2

Step 3: f (0)  00 10  0, so the y-intercept is (0, 0). x2  x  0  x 2  x  0  x  x  1  0  x 1 x  0 or x  1, so the x-intercepts are (0, 0) and (−1, 0).

Step 5: Determine whether the graph will intersect the oblique asymptote: x2  x  x2 x 1 x 2  x   x  2 x  1 

Step 4:

x2  x  x2  x  2  0  2 Thus, the graph does not intersect its oblique asymptote.

Step 6: Plot a point in each of the intervals , 1 , 1, 0 , 0,1 , and 1,   .

Step 7: Sketch the graph

Test Value Sign of Interval Point of f (x) f (x) (– ∞, –1) –4

Classroom Example 8 (page 391)

Graph Above or Below x-Axis

 125

Negative Below

(−1, 0)

 12

1 6

Positive Above

(0, 1)

1 2

 32

Negative Below

1,  

2

6

Positive Above

Step 7: Sketch the graph

2

x x x 1 Step 1: Find the vertical asymptote by setting the denominator equal to 0: x  1  0  x  1 Step 2: Since the degree of the numerator is one more than the degree of the denominator, there is an oblique asymptote. Divide the numerator by the denominator and, disregarding the remainder, set the rest of the quotient equal to y to obtain the equation of the asymptote: f  x 

51

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52

Chapter 3 Polynomial and Rational Functions

(c)

Classroom Example 9 (page 392) x2  x  6 x3 In lowest terms, the function becomes x 2  x  6  x  3 x  2   x  2, x  3 f  x  x3 x3 f  x 

The solution set of the inequality 2  x  1 x  3  0 includes the x –value for

which the graph of f  x  lies on or below the

x-axis. Because the inequality  is nonstrict, the zeros of the related graph are included. The solutions set is ,  1  3,   .

Classroom Example 2 (page 404)

Step 1 Each equation or inequality is written with an expression on one side and 0 on the other.

Step 2 Let f  x    x  x  1 and graph. Multiplying 2

the factors and identifying the term of greatest degree show that the dominating term of f  x  is  x3 . Thus, the end behavior is

Using the appropriate end behavior, draw a smooth curve that crosses the x-axis at 0 and touches the x-axis at 1, then turns and changes direction. Use this graph to determine the solution sets.

Classroom Example 10 (page 393) f  x 

(a) (b)

x2 2 1  x 

The traffic intensity x 

3.6  0.9 4.0

x2 0.92 0.81    4.05 2 1  x  2 1  0.9 0.2 The average number of vehicles waiting in line to enter the ramp is 4.05. f 0.9 

Section 3.6

Polynomial and Rational Inequalities

Classroom Example 1 (page 403)

(a)

f  x   0 for the x-values of the x-intercepts

of the graph of f  x  . The solution set of the equation  x  x  1  0 is 0, 1 .

2

(b)

f  x   0 —that is, the graph lies below the

x-axis—for 0  x  1 and x  1. The solution set of the inequality  x  x  1  0 is 2

0, 1  1,   .

(c) (a)

The solutions of the equation 2  x  1 x  3  0 are the x-intercepts of the

graph of f  x  , –1 and 3. Therefore, the solution set is{–1, 3}. (b)

.

f  x   0 —that is, the graph lies on or above

the x-axis—for the solution set , 0  1 . (d)

f  x   0 —that is, the graph lies on or below

the x-axis—for the solution set  0,   .

The solution set of the inequality 2  x  1 x  3  0 includes the x-values for

which the graph of f  x  lies above the x-axis.

The solution set is 1, 3 .

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Section 3.6 Polynomial and Rational Inequalities

Classroom Example 3 (page 405)

Classroom Example 4 (page 406)

Solve 9 x 4  10 x3  4 x 2  20 x3  14 x  3

Solve

Step 1 Subtract 20 x3  14 x  3 from 9 x 4  10 x3  4 x 2 to obtain the equivalent inequality 9 x 4  30 x3  4 x 2  14 x  3.

Step 2 Let f  x   9 x  30 x  4 x  14 x  3 and 4

3

2

53

x3  0. x 1

Step 1 The inequality is already written with 0 on one side.

Step 2 Let f  x   xx13 and graph. The vertical asymptote has equation x = 1, and the horizontal asymptote has equation y = 1. The x-intercept, found by setting the numerator equal to 0, is  3, 0 . Evaluating f 0 give

graph. By the rational zeros theorem, the possible rational zeros of the polynomial are  19 ,  13 ,  1, and 3. Use synthetic division

the y-intercept 0,  3 . The graph does not

to show that 1 is a zero.

1 9  30 4 14 3 9  21  17  3 9  21  17  3 0

intersect its horizontal asymptote because f  x   1 has no solution.

Use the results of the synthetic division to factor as follows.





f  x    x  1 9 x3  21x 2  17 x  3

The cubic polynomial has the same possible rational zeros as f  x  . Use synthetic division to show that 3 is a zero of the cubic polynomial.

3 9  21  17 27 18 9 6 1

Therefore,

3 3 0



Step 3 The solution set of the inequality xx13  0 includes the x-values for which the graph of f  x  lies on or above the x-axis. Because the

inequality is nonstrict, the zero of f  x  ,  3,



f  x    x  1 x  3 9 x  6 x  1 2

  x  1 x  33 x  1 .

2

The zeros of the polynomial are 1, 3, and  13

is included in the solution set ,  3  1,   .

Classroom Example 5 (page 406)

(multiplicity 2). The dominating term of f  x 

Solve

is 9 x 4 , indicating end behavior . The y-intercept, found by evaluating f 0 is

Step 1 Subtract x 1 2 so that 0 is on one side.

(0, 3).

3 1 .  x4 x2

3 1  0 x4 x2 3  x  2 1 x  4  0  x  4 x  2  x  2 x  4 3 x  6   x  4 0  x  2 x  4 2 x  10 0  x  2 x  4

Step 3 Referring to the figure, f  x   0, that is the graph lies on or below the x-axis, for the

 

solution set  13  1, 3 .

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(continued on next page)

54

Chapter 3 Polynomial and Rational Functions

(continued)

Classroom Example 3 (page 414)

Step 2 Let f  x 

 x 22x 10   x  4

and graph. The vertical

asymptotes have equations x = 2 and x = –4.

The horizontal asymptote has equation x = 5.

 

The y-intercept is 0, 54 , and the x-intercept is x = 5.

Step 1: Let h = the height of the cylinder, r = the radius of the cylinder, and V = the volume of the cylinder. Then, V  khr 2

 

Step 2: Substitute 785  k 10 52 to find k.

785  250k  k 

157 50

157 2 hr 50 Step 4: Now find V when r  10 and h  15 157 V 15 102  4710 50 The volume of the cylinder is 4710 cm3.

Step 3: V 

 

Classroom Example 4 (page 414)

Step 3 The solution set of the inequality 2 x 10  0 includes the x-values for which x2 x4 





the graph of f  x  lies on or above the x-axis.

Because the inequality is nonstrict, the zero of f  x  , 5, is included in the solution set

4, 2  5,   .

Section 3.7

Variation

Classroom Example 1 (page 412)

Step 1: d  kt Step 2: Substitute d  156 and t  3 to find k.

Let n = the number of children, a = the amount spent on advertising per year, and c = the weekly charge. ka . Substitute the given values for n, a, Then n  c 900k  k  8.5. Thus and c to determine k: 51  150 8.5a n . Now use the second set of values for a and c 8.5 1000  50. She should c to determine n: n  170 expect to enroll 50 children.

156  k 3  k  52

Step 3: d  52t Step 4: Now find d when t  5. d  52 5  260 The vehicle will travel 260 miles.

Classroom Example 2 (page 413)

Step 1: Let x = the number of items produced and let k y = the cost per item. Then, y  2 x k to find k.

Step 2: Substitute 1.50  1002 15, 000  k 15, 000 Step 3: y  x2 Step 4: Now find y when x  250.

15, 000 y  0.24 2502 The cost per item will be $0.24.

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Chapter 4 Inverse, Exponential, and Logarithmic Functions f  x   3 x  7

 f  g  x   2  12 x  5  5  x  10  5  x  5  g  f  x   12 2 x  5  5  x  52  5  x  52 Because  f  g  x    g  f  x   x, g is not the

Using the definition of a one-to-one function we have f  a   f b   3a  7  3b  7 

Classroom Example 4 (page 438)

Section 4.1

Inverse Functions

Classroom Example 1 (page 434) (a)

3a  3b  a  b. So, the function is one-toone. (b)

inverse function of f.

(a)



If we choose a = 7 and b = −7, then 7  7 , (b)

f  7   49   7   0 . So, even though 2







G   2, 5 ,  1, 2 , 0,1 , 1, 2 ,  2, 5

G is not one-to-one because the y-values 2 and 5 each correspond to two different x-values.

Thus, G does not have an inverse.

7  7 , f 7   f  7   0 . By the definition,

f is not a one-to-one function.



F is one-to-one and has an inverse because each y-value corresponds to only one x-value.

F 1   8, 2 ,  1, 1 , 0, 0 , 1,1 , 8, 2

f  x   49  x 2

but f 7   49  7 2  0 and



F   2, 8 , 1, 1 , 0, 0 , 1,1 ,  2,8

(c)

Classroom Example 2 (page 435) (a)

h is not one-to-one if the number of hurricanes for 2018 is decreased by 1 because the y-value 7 corresponds to two different x-values. Thus, h does not have an inverse.

Classroom Example 5 (page 439) (a)

f  x  x

If x  3, f  x   3  3 . If

This function is one-to-one because every horizontal line will intersect the graph in exactly one point.

x  3, f  x   3  3 . Thus f  x   x is not

one-to-one. (b)

g  x  4x  7

The graph of g  x   4 x  7 nonhorizontal

(b)

linear function, so g is a one-to-one function.

This is not a one-to-one function because there is a horizontal line that intersects the graph in more than one point.

[–10, 10] × [–10, 10] Step 1: Interchange x and y: y  4x  7  x  4 y  7

Classroom Example 3 (page 437) f  x   2 x  5; g  x   12 x  5

f  x   2 x  5 is a nonhorizontal linear function and

(continued on next page)

is one-to-one, so it has an inverse.

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55

56

Chapter 4 Inverse, Exponential, and Logarithmic Functions

(continued)

Classroom Example 7 (page 441)

Step 2: Solve for y: x  4y  7  x  7  4y  y  14 x  74

x7  y 4

Step 3: Replace y with g 1  x  : g 1  x   14 x  74

(c)

h  x   x3  5

The graph of a cubing function shows that the function is one-to-one.

f and g are not inverses because the graphs are not reflections across the line y = x.

Classroom Example 8 (page 441) f  x   x 2  4, x  0

Because the domain is restricted, the function is oneto-one and has an inverse.

[–10, 10] × [–10, 10] Step 1: Interchange x and y: y  x3  5  x  y 3  5 Step 2: Solve for y: x  y3  5  x  5  y3  3 x  5  y Step 3: Replace y with h 1  x  : h 1  x   3 x  5

Classroom Example 6 (page 440)

Step 1: Interchange x and y: y  x 2  4  x  y 2  4

3 x  1 , x5 x5 To find the inverse, start by replacing f(x) with y. 3 x  1 y x5 3 y  1 , y5 Step 1: Interchange x and y: x  y5 Step 2: Solve for y: 3 y  1 x  xy  5 x  3 y  1  y5 xy  3 y  5 x  1  y  x  3  5 x  1  5x  1 y x3 Step 3: Replace y with f 1  x  :

Step 2: Solve for y. We must consider that the domain of f is the range of f 1

f  x 

f 1  x  

x  y2  4  x  4  y2   x  4  y

Step 3: Replace y with f 1  x  : f 1  x    x  4, x  4

5x  1 , x  3 x3

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Section 4.2 Exponential Functions

Classroom Example 9 (page 442)

Classroom Example 2 (page 451)

Given f  x   3 x  1, find f 1  x.

x

Step 1: Interchange x and y: y  3 x  1  x  3 y  1 Step 2: Solve for y. x 1 x  3y  1  x  1  3y   y 3 y  13 x  13

1 f ( x)    2 The y-intercept is (0, 1), and the x-axis is a horizontal asymptote. Make a table of values.

Step 3: Replace y with f 1  x  : f 1  x   13 x  13

x

f  x

−3 −2 −1 0 1

8 4 2 1

11: f 1 11  13 11  13

44; f 1  44  13  44  13

 45  15; O 3

2

74; f 1 74  13 74  13

44; f 1  44  13  44  13

3

 123  4; D

 75  25; Y 3

59; f

1

59

 13 59  13  60  20; T 3

 

14; f

1

14

 13 14  13  153  5; E

 

57

 45  15; O 3

68; f

1

68

 13 68  13  69  23; W 3

 

14; f

1

14

 13 14  13  153  5; E

 

1 2 1 4 1 8

The function has domain  ,   and range 0,   .

Classroom Example 3 (page 452)

For this example, refer to this graph of f  x   3x .

59; f 1 59  13 59  13

 60  20; T 3

The message is DO YOU TWEET.

Section 4.2

Exponential Functions

Classroom Example 1 (page 448) f  x  4x

(a) 1 1  42 16

The graph of f  x   3x is obtained by reflecting the graph of f  x   3x across the

(a)

f  2  42 

(b)

f 5  4  1024

(c)

13 2 f    42 3  42  161 3  3 16 3

x-axis. The domain is  ,   and the range is

5

, 0 .

 

 3 2  23  2 3 2

(d)

f  2.15  19.698

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58

Chapter 4 Inverse, Exponential, and Logarithmic Functions

(b)

The graph of f  x   3x  2 is obtained by shifting the graph of f  x   3x two units to the right. The domain is  ,   and the range is 0,   .

Check: ?

1252 3  25

?

1252 3  25

 125   25  125   25 3

2 ?

2 ?

3

?

?

52  25

52  25

25  25 25  25 Both are true statements.

Solution set: 125

Classroom Example 7 (page 454) (a)

Use the compound interest formula to find the



future value, A  P 1  nr (c)

 , given tn

n  2, P  2500, r  0.03, and t  10.

The graph of f  x   3x  2  2 is obtained by

tn

10(2)

 r  0.03  A  P 1     2500 1    n  2 

shifting the graph of f  x   3x two units left and two units down. The domain is  ,  

  25001.015

and the range is  2,   .

20

 3367.14

Rounding to the nearest cent, there will be $3367.14 in the account after 10 years. (b)

The amount of interest is $3367.14 – $2500 = $867.14

Classroom Example 8 (page 455) (a)

Use the compound interest formula to find the



present amount, A  P 1  nr

 , given tn

A = 15,000, n = 1, r = 0.24, and t = 8.



Classroom Example 4 (page 452)

15, 000  P 1  0.024 1

Classroom Example 5 (page 453) 3x 1  9 x  3

(b)

 

2 x 3

3x 1  3

A = 15,000, P = 12,025, n = 1, and t = 8. We must find r: 8(1)

15, 000  12, 025 1  r  15, 000 600 8   1  r  12, 025 481 18  600     1  r 481

8

Classroom Example 6 (page 453) x 2 3  25

 x    25  x    25    5  125 32

3

8

 r 15, 000  12, 025 1    1

3x 1  32 x  6 x  1  2x  6  7  x Solution set: {7}

23 32

8(1)

15, 000  P 1.024 15, 000  P  P  12, 407.71 1.0248 Rounding to the nearest cent, Brendan must deposit $12,407.71.

 

1 1  5 x  1251  5 x  53  125 5 x  53  x  3 Solution set: {−3}

5x 



18

3

Recall from Chapter 1 that it is necessary to check all proposed solutions in the original equation when you raise both sides to a power.

 600     1  r  r  0.028 481  An interest rate of about 2.80% will produce enough interest to increase $2,025 to $15,000 by the end of 8 years.

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Section 4.3 Logarithmic Functions

Classroom Example 9 (page 456)

Section 4.3

Use the continuous compounding interest formula to find the total amount on deposit, A  Pert , given P = 8000, r = 0.05, and t = 6.

Classroom Example 1 (page 464)

Logarithmic Functions

Logarithmic Form Exponential Form (a)

log3 81  4

34  81

(b)

log1 2 8  3

 12   8

(c)

log10 1000  3

103  1000

(d)

1 log5 125  3

1 53  125

(e)

and the continuous compounding interest formula A  Pert , given P = 2500, r = 0.03, and t = 10.

log12 12  1

121  12

(f)

log 6 1  0

60  1

Compounded

Classroom Example 2 (page 464)

A  Pert  8000e    8000e0.3  10, 798.87046 Rounding to the nearest cent, the total amount will be $10,798.87.

0.05 6

Classroom Example 10 (page 457)



Use the compound interest formula A  P 1  nr



tn

Formula

Quarterly

A  2500 1  0.03 4





 $3370.87

Monthly

A  2500 1  0.03 12





 $3373.38

Daily

A  2500 1  0.03 365





 $3374.61

Continuously

A  2500e0.03(10)  $3374.65

10(4)

10(12)

10(365)

(a)

Classroom Example 11 (page 458) (a)

f  2025  0.001942e

0.00609  2025

16  2 9 16 x 2  9 9 2 x  16 9 3 x  16 4

3 Solution set:   4

0.00609 x

y  0.001942e In 2025, the atmospheric carbon dioxide level will be

log x

(b)

 441 ppm .

3 4 163 4  x

log16 x 

16   x 14 3

(b)

23  x 8 x Solution set: {8}

(c)

log 36 6  x 36 x  6

6   6 2 x

The carbon dioxide level will be triple the level that it was in 2000 by 2179.

12

62 x  61 2 1 2x  2 1 x 4 1 Solution set:   4

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3

59

60

Chapter 4 Inverse, Exponential, and Logarithmic Functions

units to the left. The graph has a vertical asymptote at x = −2. Domain: 2,   .

Classroom Example 3 (page 467) (a)

We can write the exponential form of

Range:  ,  

y

1 f  x   y  log1 3 x as x    to find 3 ordered pairs that satisfy the equation. It is easier to choose values for y and then find the corresponding values of x. Make a table of values.



y

f  x   log1 3 x

x

x  13

9 3 1

−2 −1 0

−2 −1 0

1 3

1

1

1 27

3

3

1 19,683

9

9

(b)

by translating the graph of y  log 4 x two units up. The graph has a vertical asymptote at x = 0. Domain: 0,   . Range:  ,  

The graph can also be found by reflecting the x

1 graph of f ( x)    about the line y = x. The 3 graph has the y-axis as a vertical asymptote.

(b)

The graph of f  x   log 4 x   2 is obtained

(c)

The graph of f  x   log 2  x  3  2 is obtained by translating the graph of y  log 2 x three units right and two units up.

The graph has a vertical asymptote at x = 3. Domain: 3,   . Range:  ,  

f  x   log5 x

Because f  x   y  log 5 x, we can write the exponential form as x  5 y to find ordered pairs that satisfy the equation. It is easier to choose values for y and find the corresponding values of x. Make a table of values. y  log5 x x 1  0.04 25

2

1  0.2 5

1

1 0 5 1 25 2 The graph can also be found by reflecting the graph of f ( x)  5 x about the line y = x. The graph has the y-axis as a vertical asymptote.

Classroom Example 5 (page 469) (a)

log 7 8  6  log 7 8  log 7 6

(b)

log 6

(c)

log 2 3 9  log 2 91 3  13 log 2 9

(d)

log a 5 r 3  log a r 3 5  53 log a r

(e)

logb

12  log 6 12  log 6 5 5

 

rs 2t  logb rs 2t  logb u 3v5 u 3v 5  logb r  log b s 2  log b t







 log b u 3  logb v5

Classroom Example 4 (page 468) (a)

 

The graph of f  x   log 3  x  2 is obtained by translating the graph of y  log 3 x two Copyright © 2021 Pearson Education, Inc

 logb r  2 log b s  log b t  3logb u  5 logb v



Section 4.4 Evaluating Logarithms and the Change-of-Base Theorem 1m

(f)

log a

m

 r 3s2   r 3s2  r 3s2 1   log log a  4  a  4  m t4  t   t  1 1  log a r 3 s 2  log a t 4 m m 1 1 1 3  log a r  log a s 2  log a t 4 m m m 3 2 4  log a r  log a s  log a t m m m

(a)

(b)

4.3   log  H3 0 

(c)

 4.3  log  H3 0 

 H3 0   104.3  5.0  105  

Classroom Example 3 (page 479)



pH   log  H3 0    log 4.5  103



  log 4.5  log10

xz log 4 x  log 4 y  log 4 z  log 4 y 4 logb r  5 log b s  log b

pH   log  H3 0 

(b)

 

Classroom Example 6 (page 471)

61

3



   0.6532  3  2.3

The pH level is below 3, so this wetland is a bog.

Classroom Example 4 (page 479)

4

r s5

d  10 log II , where d is the decibel rating.

0

1 2 log a x  log a y  log a xy 3 3  log a 3 x  log a 3 y 2  log a xy  log a

3 xy 2

xy

10, 000, 000 I 0 I d  10 log  10 log I0 I0  10 log10, 000, 000  10 7   70

Classroom Example 5 (page 480)

Classroom Example 7 (page 471) (a)

log10 49  log10 7 2  2 log10 7  2 0.8451  1.6902

(b)

log10 70  log10 7  10  log10 7  log10 10  0.8451  1  1.8451

Section 4.4 Evaluating Logarithms and the Change-of-Base Theorem

Classroom Example 6 (page 480) (a)

Because A = K, KA  1 .



Classroom Example 1 (page 478)

t  1.26  109

  4.06  109  ln1ln8.33(1) 2

The rock sample is approximately 4.06 billion years old. (b)



t  1.26  109

  2.38  109  ln18.33(0.325) ln 2

The rock sample is approximately 2.38 billion years old.

Classroom Example 7 (page 481)

Classroom Example 2 (page 478) (a)



pH   log  H3 0    log 6.8  10 8



  log 6.8  log10  0.8325  8  7.2

8



   0.8325  8

R  k ln

(a)

C , C  2C0 , and k = 14.

C0

Because C  2C0 , CC  2 and 0

R  14 ln 2  9.7 watts per m 2 .

(b)

T  R   1.03R  T 9.7   1.03 9.7   10F

The average global temperature increase will be about 10ºF.

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62

Chapter 4 Inverse, Exponential, and Logarithmic Functions

Classroom Example 8 (page 482)

Classroom Example 3 (page 491)

ln 20  2.1610 or ln 4 log 20 log 4 20   2.1610 log 4

(b)

x

ln 0.7  0.5146 or ln 2 log 0.7 log 2 0.7   0.5146 log 2

ln e  ln 50 x  ln 50 x  3.912 Solution set: 3.912

log 2 0.7 

(b)

Classroom Example 9 (page 483) H    P1 log 2 P1  P2 log 2 P2    Pn log 2 Pn  There are 60 + 140 = 200 members in the 60  0.3 and P2  140  0.7. community, so P1  200 200

Thus, H    0.3log 2 0.3  0.7 log 2 0.7  ln 0.7   ln 0.3   0.3  0.7 ln 2 ln 2    0.3ln 0.3  0.7 ln 0.7      0.881 ln 2 

Section 4.5 Exponential and Logarithmic Equations Classroom Example 1 (page 490) 8 x  21 ln 8 x  ln 21 x ln 8  ln 21 x  lnln21  1.464 8

x

e  50

(a)

log 4 20 

(a)

e4 x  e x 1  5e e4 x  x 1  5e e5 x 1  5e e5 x  2  5 ln e5 x  2  ln 5 5 x  2  ln 5 ln 5  2 x  0.722 5 Solution set: {0.722}

Classroom Example 4 (page 491) e 2 x  6e x  5  0 Let u  e x . Then we have u 2  6u  5  0 u  1u  5  0  u  1, 5

Now substitute these values for u and solve for x. u 1 u5 ex  1 ex  5 ln e x  ln1 ln e x  ln 5 x0 x  ln 5 Both values check, so the solution set is {0, ln 5}.

Solution set: {1.464}

Classroom Example 5 (page 492)

Classroom Example 2 (page 490)

(a)





 

2 x  3 ln 5   x  1 ln 8

2 x ln 5  3ln 5  x ln 8  ln 8 2 x ln 5  x ln 8  3ln 5  ln 8 x  2 ln 5  ln 8  3ln 5  ln 8 3ln 5  ln 8 x  6.062 2 ln 5  ln 8 Solution set: {6.062}

 

Solution set: e9

52 x  3  8 x 1

ln 52 x  3  ln 8 x 1

4 ln x  36  ln x  9  x  e9

(b)





log3 x3  5  1  x3  5  31  3  3

x 8 x  2 Solution set: {2}

Classroom Example 6 (page 492) log  2 x  1  log x  log  x  8 log  x  2 x  1  log  x  8





log 2 x 2  x  log  x  8 2

2x  x  x  8 2 x 2  8  x 2  4  x  2 The negative solution is not in the domain of log x in the original equation, so the only valid solution is x = 2. Solution set: {2}

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Section 4.6 Applications and Models of Exponential Growth and Decay

Classroom Example 7 (page 493) log3  4 x  1 x  1  3

4 x  1 x  1  33  27

4 x 2  5 x  1  27 4 x 2  5 x  26  0 4 x  13 x  2  0  x   134 , 2

Check to be sure that the argument of the logarithm in the original equation is positive. If x   134 , then

 T  T0  ln    kt  C  1  T  T0  ln  k  C  t

Classroom Example 11 (page 495) f t   23.19 ln t  75.19, t  1

(a)

4 x  1 x  1  27, and log3 27  3. Similarly, if x = 2, then  4 x  1 x  1  27.





(b)

2 x 2  11x  15  23  8 2 x 2  11x  7  0 Use the quadratic formula with a = 2, b = −11, and c = 7 to solve for x:

x

112  4 27 11  65  2  2 4

The negative solution makes 2x − 5 in the original equation negative, so we reject that solution. 11  65  Solution set:   4   Classroom Example 9 (page 494) ln eln x  ln  x  4  ln 5 ln x  ln  x  4  ln 5 x ln  ln 5 x4 x 5 x4 x  5 x  20 4 x  20 x5 Be sure to check the solution.

Solution set: {5}

Classroom Example 10 (page 495) T  T0  Ce  kt T  T0  Ce  kt T  T0  e  kt C  T  T0  ln   ln e  kt  C 

Let f t   131 and solve for t:

f t   131 23.19 ln t  75.19  131 23.19 ln t  55.81 55.81 ln t  23.19 t  e55.81 23.19  11.10 Adding 11 to 2012 gives the year 2023. Based on this model, the number of U.S. residents who watch video content on a tablet will reach 131 million in 2023.

log 2  2 x  5  log 2  x  3  3 log 2  2 x  5 x  3  3

  11 

The year 2018 is represented by t = 6. f 6  23.19 ln 6  75.19  116.7 Based on this model, 116.7 million U.S. residents watched video content on a tablet in 2018. This is close to the actual figure of 116.5 million.

Solution set:  134 , 2

Classroom Example 8 (page 493)

63

Section 4.6 Applications and Models of Exponential Growth and Decay Classroom Example 1 (page 502) (a)

The equation will take the form y  y0 e kt . We must find the values of y0 and k. Let the year 2000 correspond to t = 0. Then the year 2175 corresponds to t = 175. Because y0 is the initial amount, y0  375 when t = 0. Thus, the equation is y  375e kt . When t = 175, y = 1090. Substitute these values into the equation and solve for k.

1090  e175k  1090  375e175k  375 ln 1090  1090  375   k k ln  175  375  175 0.00610  k The equation is y  375e0.00610 x .

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64

(b)

Chapter 4 Inverse, Exponential, and Logarithmic Functions

Let y = 3(280) = 840, then solve for t: 840  e0.00610t  840  375e0.00610t  375 ln 840  840  375   t t ln  0.00610  375  0.00610 132  t Because t = 0 corresponds to the year 2000, t = 132 corresponds to the year 2132. Thus, the carbon dioxide level will triple from the preindustrial level of 280 ppm during 2132.

Classroom Example 2 (page 502)

Use the continuously compounding formula A  Pert with A = 2P and r = 0.0225 to solve for t: 2 P  Pe0.0225t  2  e0.0225t  ln 2 ln 2  0.0225t  t   30.8 0.0225 It will take about 31 years for the money to double.

Classroom Example 3 (page 503) f t   7.504e0.00936t , with t = 10 corresponding to

(b)

Classroom Example 5 (page 504) y  y0 e 0.0001216t . Letting y 

3 3 y0  y0 e0.0001216t   e0.0001216t  4 4 ln 34 3 ln    0.0001216t   2365.8  t 4 0.0001216 The woman died approximately 2366 years before 2005, in 361 BCE.

(a)

f 0  350  Ce k (0)  34  350  C  C  316 Now solve for k: f 1  350  316e  k (1)  70  350  316e  k (1)  280 280  316e  k   e k  316  280   ln   0.1210  k  316 

8  7.504e0.00936t 8  e0.00936t 7.504  8  ln   0.00936t  7.504  8 ln 7.504 t  6.8 0.00936 The world population will reach 8 billion during the 6th year after 2018, or during the year 2024.

Classroom Example 4 (page 504)

Thus, the equation is f t   350  316e0.121t . (b)

800  y0 e

 800  y0 . Thus, the

exponential equation becomes y  800ekt .

Now, let y = 400 and t = 2.5, then solve for k:

f 1.5  350  316e 0.121(1.5)  86.5

After 90 minutes (1.5 hours), the temperature is about 86.5ºF.

kt

The equation will take the form y  y0 e .

Letting y = 800 and t = 0, solve for y0 :

From the given information, when t = 0, f 0  34 , and T0  350 . When t = 1, f 1  70 . First solve for C:

f t   8 . Solve for t:

k (0)

3 y , solve for t: 4 0

f t   T0  Ce  kt

According to the model, the population will be 7.646 billion at the end of 2020.

(a)

Letting t = 4, we have y  800e 0.277(4)  264.2 .

There will be about 264 g of the substance left after 4 years.

Classroom Example 6 (page 506)

the year 2000. (a) 2020 corresponds to t = 2. f  2  7.504e0.00936(2)  7.646

(b)

1  e 2.5k  2 ln 1 1 ln    2.5k  2  0.277  k 2 2.5 Thus, the exponential equation is y  800e0.277t .

400  800e 2.5k 

(c)

145  350  316e0.121t  205 205  316e0.121t   e0.121t  316 205 ln 316  205   0.121t   3.576  t ln    316  0.121 The temperature will be 145ºF after about 3.576 hours or about 3 hours 35 minutes.

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Chapter 5 Systems and Matrices Section 5.1

Systems of Linear Equations

Classroom Example 1 (page 527) 4 x  3 y  14 (1) x  2 y  1 (2) Solve equation (2) for x: x  1 2y Replace x in equation (1) with 1 + 2y and then solve for y: 4 1  2 y   3 y  14 4  8 y  3 y  14 5 y  10  y  2 Replace y in equation (2) with 2, then solve for x: x  2 2  1  x  4  1  x  5

Check: 4 x  3 y  14 (1) x  2 y  1 (2) 4 5  3  2  14 ? 5  2  2  1 ?

20  6  14 5 4 1 14  14 True 1  1 True Solution set: 5, 2 Classroom Example 2 (page 528) 2 x  3 y  1 (1) 3x  2 y  18 (2) Multiply both sides of equation (1) by 2, and then multiply both sides of equation (2) by 3. Add the resulting equations to eliminate y and solve for x: 4 x  6 y  2 9 x  6 y  54 13x  52  x  4 Substitute 4 for x in equation (1), then solve for y: 2 x  3 y  1 2  4  3 y  1 8  3 y  1  3 y  9  y  3 Check: 2 x  3 y  1 (1) 3 x  2 y  18 (2) 2  4  3  3  1 ? 3  4  2  3  18 ?

8  9  1 12   6  18 1  1 True 18  18 True

Solution set:  4, 3

Classroom Example 3 (page 528) 7 x  3 y  5 (1) 14 x  6 y  10 (2) Multiply equation (1) by 2, then add the resulting equation to equation (2) and solve for y: 14 x  6 y  10 14 x  6 y  10 0  20 This is a false statement. The solution set is , and the system is inconsistent.

Classroom Example 4 (page 5290) 9 x  3 y  24 (1) 3x  y  8 (2) Multiply equation (2) by 3, then add the resulting equation to equation (1) and solve for x: 9 x  3 y  24 9 x  3 y  24 0 0 This is a true statement. There are infinitely many solutions. We will now express the solution set with x as the arbitrary variable. Solve equation (2) for y.

3x  y  8  3x  8  y Solution set: {(x, 3x − 8)}

Classroom Example 5 (page 530) Let x = the salary in Boston.

Let y = the salary in Indianapolis. Then, we have the system: x y  76, 553.50 (1) 2 x  y  11, 989 (2) Multiply both sides of equation (1) by 2 to eliminate the fraction. x y  76, 553.50  x  y  153,107 (3) 2 We now have the system x  y  153,107 (3) x  y  11, 989 (2) Add the two equations to eliminate y. x  y  153,107 (3) x  y  11, 989 (2) 2 x  165, 096 x  82, 548 (continued on next page)

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Chapter 5 Systems and Matrices

(continued)

Classroom Example 7 (page 533)

Substitute 82,548 for x in equation (2) and solve for y.

82, 548  y  11, 989 y  70, 559 Verify that (82548, 70559) satisfies both equations in the original system. The median salary in Boston is $82,548, and the median salary in Indianapolis is $70,559.

3x  y  2 z  7 (1) 5 x  2 y  z  6 (2) Multiply equation (1) by −2, then add the result to equation (2) to eliminate y: 6 x  2 y  4 z  14 5 x  2 y  z  6 x 5 z  8 (3) Now solve equation (3) for x:  x  5 z  8   x  5 z  8  x  5 z  8 Substitute 5z − 8 for x in equation (2), then solve for y: 5 5 z  8  2 y  z  6 25 z  40  2 y  z  6 2 y  26 z  34 2 y  26 z  34 y  13z  17 With z arbitrary, the solution set is 5 z  8, 13z  17, z  .

Classroom Example 6 (page 532) 3x  4 y  2 z  14 (1) 2 x  y  2 z  9 (2) x  y  z  9 (3) Eliminate y by adding equations (2) and (3) to obtain 3x  3z  18  x  z  6 (4) . Now multiply equation (3) by 4, then add the result to equation (1) to eliminate y: 3x  4 y  2 z  14 4 x  4 y  4 z  36 7x  2 z  22 (5) Multiply equation (4) by −2, then add the result to equation (5) to solve for x: 2 x  2 z  12 7 x  2 z  22  10  x  2 5x Substitute −2 for x in equation (4) to solve for z: x  z  6 2  z  6  z  4 Substitute −2 for x and −4 for z in equation (3) to solve for y: x  y  z  9 2  y  4  9   y  3  y  3 Thus, x = −2, y = 3, and z = −4.

Verify that the ordered triple  2, 3, 4 satisfies all

three equations in the original system.

3x  4 y  2 z  14 (1) 3  2  4 3  2  4  14 ? 6  12  8  14 14  14 True 2 x  y  2 z  9 (1) 2  2  3  2  4  9 ? 4  3  8  9 9  9 True x  y  z  9 (1) 2  3  4  9 ? 9  9 True Solution set:  2, 3, 4

Classroom Example 8 (page 534) Because y  ax 2  bx  c and the parabola passes through the points (−5, 7), (–1, −2), and (3, 5), we have the following equations: 7  a  5  b  5  c  25a  5b  c  7 (1) 2

2  a  1  b  1  c  2

5  a 3  b 3  c 2

a  b  c  2 (2)

 9a  3b  c  5 (3)

Multiply equation (2) by −1, then add the result to equation (1) to eliminate c: 25a  5b  c  7 a  b  c  2 24a  4b  9 (4) Multiply equation (2) by −1, then add the result to equation (3) to eliminate c: a  b  c  2 9a  3b  c  5 8a  4b  7 (5) Now, add equations (4) and (5) and solve for a: 24a  4b  9 8a  4b  7 32a

 16  a  12

Substitute 12 for a in equation (4) to solve for b:



24 12  4b  9 12  4b  9  4b  3  b  34

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Section 5.2 Matrix Solution of Linear Systems

(continued) Now substitute 12 for a and 34 for b in equation (2) to solve for c: 1  34  c  2   14  c  2  c   74 2 The equation of the parabola is y  12 x 2  43 x  74 or y  0.5 x 2  0.75 x  1.75 .

Classroom Example 9 (page 535) Let x = the number of units of corn, y = the number units of soybeans, and z = the number of units of cottonseed.

The total amount of protein is to be 35 units, so the first row of the table is 0.25 x  0.4 y  0.2 z  35 (1) The second row of the table is 0.4 x  0.2 y  0.3 z  38 (2) and the third row of the table is 0.3 x  0.2 y  0.1z  28 (3) Multiply both sides of equation (1) by 100 and both sides of equations (2) and (3) by 10 to obtain the equivalent system 25 x  40 y  20 z  3500 (4) 4 x  2 y  3z  380 (5) 3x  2 y  z  280 (6) Multiply equation (5) by −20, then add the result to equation (4) to eliminate y: 25 x  40 y  20 z  3500 80 x  40 y  60 z   7600 55 x  40 z  4100 (7) Multiply equation (6) by −20, add the result to equation (4) to eliminate y and z, then solve for x: 25 x  40 y  20 z  3500 60 x  40 y  20 z   5600 35 x  2100  x  60 Substitute 60 for x in equation (7) and solve for z: 55 60  40 z  4100 3300  40 z  4100  40 z  800  z  20 Now substitute 60 for x and 20 for z into equation (6) to solve for y: 3 60  2 y  20  280 200  2 y  280  2 y  80  y  40 Verify that the ordered triple (60, 40, 20) satisfies the system formed by equations (1), (2), and (3).

The feed should contain 60 units of corn, 40 units of soybeans, and 20 units of cottonseed.

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Section 5.2 Matrix Solution of Linear Systems Classroom Example 1 (page 545) 2x  3y  7 3x  4 y  32 This system has the augmented matrix 3 7 2  3 4 32 .   7 1 1 3 R1 2 2 2   3 4 32 3 7 1 2 2  17 85   0  2  2  3R1  R2 1 3 7  2 2   2  0 1 5  17 R2  1 0 4  32 R2  R1  0 1 5   Verify that the ordered pair (−4, 5) satisfies the original system.

Solution set:  4, 5

Classroom Example 2 (page 547) x  y  4 z  10 2x  3 y  z  7 3x  y  z  12 This system has the augmented matrix  1 1 4 10  2 3 1 7  .   3 1 1 12 1 0   3 1 0   3 1 0   0 1  0 0 

1 4

10 5 9 13 2R1  R2 1 1 12 1 4 10   1 R2 1  95 13 5 5 1 1 12 1 4 10 13  1  95 5 4 11 18 3R1  R3 1 4 10 9 13  1 5 5  4R2  R3 0 195  38 5 

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68

Chapter 5 Systems and Matrices

(continued)

Classroom Example 5 (page 550)

1 4 10 1  95 135  0 1 2 195 R3 1 4 10 1 0 1 95 R3  R2 0 1 2 0 4 11 R1  R2  1 0 0 3 R1  4R3   1 0 1  0 1 0 1 0 0 1 2 0 1 2 Verify that the ordered triple (3, −1, −2) satisfies the original system.

Solution set: 3, 1, 2

x  2y  z  5 2 x  y  z  5 3x  y  2 z  8 This system has the augmented matrix 1 5  1 2  2 1 1 5 .    3 1 2 8

1 0  0 1 0  0 1 0  0

Classroom Example 3 (page 548) 2x  3y  7 6 x  9 y  0 This system has the augmented matrix  2 3 7   6 9 0 .    0 0 21 3R1  R2  6 9 0   The first row of the augmented matrix corresponds to the equation 0 x  0 y  21, which has no solution.

Thus, the solution set is .

Classroom Example 4 (page 549) 2x  y  z  5 3 x  2 y  z  8 This system has the augmented matrix  2 1 1 5  3 2 1 8 .   5 2 1 1  0 1 5 31 3R1  2R2    2 0 6 36 R1    R2  0 1 5 31   3 18 12 R1 1 0  0 1 5 31   The equations that correspond to the final matrix are x  3 z  18 and y  5 z  31. Solve these equations for x and y in terms of z: x  3 z  18  x  3z  18 and y  5 z  31  y  5 z  31.

1 0   0 1 0   0 1  0 0 

5 5 1 15 2R1+R2 7 1 23 3R1+R3 2 1 5 1 15 3  15 R2 7 1 23 2 1

2 1 0

1 5  3 2 7R2  R3

1 5  52

 1 2 1 5 0 1 1 3 5   0 0 1 5  52 R3 The corresponding equations are x  2y  z  5 y  15 z  3

z  5 The value of z is given. The values of y and x can now be found in reverse order by back-substitution. y  15  5  3 y 1  3 y4

x  2  4   5  5 x35 x2 The solution set is  2, 4,  5 . Check the solution



in the original system.

Solution set:  3 z  18, 5 z  31, z 

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

Section 5.3 Determinant Solution of Linear Systems

Section 5.3 Determinant Solution of Linear Systems Classroom Example 1 (page 559)

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2 3 5 1 4 5  a31  A31  a32  A32  a33  A33 0 2 6  2  15  6  11  36

Algebraic solution: 5 8 2 4  5  4  2  8  4

Classroom Example 4 (page 563)

Graphing calculator solution:

(a)

2 1

7

4 0

8

6 5 21

Use determinant theorem 6 to obtain a 0 in row 2, column 3.

2 1 11 4 0 0 2C1  C3 6 5 33 Now expand by row 2 2 1 11 1 11 2 11 2 1 4 0 0    4 0 0 5 33 6 33 6 5 6 5 33  4 1  33  11  5  0  0  4 33  55  4  22  88

Classroom Example 2 (page 561)

(a)

 5 1 6  0 2 9  4 1 3 To find the cofactor of 5, we have i = 1, j = 1, 2 9 and M11  1 3  15 . Thus, the cofactor is

111 15  15. (b)

1 2 5 8 0 4 1 7 0 0 2 5 0 0 0 9 User Determinant Theorem 7 to find the determinant of this triangular matrix by multiplying the elements on the main diagonal.

1 2 5 8 0 4 1 7  14 29  72 0 0 2 5 0 0 0 9

(b)

To find the cofactor of 0, we have i = 2, j = 1, 1 6 and M 21  1 3  9 . Thus, the cofactor is

12 1 9  9. (c)

To find the cofactor of 3, we have i = 3, j = 3, 5 1 and M 33  0 2  10 . Thus, the cofactor is

13 3 10  10.

Classroom Example 3 (page 562)

Classroom Example 5 (page 565)

2 3 5 Evaluate 1 4 5 by expanding on the third row.

0 2 6

4x  3 y  2 x  2y  5

2 3 5 1 4 5  a31  A31  a32  A32  a33  A33 .

0 2 6 However, we do not need to calculate A31 because a31  0.

5  1 15  15   1 5 3 3 3 3 2 3  11 A33   1 M 33  1 1 4 3 2

A32   1

3 2 2

M 32   1

Algebraic solution: 4 3 2 3 D  11; Dx   19; 1 2 5 2 4 2  18 Dy  1 5 D

D

x  Dx  19  19 and y  Dy  1811   18 .

11 11 11

Solution set:

 ,   19 11

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18 11

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70

Chapter 5 Systems and Matrices

(continued)

Section 5.4 Partial Fractions

Graphing calculator solution:

Classroom Example 1 (page 573) 3 x3  3 x 2  7 x  4 x2  x The degree of the numerator is greater than the degree of the denominator, so first find the quotient.

3x x 2  x 3 x3  3 x 2  7 x  4 3 x3  3 x 2 7x  4

The quotient is

Classroom Example 6 (page 566) 2 x  4 y  z  19  0  2 x  4 y  z  19 4 x  6 y  z  15  0  4 x  6 y  z  15 x  y  2 z  11  0  x  y  2 z  11

2 4 1 Using a calculator, we have D  4 6 1  60 , 1 1 2 19 4 1 2 19 1 Dx  15 6 1  90, D y  4 15 1  150, 11 1 2 1 11 2 2 4 19 and Dz  4 6 15  360 . Thus, we have 1 1 11 Dy

D

150 x  Dx  90  23 , y  D  60   52 , and 60 D

z  Dz  360  6.

60

Solution set:

 ,  , 6 3 2

5 2

3 x3  3 x 2  7 x  4 7x  4 .  3x  2 2 x x x x

Now find the partial fraction decomposition for 7x  4 x2  x A B 7x  4 7x  4     2  1 x x x x 1   x x 7 x  4  A  x  1  Bx 7 x  4   A  B x  A  7  A  B and  4   A  4  A and 3  B 3 x3  3 x 2  7 x  4 4 3 Thus,  3x   2 x x 1 x x Classroom Example 2 (page 574) x2  4 x  7

 x  33 This is a proper fraction, so x2  4 x  7 A B C     3 2 x  3  x  3  x  3  x  33 x 2  4 x  7  A  x  3  B  x  3  C (1) 2

Classroom Example 7 (page 566)

Substituting 3 for x in equation (1) gives

x  3y  2z  5 3x  y  4 z  8 9 x  3 y  12 z  24 We need to show that D = 0. Expanding about column 1 gives 1 3 2 4 3 2 3 2 D  3 1 4  1 31  12  3 3 12  9 1 4 9 3 12  0  3  30  9 10  0

C  32  4  3  7  4 . Substituting 4 for C and 1 for x in equation 1 gives

Thus, Cramer’s rule does not apply to the system.

12  4 1  7  A 1  3  B 1  3  4 4  4 A  2B  4 0  4 A  2 B (2) Substituting 4 for C and −1 for x in equation (1) gives 2

12  4 1  7  A 1  32  B 1  3  4 12  16 A  4 B  4 8  16 A  4 B 2  4 A  B (3)

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Section 5.5 Nonlinear Systems of Equations

(continued)



Multiplying both sides by x x 2  2

71

 gives 2

Now solve the system consisting of equations (2) and (3) by subtracting equation (3) from equation (2): 4 A  2B  0 and 4 A  B  2   B  2  B  2 4 A  2  2  4 A  4  A  1 . Thus, x2  4 x  7 1 2 4    3 2  x 3  x  3  x  3  x  33

If x = 0, then equation (1) gives 4  4 A  A  1 .

Substituting 1 for A in equation (1) and expanding and combining terms on the right gives x 4  3x3  4 x 2  2 x  4

Classroom Example 3 (page 575)

 x2  2

8 x 2  10 x  3

 x 4  4 x 2  4  Bx 4  Cx3  2 Bx 2  2Cx  Dx 2  Ex  1  B  x 4  Cx3  4  2 B  D  x 2  2C  E  x  4 (2)

 x  4  x 2  1 This is a proper fraction, so 8 x 2  10 x  3 A Bx  C   2 2  x  4 x  1 x  4 x  1









Multiply both sides by the  x  4 x 2  1 to obtain





8 x 2  10 x  3  A x 2  1   Bx  C  x  4 (1) .

Letting x = −4, we have 8  4  10  4  3 2  A  4  1   B  4  C   4  4   85  17 A  A  5 Replace A with 5 in equation (1), and substitute 0 for x: 2 2 8 0  10 0  3  5 0  1   B 0  C  0  4   3  5  4C  C  2 Now replace A with 5 and C with −2 in equation (1), and substitute 1 for x to solve for B: 2 2 8 1  10 1  3  5 1  1   B 1  2 1  4   15  10  5B  10  B  3 2

Thus,

8 x 2  10 x  3

 x  4  x  1 2



5 3x  2  2 x  4 x 1

x 4  3x3  4 x 2  2 x  4





2

This is a proper fraction that has both a linear factor and a repeated quadratic function, so x 4  3x3  4 x 2  2 x  4 A Bx  C Dx  E   2  2 2 2 x 2 x  x x 2 x2  2







 A x2  2



   Bx  C  x  x  2   Dx  E  x (1) 2

2

   Bx  C  x  x  2   Dx  E  x 2

2

Equate the coefficients of like powers of x on the two sides of equation (2): 1  1 B  B  0 3C 4  4  2B  D (3) 2  2C  E (4) Substituting 0 for B in equation (3) gives D = 0.

Substituting 3 for C in equation (4) gives E = −4.

Thus, x 4  3x3  4 x 2  2 x  4 1 3 4   2  2 2 x x 2 x x2  2 x2  2









Section 5.5 Nonlinear Systems of Equations Classroom Example 1 (page 580) x2  y  6 (1) x  y  4 (2)

Algebraic solution: Solve equation (2) for y, then substitute that expression into equation (1) and solve for x: x  y  4  y  x  4 x 2   x  4  6  x 2  x  2  0   x  2 x  1  0  x  2 or x  1

Classroom Example 4 (page 576)

x x2  2

x 4  3 x3  4 x 2  2 x  4



Substituting −2 for x in equation (2) gives y = 2.

Substituting 1 for x in equation (2) gives y = 5. Verify that the ordered pairs (−2, 2) and (1, 5) satisfy the equations. The solution set is  2, 2 , 1, 5



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(continued on next page)

72

Chapter 5 Systems and Matrices

(continued) Graphing calculator solution:

Classroom Example 3 (page 582) x 2  xy  y 2  21 (1)

Classroom Example 2 (page 581) 2

2

x  y 9 (1) 9 x 2  4 y 2  36 (2)

Algebraic solution: Using the elimination method, we multiply equation (1) by −4, then add the resulting equation to equation (2): 2

2

x 2  xy  y 2  9 (2) Subtract equation (2) from equation (1) to obtain 6 2 xy  12  xy  6  y  (3) x Now substitute 6x for y into equation (1) to solve

for x:

     21  x2  6  36x  21  2

4 x  4 y  36 9 x 2  4 y 2  36

x 2  x 6x  6x

5x2 0 x0 Substitute 0 for x in equation (1), then solve for y:

 x  3 x  12  0  x  3  x   3 or

2

2

y  9  y  3 Solution set: {(0, −3), (0, 3)} Graphing calculator solution: To solve the system graphically, solve equation (1)

for y to get y   9  x 2 . Solve equation (2) for y to 9x2 x2 . Graph these four  3 1  4 4 functions in the window [–6.6, 6.6] × [–4.1, 4.1] to find the two solutions.

get y   9 

y1  9  x y3  3 1 

2

y2   y1   9  x 2

x 4

y4   y3  3 1 

2 2

x 4

2

x  15  362  0  x 4  15 x 2  36  0  x 2 2 2

x 2  12  x  2 3 Substitute these values into equation (3) to find the corresponding values of y: 6 6 3 x 3 y  2 3 3 3

6 6 3   2 3 3 3

x 3 y

x2 3 y

6 2 3



3 3  3 3

6 3 3   3 3 2 3 Verify these solutions in the original system. x  2 3  y  

 3, 2 3  ,  3, 2 3  , 2 3, 3  , 2 3,  3 

Solution set:

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Section 5.6 Systems of Inequalities and Linear Programming

Verify the solutions in the original system.

Solution set:

Classroom Example 4 (page 583)

2i, 10  , 2i, 10  , 2i,  10  , 2i,  10 

x 2  y 2  4 (1)  x  y  0 (2)

Algebraic solution: Solving equation (2) for x gives x  y . Because 2

2

x  0 for all x, y  0 . In equation (1), x  x , so y 2  y 2  4  y 2  2  y  2 . (Reject the

negative solution.) Substituting 2 for y in equation (2), we have  x  2 0 x  2  x  2  x  2 or x   2 Verify the solutions in the original system.

Solution set:

 2, 2  ,  2, 2 

Graphing calculator solution: Graph the following in the window [–6.6, 6.6] × [–4.1, 4.1] to find the two solutions. y1  4  x 2

73

y2   y1   4  x 2

y3  x

Classroom Example 6 (page 584)

Let x = the length and width of the square base, and let y = the height. Then x 2 y  384

(1)

2

x  4 xy  256 (2) Solve equation (1) for y, then substitute that expression into equation (2) to solve for x: 384 x 2 y  384  y  2 x 384 1536    256  x 2  4 x  2   256  x 2   x  x

x3  1536  256 x  x3  256 x  1536  0 Values for x are restricted to positive values, and should be relatively small considering the nature of the problem. Apply the rational zeros theorem to find the rational solutions.

Using synthetic division, we see that 8 is a solution.

81 1

0 8 8

 256 1536 64  1536  192 0

 6.

Therefore, one value of x is 8 and y  384 2 8

2

Now solve x  8 x  192  0 for any other positive solutions. 8  82  4(1)(192) 8  832  2(1) 2  10.422 or  18.422 Reject the negative solution. If x = 10.422, then x

y

Thus, there are two possible answers.

Answer 1: length = width = 8 m; height = 6 m Answer 2: length = width ≈ 10.422 m height ≈ 3.535 m

Classroom Example 5 (page 584) x 2  y 2  6 (1) 3x 2  2 y 2  8 (2)

Multiply equation (1) by −3, then add the resulting equation to equation (2) to obtain  y 2  10  y 2  10  y   10 . Find the corresponding values of x by substituting the values of y into equation (1):

 10   6  x  4  x  2i 2 y   10  x 2    10   6  x 2  4  x  2i 2

y  10  x 

2

2

384  3.535.

10.4222

Section 5.6 Systems of Inequalities and Linear Programming Classroom Example 1 (page 591) 4 x  3 y  12

The boundary of the graph is the line −4x −3y = 12 which can be graphed using the x-intercept −3 and the y-intercept −4. The boundary is included in the graph, so draw a solid line. Solving for y, we have 4 x  3 y  12  3 y  4 x  12  y   43 x  4 .

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74

Chapter 5 Systems and Matrices

(continued) (Note that the direction of the inequality changes due to division by −3.) Thus, the graph of the solution set is the half-plane below the boundary. As a check, use (0, 0) as a test point. Because 4 0  3 0  12  0  12 is a false statement, shade the side of the graph not containing the test point (0, 0) (the half-plane below the boundary).

(b)

y 1 x0 y  2 x 1 y  1  y  1 or y  1 . Graph solid lines

and shade the region between the two lines.

The set of points that satisfies x  0 includes the points on the y-axis and all points to the right of the y-axis. The set of points that satisfies y  2 x  1 includes all points

Classroom Example 2 (page 592) y   x  2  1 2

The boundary of the graph is the parabola

between the rays that make up the graph of y  2 x 1.

y   x  2  1. This parabola has vertex 2, 1 and 2

opens downward. The boundary is not included in the graph, so draw a dashed line. Because the inequality y   x  2  1 is already solved for y, we determine 2

from the presence of the < symbol that the solution set is the region below the boundary. As a check, use (0, 0) as a test point. Because 0  0  2  1 is a 2

true statement, shade the side of the graph containing the test point (0, 0) (the half-plane below the boundary).

The solution sets of y  1 and y  2 x  1 have no points in common, so the solution set of the system is  .

Classroom Example 3 (page 593) (a)

y  4  x2 x  y 1

Graph y  4  x 2 as a dashed parabola with vertex (0, 4) and x-intercepts (−2, 0) and (2, 0).

Shade the region inside the parabola. Graph x  y  1 as a dashed line with x-intercept −1 and y-intercept 1. Shade the region above the line. The solution set of the system is the common region, which is shaded in the final graph.

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Section 5.7 Properties of Matrices

Classroom Example 4 (page 595)

Let x = the number of video games to be produced daily, and let y = the number of controllers to be produced daily. The company must produce at least 10 MP3’s, so x  10. No more than 50 video games may be produced, so x  50. No more than 60 controllers can be produced, so y  60. The number of video games cannot exceed the number of controllers, so x  y . The number of video games and controllers cannot be negative, so x  0 and y  0 . Thus, the constraints are x  10, x  50, y  60, x  y, x  0, and y  0.

Each video game gives a profit of $30, so the daily profit from production of x video games is 30x. The profit from production of y controllers is 70y. Thus, the total daily profit is 30x + 70y. This is the objective function.

To find the maximum possible profit, graph each constraint. The graph of the feasible region is the intersection of the regions that are the graphs of the individual constraints.

75

Robin gets 2x units of vitamin C from the capsules and 7y units of vitamin C from the chewable tablets.

She wants at least 20 units of vitamin C per day, so 2 x  7 y  20 . Also, x  0 and y  0 because she cannot buy a negative number of pills.

The intersection of the graphs is shown below:

From the graph, we see that the vertices are (0, 8), (1, 4), (3, 2), and (10, 0). Evaluate the objective function at each vertex.

Point Cost = 20x + 10y (0, 8) 20(0) + 10(8) = 80  Minimum (1, 4) 20(1) + 10(4) = 60 (3, 2) 20(3) + 10(2) = 80 (10, 0) 20(10) + 10(0) = 200 The minimum cost of 60¢ will be obtained when she takes 1 capsule and 4 chewable tablets per day.

Section 5.7 Properties of Matrices Classroom Example 1 (page 603) (a)

From the graph, we see that there are four vertices (10, 10), (10, 60), (50, 60), and (50, 50). Evaluate the objective function at each vertex.

Point Profit = 30x + 70y (10, 10) 30(10) + 70(10) = 1000 (10, 60) 30(10) + 70(60) = 4500 (50, 60) 30(50) + 70(60) = 5700  Maximum (50, 50) 30(50) + 70(50) = 5000 The maximum profit of $5700 will be reached when 50 video games and 60 controllers are produced.

(b)

x

3 y    5  

3 y  is a 1  2 matrix and  5 is   a 2  1 matrix, the statement cannot be true.

Because  x

Classroom Example 2 (page 603)

Classroom Example 5 (page 597)

Let x = the number of capsules, and let y = the number of chewable tablets. Then, the cost per day (in cents) is 20x + 10y. This is the objective function.

She gets 8x units of vitamin A from the capsules and 2y units of vitamin A from the chewable tablets. She wants at least 16 units of vitamin A per day, so 8 x  2 y  16 . Each capsule and each chewable tablets supplies 1 unit of vitamin B. Robin wants at least 5 units per day, so x  y  5 .

 a b   3 9   5 0   c d      Corresponding elements are equal, so a  3, b  9, c  5, d  0

(a)

 3 8  7 5 10 13  4 6  10 6   6 0     

(b)

 9  9  18  7    5   12  6  6  0  3  10  7 

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76

(c)

Chapter 5 Systems and Matrices

3 1 6 Because A  0 7 2 is a 2  3 matrix  

Classroom Example 4 (page 606)

 4 8 and B   3 1 is a 3  2 matrix, A and B  5 2 cannot be added.

(a)

 2 5  6 15 3     4 0  12 0

(b)

8 4  25 10  20    5  15 45  12 36

Classroom Example 3 (page 605) (a)

 8 9  8 4 16 13  6 2   7 11   1 9      

(b)

 9  18  9  3   12   15  6  6  12

(c)

Classroom Example 5 (page 608) C is a 2  5 matrix and D is a 4  2 matrix (a)

CD cannot be calculated.

Matrix C Matrix D 25 42  different

4 5 0 Because A   2 3 1 is a 2  3 matrix   4 5 and B   3 2 is a 2  2 matrix, the   difference A − B does not exist.

(b)

CD cannot be calculated.

(c)

DC can be calculated.

Matrix D Matrix C 42 25  matches   size of DC 45

(d)

The result will be a 4  5 matrix

Classroom Example 6 (page 608) 3 4 5 4 1 A   1 0 6 , B   3 8    

(a)

AB cannot be calculated.

Matrix A Matrix B 23 22  different

(b)

A 2  2 matrix multiplied by a 2  3 matrix results in a 2  3 matrix.  4 1 3 4 5  4 3  1 1 4  4  1 0 4 5  1  6  13 16 14  3 8  1 0 6   3 3  8 1 3 4  8 0 3 5  8 6    1 12 63                 

Classroom Example 7 (page 609) 2 5  3 4 C   6 1 , D   2 3 

(a)

2 5  3 4  2 3   5 2 2  4   53  4 23   CD   6 1  2 3  6 3  1 2  6  4  13   16 27 

(b)

3  2  46 3  5   41  30 19 DC   23 43  26 51         2  2  3  6 2  5  3 1   14 7 

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Section 5.8 Matrix Inverses

77

Classroom Example 8 (page 610) Concrete

Colonial Ranch

Model A  10 30 Model B  15 25  M Model C  25 45

(a)

Lumber

Brick

Shingles

2 1

0 20

2 Q 2

Colonial 10 Ranch 50

Cost per unit

Concrete  20  Lumber 180 Brick  60   R Shingles  25 

To calculate the total cost of material for all houses of each model, first find MQ, which will show the total amount of each material needed for all houses of each model.  10 10  30 50 10  2  30 1 10 0  30  20 10  2  30  2 10 30 MQ   15 25 10 2 0 2   15 10  25 50 15  2  25 1 15 0  25  20 15  2  25  2 50 1 20 2  25 10  45 50 25  2  45 1 25 0  45  20 25  2  45  2  25 45   Concrete Lumber Brick Shingles

 1600 50 600 80   1400 55 500 80  2500 95 900 140 Multiplying MQ and the cost matrix R gives the total cost of material for each model.  20   1600 50 600 80 180  1600 20  50 180  600 60  80  25  MQ  R   1400 55 500 80  60    1400 20  55 180  500 60  80 25  2500 95 900 140  25   2500  20  95 180  900 60  140  25  79, 000    69, 900  124, 600 The cost of materials for model A is $79,000. For model B, the cost is $69,900. For model C, the cost is $124,600.

(b)

To find how much of each kind of material to order, total each column of matrix MQ. Write this as a row matrix, T  5500 200 2000 300 Thus, 5500 units of concrete, 200 units of lumber, 2000 units of brick and 300 units of shingles must be ordered.

(c)

To find the total cost of the materials, find TR.  20  180 TR   60  5500 200 2000 300   20 5500  180  200  60  2000  25 300   273, 500    25  The total cost of the materials is $273,500.

Section 5.8 Matrix Inverses Classroom Example 1 (page 618)

Algebraic solution:  3 5 8 1 0 0   B 2 0 6 and I 3  0 1 0  4 0 0 1 1 7 

1 0 0  3 5 8 13  0 2  0  4 1  5  0 0  0 1 1 8  0 6  0  7     I 3 B  0 1 0  2 0 6  0 3  1 2  0  4 0  5  10  0 1 0 8  16  0  7  0 0 1  4 1 7  0 3  0  2  1 4 0  5  0 0  11 0 8  0 6  1 7   3 0 0 5  0  0 8 8  0  0   3 5      020 000 0  6  0    2 0 6  B 0  0   4 0  0  1 0  0   7   4 1 7 

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78

Chapter 5 Systems and Matrices

(continued) Graphing calculator solution:

Classroom Example 2 (page 620)  4 2 0 Find B 1 if B   1 1 2  0 1 4 Algebraic solution:

 4 2 0 1 0 0 Write the augmented matrix  B I 3  :  1 1 2 0 1 0 .  0 1 4 0 0 1  1  12 0  14 0 0  14 R1  1  12 0  14 0 0     1 1 0 1 0  0 2  4 1 0 R1  R2   1 1 2 2 0 0 1 4 0 0 1 1 4 0 0 1   1  1  1  12 0  14 0 0 0  14 0 0 2     1 1 4  12 2 0 2R2  2  14 1 0  0 0 2 1 1 1 1 0 0 8  12 2 1 2R2  R3 0 0 1 16  R3 4 8 8 

1  1 0  1 2 4  1 0  14 0 1 0 0 1 16 

1 0 0  3 0 0 8  1 1 2  R2  4R3  0 1 0  14 1 1 1 0 0 1 4 8 16 

 3  8 Thus, B 1    14  1  16

 12 1 1 4

 12 1 1 4

1 R1  12 R2 4 1 2 1 8

1 4 1 .

2 1 8

Graphing calculator solution:

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Section 5.8 Matrix Inverses

79

Classroom Example 3 (page 622) 5  4 2 1 0 .

Find A1 given A   0  8 4 10 5 1 0 0  4 2  4 2 5 1 0 0     .

Write the augmented matrix  A I 3  :  0 1 0 0 1 0   0 1 0 0 1 0  8 4 10 0 0 1  0 0 0 2 0 1 2R1  R3

Because there is no way to convert the third element in the third row to a 1, A1 does not exist.

Classroom Example 4 (page 623) (a)

5 x  2 y  1 2 x  3 y  15 Write the system in matrix form:  5 2  x    1 , where A   5 2 , X   x  , and B   1 . Now find A1 .  A I    5 2 1 0 , 2  2 3 0 1   2 3  2 3  y  15  y  15   so we have 1 1 1 2  1 52  1 52 15 0 15 R1 0 0 5 5   115     5  5 R2  2 2  2 3 0 1 0 1  11 11  11 0 5  5 1 2R1  R2

3 2 1 0  11 R1  52 R2 11   5 2  0 1  11 11   3 2   11 Thus, A1   11 .

5 2  11    11

Now find X  A1B :

 

3 30   3  2   1  11 1   112 15   113  11  3 11   X   11       5   75  2 2 5 2 11    11  15   11  1  11 15  11  11   7 

Solution set:  3, 7 

Graphing calculator solution:

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80

(b)

Chapter 5 Systems and Matrices

4 x  2 y  12 x  y  2z  7 y  4 z  20 Write the system in matrix form:  4 2 0  x  12  4 2 0  x   12  1 1 2  y    7  , where A   1 1 2 , X   y  , and B   7  . Now find A1 .             0 1 4  z   20  0 1 4  z   20  4 2 0 1 0 0  A I 3    1 1 2 0 1 0    0 1 4 0 0 1  1  12 0  14 0 0  14 R1  1  12 0    1 1  1 2 0 1 0  0  2    2 0 0 1 4 0 0 1 1 4   1 1 1  2 0  4  1  12 0 0    2 0  12 2 1 2R2  R3  0 1 0 0   1 4 0 0 1 1  0 1 0 0  3 8  1 0 1 0  4 1  4 0 0 4

 12 1 1

1 R1  12 R2  1 4   1  0 2 1 R3  R2 0 2

 3  1 2  8 Thus A1    14 1  1 1  16 4

 14 0 0  1  4 1 0 R1  R2  0 0 1 0  14 0 0  1 0  4 1 12  12 R2  4 0 0 1

0 0  83 1 0  14 0 1 161

 12 1 1 4

1 4 1 2 1 1 R3 8 4

1 4 1  2 1 8 

Now find X  A1B :   3  1 1   12   3 12  1 7   1  20   9  7  5  3 2 4 2 4  8   8   2 2  X    14 1 12   7     14 12  17   12  20   3  7  10   0 1 1  1  1 20  16 12  14 7  18 20  34  74  52   5 4 8   16 Solution set:

3, 0, 5

Graphing calculator solution:

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Chapter 6 Analytic Geometry x

y

Classroom Example 1 (page 645)

6

0

x  4    y  2

0

2

−2

4

0

6

6

8

Section 6.1

Parabolas 2

The vertex is (4, −2). The graph opens to the left because a = −1. It is a reflection of the graph of x  y 2 across the y-axis and has a translation 4 units

to the right and 2 units down. The domain is , 4 .

The range is  ,   . The graph is symmetric about

its axis, the horizontal line y = −2. Use the vertex and axis and plot a few additional points.

Graphing calculator solution: Write two equations by solving for y using the equation found in the algebraic solution.

x

y

x  2  12  y  4  2 x  4   y  4 

−5

1

 2x  4  y  4  4  2x  4  y

0

0

4

−2

Graph y1  4  2 x  4 and y2  4  2 x  4 in the window [–3, 9] × [–3, 9]. Their union is the graph of

0

−4

x  12 y 2  4 y  6 .

−5

−5

2

Classroom Example 2 (page 646)

y1

x  12 y 2  4 y  6

y2

Algebraic solution: Complete the square on y to find the vertex and the axis. x  12 y 2  4 y  6

Classroom Example 3 (page 647)

  2 1  2  y  8 y  16  16   6  12 y 2  8 y 

6

 12  y  4  12 16  6

(a)

2

 12  y  4  2  x  2  12  y  4 2

2

2

The vertex is (−2, 4). The graph opens to the right and has the same shape as x  12 y 2 , translated 2 units to the left and 4 units up. The domain is  2,   and

The equation x 2  4 y has the form x 2  4 py , so 4p = −4, from which p = −1.

The x-term is squared, so the parabola is vertical, with focus 0, p   0, 1 and

directrix, y   p, is y = 1 . The vertex is (0, 0), and the axis of the parabola is the y-axis, x = 0. Use the vertex and axis, and plot a few additional points.

the range is  ,   . The graph is symmetric about

x

y

its axis, the horizontal line y = 4. Use the vertex and axis, and plot a few additional points.

−4

−4

−2

−1

0

0

2

−1

4

−4

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81

82

Chapter 6 Analytic Geometry

(b)

The equation y 2  12 x has the form

x

y

y 2  4 px , so 4 p  12, from which p  3.

Because the y-term is squared, the parabola is horizontal, with focus  p, 0  3, 0 and

−6

−1

−4

2

−2

3

0

2

4

−6

directrix, x   p, is x  3. The vertex is (0, 0), and the axis of the parabola is the x-axis, y = 0. Use the vertex and axis, and plot a few additional points. x

y

3

−6

1

2 3  3.5

0

0

1

2 3  3.5

3

6

Classroom Example 6 (page 649) (a)

(b)

Classroom Example 4 (page 648) (a)

 

A parabola with focus 0, 34 and vertex at the origin is a vertical parabola. The equation has the form x 2  4 py . Because p  34 is positive, it opens upward. Substituting 34 for p, we find that an equation for this parabola is



x 2  4 34 y  x 2  3 y or y  13 x 2 .

(b)

A parabola with a horizontal axis, vertex at the origin, and passing through (−4, −8) is a horizontal parabola with equation of the form y 2  4 px . Use this equation with the coordinates of the point to solve for p:

82  4 p 4  64  16 p  4  p Thus, an equation of the parabola is y 2  16 x .

Classroom Example 5 (page 648)

The vertex is (−2, 3) and the focus is (−2, 2) The focus is below the vertex, so the axis is vertical and the parabola opens downward. The distance between the vertex and the focus is 3 − 2 = 1. The parabola opens downward, so choose p = −1. The equation will have the form

11, 025  86.1 and the parabola 128 is vertical, opening upward, the equation of the directrix is y  86.1 .

Because p 

Let y = 25, then solve for x: 11, 025 x2   25  x  92.8 32 Thus, the width of the dish is 2 92.8  185.6 ft.

Section 6.2 Ellipses Classroom Example 1 (page 656) (a)

4 x 2  25 y 2  100 Rewriting the equation, we have x2 y2   1.

52 2 2 The center is (0, 0). Because 5 > 2, we have a = 5 and b = 2, and the major axis is horizontal. The vertices are (–5, 0) and (5, 0).

The endpoints of the minor axis are (0, –2) and (0, 2). The domain is [–5, 5]. The range is [–2, 2]. To find the foci, we need to find c such that c 2  a 2  b 2 . c 2  a 2  b 2  c 2  25  4  c 2  21  c  21 The major axis lies on the x-axis, so the foci are ( 21, 0) and ( 21, 0).

 x  h 2  4 p  y  k  . Substitute p = −1, h = −2, and k = 3 to find the required equation.

 x  h 2  4 p  y  k   x  (2)2  4 1 y  3   x  22  4  y  3 The domain is  ,   and the range is  , 3 . Use the vertex and axis, and plot a few additional points.

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Section 6.2 Ellipses

(b)

49 x 2  9 y 2  441 Rewriting the equation, we have x2 y 2 x2 y2   1 or 2  2  1. The center is 9 49 3 7 (0, 0). Because 7 > 3, we have a = 7 and b = 3, and the major axis is vertical. The vertices are (0, –7) and (0, 7). The endpoints of the minor axis are (–3, 0) and (3, 0). The domain is [–3, 3]. The range is [–7, 7]. To find the foci, we need to find c such that c 2  a 2  b 2 .

83

y x2   1  , the only possible values of y 3 4 y are those making  0  y  0. The domain is 3 [−2, 2] and the range is [−3, 0].

Because

c 2  a 2  b 2  c 2  49  9  c 2  40  c  40  2 10 The major axis lies on the y-axis, so the foci









are 0, 2 10 and 0, 2 10 .

Classroom Example 4 (page 658)

 x  32  y 2  1 36

9

Rewriting the equation, we have

Classroom Example 2 (page 657)

Major axis with length 18, center (0, 0), and foci (−5, 0) and (5, 0) The length of the major axis is 2a. Thus, we have 2a  18  a  9. From the foci, we have c = 5.

Solving for b 2 we have c 2  a 2  b 2  25  92  b 2  b 2  56.

The foci are on the y-axis and the ellipse is centered at 2

the origin, so the equation has the form x2  b

y2 a

2  1.

62 32 The center is (3, 0). We have a = 6 and b = 3.

Because a = 6 is associated with x 2 , the major axis is horizontal. The vertices are on the vertical line through (3, 0). The vertices are 6 units to the right and 6 units to the left of the center at (−3, 0) and (9, 0). The endpoints of the minor axis are on the horizontal line through (3, 0). The endpoints of the minor axis are 3 units below and 3 units above the center at (3, 3) and (3, −3). The domain is [–3, 9].

The range is [–3, 3]. To find the foci, we need to find c such that c 2  a 2  b 2 . c 2  a 2  b 2  c 2  36  9  c 2  27  c  27  3 3 The major axis lies on y = 0, so the foci are

3  3 3, 0 and 3  3 3, 0 .

2

2

Because a  9  a  81 and and b  56, the equation of the ellipse is

 x  32  y 2  1

x2 y2   1.

81 56

Classroom Example 3 (page 657) y x2   1 3 4 y2 x2 x2 y2  1    1, 9 4 4 9 which is the equation of an ellipse centered at the origin with y-intercepts 3 (vertices) and x-intercepts 2 (endpoints of the minor axis).

Square both sides to get

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84

Chapter 6 Analytic Geometry

Classroom Example 5 (page 659)

Also c  a 2  b 2  11  6  5. Thus,

4 x 2  24 x  y 2  8 y  12 To write the equation in standard form, first group the terms containing x and group the terms containing y.

4 x  24 x   y  8 y   12 2

2

Now factor out 4 from the first grouping.



 



4 x 2  6 x  y 2  8 y  12

Complete the square for both x and y.



 



4 x 2  6 x  9  9  y 2  8 y  16  16  12

Apply the distributive property.









4 x 2  6 x  9  36  y 2  8 y  16  16  12

Factor and combine constant terms.

4  x  3   y  4  64 2

2

e

c 5   0.67 a 11

Classroom Example 7 (page 661)

The maximum distance is a + c and the minimum distance is a − c. a  c  507.4  c  507.4  a c Using e  , we have a 507.4  a  0.0489  507.4  a  0.0489a  a 1.0489a  507.4  a  483.75 Thus, c  507.4  483.75  23.65 and a  c  483.7  23.65  460.1 The closest distance Jupiter comes to the sun is about 460.1 million miles.

Divide by 64, then simplify.

Classroom Example 8 (page 662)

4  x  3  y  4  64  64 64 64 2 2  x  3   y  4  1 16 64 This is the standard form of the equation of an ellipse with center is  3, 4 . The major axis is vertical

x2 y 2  1 40 24 The kidney stone and the source of the beam must be placed at the foci. a 2  40 and b 2  24 , so

2

2

because a 2  64 is associated with y 2 . The vertices are located a distance of a = 8 units directly above and below the center, at  3, 4 and 3, 12 .

Because b 2  16 is associated with x 2 , the endpoints of the minor axis are located a distance of b = 4 units to the right and left of the center, at 7, 4 and 1, 4 .

Classroom Example 6 (page 660) 2

(a)

2

x y  1 81 100 Because 100 > 81, a 2  100, which gives a  10. Also

c  a 2  b 2  100  81  19. Thus, e

(b)

c 19   0.44 a 10

x2 y2  1 11 6 Because 11 > 6, a 2  11, which gives a  11.

c  a 2  b 2  40  24  16  4 . Thus, the foci are at (−4, 0) and (4, 0), so the kidney stone and the source both must be placed on a line 4 units from the center.

Section 6.3 Hyperbolas Classroom Example 1 (page 669) x2 y 2  1 4 25 Algebraic solution:

This equation may be written as 2

has the form x 2  a

y2 b2

x2 y 2   1 which 2 2 52

 1. The hyperbola is centered

at (0, 0) with branches opening to the left and right.

The graph has vertices at (– 2, 0) and (2, 0) (also x-intercepts  2). There are no y-intercepts. The domain is (– ∞, –2]  [2, ∞). The range is (– ∞, ∞).

The foci are on the x-axis. c 2  a 2  b 2  c 2  4  25  c 2  29  c  29, so the foci are

 29, 0 and  29, 0 . Because a = 2 and b = 5,

the asymptotes are y   ba x   52 x. The corners of the fundamental rectangle are (2, 5), (−2, 5), (−2, −5), and (2, −5). (continued on next page)

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Section 6.3 Hyperbolas

85

(continued)

Graphing calculator solution: 2

Classroom Example 3 (page 670)

y2

Solve x4  25  1 for y to get the equations of the two

functions y1  52 x 2  4 and y2   52 x 2  4.

2

y

2

 x  2 2  y  4 2  1   16

 x  h 2



 y  k 2

 1 where

Their union is the graph of x4  25  1. Graph both

The equation has the form

equations along with the asymptotes y3  52 x and

h = 2, k = −4, a = 4, and b = 1. The center is (2, −4).

The vertices are 4 units left and right of the center (2, −4). These points are (6, −4) and (−2, −4). The domain is ,  2   6,   . The range is ,   .

y4   52 x, in the window [–6.6, 6.6] × [–4.2, 4.2].

a

2

b2

Because c 2  a 2  b 2  c 2  16  1  c  17 , the

y1

foci are 17 units left and right of the center (2, −4).









Thus, the foci are 2  17 , 4 and 2  17 , 4 .

Because h = 2, k = −4, a = 4, b = 1, and y  k  ba  x  h  , the asymptotes are

y2 y3

y4

y  4  14  x  2 .

Classroom Example 2 (page 669) 16 y 2  25 x 2  400 This equation may be written as y2 x2 y2 x2   1  2  2  1, which has the form 25 16 5 4 y2 a2

2

 x2  1. The hyperbola is centered at (0, 0) with b

branches opening upward and downward. The graph has vertices and y-intercepts at (0,– 5) and (0, 5).

There are no x-intercepts. The domain is (–∞, ∞).

The range is  , 5  5,   . The foci are on the y-axis. c 2  a 2  b 2 , so we have c 2  a 2  b 2  c 2  25  16 c 2  41  c  41







The foci are 0,  41 and 0,



41 . Because a = 5

and b = 4, the asymptotes are y   ba x   54 x.

Classroom Example 4 (page 671) 9 y 2  90 y  25 x 2  150 x  900 To write the equation in standard form, first group the terms containing x and group the terms containing y.

9 y  90 y   25x  150 x  900 2

2

Now factor out 9 from the first grouping and 25 from the second grouping.









9 y 2  10 y  25 x 2  6 x  900

Complete the square for both x and y.









9 y 2  10 y  25  25  25 x 2  6 x  9  9  900

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86

Chapter 6 Analytic Geometry

Section 6.4 Summary of the Conic Sections

(continued) Apply the distributive property.









9 y 2  10 y  25  225  25 x 2  6 x  9  225  900

Classroom Example 1 (page 680)

Factor and combine constant terms.

(a)

9  y  5  25  x  3  900 2

2

 x  4 x  4   y  2 y  1  5  4  1 2

Divide by 900, then simplify.

2

 x  22   y  12  0

9  y  5 25  x  3 900   900 900 900  y  52   x  32  1 100 36 This is the standard form of the equation of a hyperbola with center 5, 3 . The transverse axis is 2

x 2  4 x  y 2  2 y  5 Complete the squares on x and y:

2

This is the equation of a circle centered at (2, −1) with radius 0, that is the point (2, −1).

vertical because the leading term of the hyperbola written in standard form contains the y 2‐term.

Classroom Example 5 (page 671)

(b)

y 2 x2  1 100 81 a 2  100  a  10 and c 2  a 2  b 2  2

c  100  81  c  181. Thus, e  ac 

The equation is of the form 181  1.3.

10

Foci at (–2, 5), (–2, –3); eccentricity 3 The foci lie on a vertical transverse axis, so the  y  k 2 a

2



 x  h 2 b2

 x  h 2 a2



 y  k 2 b2

1

with a  4, b  3, h  0, and k  0, so the graph of the given equation is an ellipse with center (0, 0), x-intercepts (−4, 0) and (4, 0), and y-intercepts (0, −3) and (0, 3).

Classroom Example 6 (page 672)

equation is of the form

9 x 2  144  16 y 2  9 x 2  16 y 2  144  x2 y2  1 16 9

 1. The center

is halfway between the foci:



( 2)  ( 2) 5  ( 3) , 2 2

   ,   2,1. The distance 4 2

2 2

from the center to each focus is 4, so c = 4. Use the eccentricity to find a: e  ac  3  a4  3a  4  a  43 . Now find the value of b 2 given

c 2  a 2  b 2 , a  43 , and c  4.

 43   b2  2

(c)

2 x  y 2  10 y  17  0  y 2  10 y  2 x  17 Complete the square on y: y 2  10 y  25  2 x  17  25 

16  169  b 2  b 2  16  16  128 9 9

 y  52  2 x  8   y  52  2  x  4  2  12  y  5  x  4

a  43  a 2  16 , b 2  128 , h  2, and k  1, so we 9 9

The equation is of the form x  h  a  y  k 

c 2  a 2  b 2  42 

have

 y  12   x  22  1 or 16 9 2

128 9 2

9  y  1 9  x  2   1.

16 128

2

with a   12 , h  4, and k  5, so the graph of the given equation is a parabola with vertex (4, 5) and horizontal axis y = 5. Use the vertex and axis and plot a few additional points. (continued on next page)

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Section 6.4 Summary of the Conic Sections

(continued)

(d)

87

Classroom Example 2 (page 681)

x

y

−4

1

2

3

4

5

2

7

−4

9



9 x 2  72 x  25 y 2  100 y  19



2

2

 x  42   y  22  1

25 9 This is the equation of an ellipse with center (4, 2), major axis endpoints (−1, 2) and (9, 2) and minor axis endpoints (4, −1) and (4, 5).







4 x 2  6 x  9  25 y 2  4 y  4  36  36  100 4  x  3  25  y  2  100 2

2

 y  22   x  32  1 4

25

The equation is of the form

 y  k 2 a

2



 x  h2 b2

1

with a  2, b  5, h  3, and k  2, so the graph of the given equation is a hyperbola with center (−3, −2), vertices (−3, 0) and (−3, −4), asymptotes y  2  52  x  3 , and





9  x  4  25  y  2  225

4 x 2  24 x  25 y 2  100 y  36 Complete the squares on x and y: 4 x 2  24 x  25 y 2  100 y  36





9 x 2  8 x  16  25 y 2  4 y  4  19  9 16  25  4







foci 3, 2  29 and 3, 2  29 .

Copyright © 2021 Pearson Education, Inc.

Chapter 7 Further Topics in Algebra Section 7.1

Sequences and Series

(b)

Classroom Example 1 (page 693) (a)

(b)

an  5n  2 Replace n with 1, 2, 3, 4, and 5: a1  5 1  2  3 a2  5  2  2  8 a3  5 3  2  13 a4  5  4  2  18 a5  5 5  2  23 an   1

n 1

a1  5, an  an 1  2n, for n > 1 This is an example of a recursive definition.

We know a1  5. Because an  an 1  2n,

n  2 : a2  a2 1  2  2  a1  4  5  4  9 n  3 : a3  a3 1  2 3  a2  6  9  6  15 n  4 : a4  a4 1  2  4  a3  8  15  8  23

The first four terms are 5, 9, 15, and 23.

Classroom Example 3 (page 695) a1  5; an  2.6an 1  0.25an 12 , for n  1

(a)

a1  5 thousand per acre

 n2

a2  2.6a1  0.25a12  2.6 5  0.25 5  6.75 thousand per acre

Replace n with 1, 2, 3, 4, and 5: a1   1

11

 12  1

a2   1

2 1

 2  4

3 1

 32  9

a3   1

a4   1

 42  16

5 1

 52  25

a5   1

(c)

a3  2.6a2  0.25a2 2  2.6 6.75  0.25 6.75  6.16 thousand per acre

2

2

4 1

2

a4  2.6a3  0.25a32  2.6 6.16  0.25 6.16  6.53 thousand per acre

a5  2.6a4  0.25a4 2  2.6 6.53  0.25 6.53  6.32 thousand per acre

3n  2 n2  3 Replace n with 1, 2, 3, 4, and 5: 3 1  2 5  a1  2 4 1 3 3  2  2 8  a2  2 7 2 3 3 3  2 11  a3  2 12 3 3 3  4  2 14  a4  2 19 4 3 3 5  2 17  a5  2 28 5 3

a6  2.6a5  0.25a52  2.6 6.32  0.25 6.32  6.45 thousand per acre

an 

(b)

As n increases, the population density oscillates above and below 6.4 thousand insects per acre. It converges to near 6.4 thousand per acre. The sequence is graphed below in the window [0, 10] × [0, 10].

Classroom Example 2 (page 694) (a)

a1  2, an  3  an 1  1, for n > 1 This is an example of a recursive definition. We know a1  2. Because an  3  an 1  1, n  2 : a2  3  a2 1  1  3  a1  1  3  2  1  5 n  3 : a3  3  a3 1  1  3  a2  1  3  5  1  14 n  4 : a4  3  a4 1  1  3  a3  1  3  14  1  41 The first four terms are 2, 5, 14, and 41.

Classroom Example 4 (page 696) Algebraic solution: 5

  j  j  1  1  1  1   2  2  1  3  3  1   4  4  1  5  5  1 2

j 1

2

2

2

2

2

 3  7  13  21  31  75

(continued on next page) 88

2

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2

2

Section 7.2 Arithmetic Sequences and Series

(continued)

89

Classroom Example 7 (page 699)

Graphing calculator solution:

5

5

5

5

i 1 5

i 1 5

i 1 2

i 1

 i  2i  8   i   2i   8 3

i 1

i 1 5



3

 i  2 i   8 3

i 1

52 5  1 5 5  1   2  5 8 4 2  225  30  40  215

Section 7.2

Classroom Example 5 (page 697) 5

(a)

b  b  b  b  b  b

Classroom Example 1 (page 704)

4

20, 13, 6, −1, −8, … d  a2  a1  13  20  7

i

1

2

3

4

5

i 1

(b)

 2 x  5, if x  3, x  5, x  7, x  9 1

j

2

3

4

j 1 4

  2 x  5 j

Classroom Example 2 (page 704) (a)

a1  14 and d = 6 Starting with a1  14 , add d = 6 to each term to get the next term. a2  14  6  8; a3  8  6  2 a4  2  6  4; a5  4  6  10 The first five terms are −14, −8, −2, 4, and 10.

(b)

a1  1.5 and d = −0.5

j 1

  2 x1  5   2 x2  5   2 x3  5   2 x4  5   2 3  5   2 5  5   2 7   5  1  5  9  13  28

  2 9  5

3

(c)

1 f  xk  x, if f  x   , x1  1, x2  3, x3  5, x k 1 and x  4

 3

 f  x  x k

k 1

 f  x1  x  f  x2  x  f  x3  x 1 1 1 4 4 92  4 4 4  4   1 3 5 3 5 15

50



23, 20, 17, 14, … d  a2  a1  20  23  3 an  a1   n  1 d , so

12  50 12  600

a16  23  16  1 3  23  15  3  22

i 1

(b)

35

 10

(c)

an  23   n  1 3  23  3n  3  3n  26

35 35  1  1890 3i  3 i  3  2 i 1 i 1 35



10

10

10

i 1

i 1

Classroom Example 4 (page 705) 10

 3i  5   3i    5  3 i   5 i 1

2

i 1

2

Starting with a1  1.5 , add d = −0.5 to each term to get the next term. a2  1.5  0.5  1.0; a3  1.0  0.5  0.5 a4  0.5  0.5  0; a5  0  0.5  0.5 The first five terms are 1.5, 1.0, 0.5, 0, and −0.5.

Classroom Example 3 (page 705)

Classroom Example 6 (page 698) (a)

Arithmetic Sequences and Series

2

i 1

10 10  1 2  10  1  3  10 5 6  1205

a4  18, a5  22

First find d: d  a5  a4  22  18  4.

Now find a1 : an  a1   n  1 d 

a5  a1  5  1 4  22  a1  16  6  a1

Therefore, a10  a1  10  1 d  a10  6  9  4  a10  42.

an  6   n  1 4  6  4n  4  4n  2.

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90

Chapter 7 Further Topics in Algebra

Sn  n2  2a1   n  1 d  , we have

Classroom Example 5 (page 705) a6  9, a15  36

12

 5i  10  S  2 15  7 5

a15  a6  9d , so 36  9  9d 

8

i 5

45  9d  d  5 a6  a1  6  1 d  9  a1  5 5  (b)

Classroom Example 6 (page 707)



were connected, they would form a line with slope y y

m  x2  x1  32 11  2. Using the point-slope form of a 1

line, we have y  y1  m  x  x1  

y  1  2  x  1  y  1  2 x  2  y  2 x  1

Thus, the nth term of the sequence is determined by an  2n  1. The domain is {1, 2, 3, 4, 5} and the range is {1, 3, 5, 7, 9}.

Classroom Example 7 (page 708) (a)

48, 44, 40, 36, … d  a2  a1  44  48  4

n  1 d  , so  2 48  21  1 4  168 S21  21 2     Sn  n2  2a1 

(b)

 8  2k  k 5

The sequence is comprised of the points 1, 1 , 2, 3 , 3, 5 , 4, 7 , 5, 9 . If the points

2

 82 30  35  4 65  260

10

9  a1  25  a1  34



8 2

1, 2, 3, 4, … d=1 Sn  n2  2a1   n  1 d  , so

The first few terms of the series are 8  2 5  8  2 6  8  2 7   

 2   4   6   Thus, this is the sum

of an arithmetic sequence with d  2 and a1  2. If i started at 1, there would be 10 terms. Four terms are missing, so n = 6. Using the formula Sn  n2  2a1   n  1 d  , we have 10

 8  2k   S  2 2  5 2 6

k 5

Section 7.3

6 2

 62  4  10  3  14  42

Geometric Sequences and Series

Classroom Example 1 (page 713) a1  50, r  2 . Find a12 . a12  a1r12 1  a12  50  211  50  2048  102, 400 You will receive $102,400.

 2 1  200  11  20,100 S200  200 2    

Classroom Example 8 (page 708) S15  345, a15  65

S15  152  a1  a15   345  152  a1  65  46  a1  65  a1  19

Classroom Example 2 (page 714)

6400, 1600, 400, 100, … a 1600 1 r 2   a1 6400 4 4

1  1  a5  a1r 5 1  a5  6400    6400   25 4  256 

a15  a1  15  1 d  65  19  14d  84  14d  d  6

1 an  a1r n 1  an  6400   4

Classroom Example 9 (page 709)

Classroom Example 3 (page 714)

(a)

12

a2  18, a5  486

i 5

a2  a1r1  18  a1r  r 

 5i  10 The first few terms of the series are 5 5  10  5 6  10  5 7   10  

n 1

18 a1

a5  a1r 4  486  a1r 4

 15  20  25   Thus, this is the sum of an arithmetic sequence with d  5 and a1  15. If i starts at 1, there will be 12 terms. Four terms are missing, so n = 8. Using the formula

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(continued on next page)

Section 7.3 Geometric Sequences and Series

(continued)

91

The graph of the first six values of Sn in the window [0, 6] × [0, 10] suggests that its value is approaching 5.

4

 18   Substituting, we have 486  a1   a1 

184  a 3  184  a 3  216  a  6

486 

1

a13

1

486

1

Now find r: a5  a2 r 3  486  18r 3  r3 

486  27  r  3 18

Classroom Example 4 (page 714)

Classroom Example 8 (page 718)

a1  250, r  1.75 a6  a1r

6 1



(a)

 a6  250 1.75  4103 5

2 1 and r   .

3 5 Because 1  r  1, this series converges. We a 25 25 3 have S  1    .

1  r 1    1 3 4 3 10

For this geometric series, a1 

Classroom Example 5 (page 715)

S12 



a1 1  r12

  S  50 1  2   204, 750 12

12

1 r 1 2 You will receive a total of $204,750.



(b)

Classroom Example 6 (page 716) 8

i

i 1

This series is the sum of the first 8 terms of a geometric sequence having a1  4  51  20 and r = 5.



a1 1  r 8 1 r

  S  20 1  5   1, 953,120 8

8

1 5

Classroom Example 7 (page 717) 4 4   5 25 a 45 1 r 2   and a1  4. a1 4 5

Evaluate 4 

Because Sn 



a1 1  r n 1 r

 , we have



 0.9

j

j 1

45

S8 



For this geometric series, a1  0.9 and r  0.9. Because 1  r  1, this series converges. We have a 0.9 0.9 S  1    9.

1  r 1  0.9 0.1 Classroom Example 9 (page 718)

The future value of an annuity uses the formula a (1  r n ) Sn  1 , where r = 1 + interest rate.

1 r The payments are $3500 for 6 years at 3% compounded annually, so a1  3500 , r = 1.03, and n = 6.

6 3500 1  1.03     22, 639.43 S6  1  1.03 The future value is $22,639.43.

n n 4 1  15  4 1  15      Sn   .

1 4 1 5 5

   0. Thus, we obtain

lim 1 n  5

n



n 4 1  15    4 1  0 4 lim Sn  lim   4  5.

4 4

n 

n 

5

5

i 1

i 1

There are 4103 insects in the sixth generation. a1  50, r  2

2 1

  5    3 

5

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92

Chapter 7 Further Topics in Algebra

Section 7.4

The Binomial Theorem

Classroom Example 1 (page 727) (a)

9!

9  8  7  6  5  4  3  2 1 9  8  7  6 9    126  5   5!4! 5  4  3  2  1  4  3  2  1 4  3  2  1

(b)

25!

25!

25!  25    1  25   25! 25  25! 25!0! 25! 1

(c)

16!

16! 16   1  0   0!16  0! 1  16!

(d)

15 C8 

15!

15! 15  14  13  12  11  10  9    6435 8!15  8! 8!7!

7  6  5  4  3  2 1

Classroom Example 2 (page 729)

 x  y 8  x8  18  x7 y   82  x6 y 2   83  x5 y3   84  x 4 y 4   85  x3 y 5   86  x 2 y 6   87  xy 7  y8 8! 7 8! 6 2 8! 5 3 8! 4 4 8! 3 5 8! 2 6 8! x y x y  x y  x y  x y  x y  xy 7  y8 1!7!

2!6!

3!5!

4!4!

5!3!

6!2!

7!1!

87 6 2 876 5 3 8765 4 4 876 3 5 87 2 6  x8  8 x 7 y  x y  x y  x y  x y  8 xy 7  y8 x y  2 1 3  2 1 4  3  2 1 3  2 1 2 1  x8  8 x 7 y  28 x 6 y 2  56 x5 y 3  70 x 4 y 4  56 x3 y 5  28 x 2 y 6  8 xy 7  y8  x8 

Classroom Example 3 (page 729) 4

4

m  m    n       n  3 3  4 3 2  m   4  m  2 3 4 4  m  4  m           n       n       n   n 2  3  3   3   3  1   3  m4 4! m3n 4! m 2 n 2 4! mn3         n4 81 1!3! 27 2!2! 9 3!1! 3 m4 4 3 4  3 m2n2 4 3 m n     mn  n 4 81 27 2 1 9 3 4 3 2 2 2 4 3 m4 m n  m n  mn  n 4   81 27 3 3

Classroom Example 4 (page 730) 3

3

2

2

2 4   2   3  2   4   3   2   4   4   3     3   1   3         r r r r r   2  r3   r   r  8 3! 4 4 3! 2 16 64  9   6   3   32 1!2!

2!1! r r r r r r 8 48 96 64 8 48 96 64  9  6  4  3 2  9  13 2  4  3 2 r r r r r r r r r

3

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Section 7.5 Mathematical Induction

93

Classroom Example 5 (page 730)

2c  d 12  n  n  ( k 1) k 1 x y with n = 12, k = 4, k – 1 = 3, and n – (k – 1) = 9. The fourth term of the Use the formula   k  1



 





12! 9 9 12  11  10 9 3 12  expansion is    2c    d   2 c d 3  29 c 9 d 3  112, 640c9 d 3 3 3!9!

3  2 1

Section 7.5

Mathematical Induction

Classroom Example 1 (page 734)

Let S n be the statement 2  5  8    3n  1 

n 3n  1 .

2

Prove that Sn is true for every positive integer n.

Step 1: Verify the statement for n  1.

13  1  1 S1 is the statement 2   2  2, which is true.

2 Step 2: Show that Sk implies Sk+1, where Sk  2  5  8    3k  1  Sk 1  2  5  8    3k  1  3  k  1  1 

k  1 3 k  1  1

k 3k  1 , and 2

.

2 Start with Sk and assume it is a true statement. Add the k  1 st term, 3  k  1  1 , to both sides of this equation.

k 3k  1 2 k 3k  1 2  5  8    3k  1  3  k  1  1   3  k  1  1 2 k 3k  1  2 3  k  1  1  2 3k 2  7k  4  k  13k  4   2 2 k  1 3 k  1  1  2 This final result is the statement for n = k + 1. Thus, if Sk is true, then Sk+1 is also true. Therefore, the statement Sn is true for every positive integer value of n.

2  5  8    3k  1 

Classroom Example 2 (page 735) n

an a Let S n be the statement    n , assuming a and b are constants, with b  0. b b 1

a1 a a Step 1: Show that the statement is true when n = 1: S1 is the statement if b  0,    1  which is true. b b b k

ak a Step 2: Show that S k implies Sk 1 , where Sk is the statement: If b  0 then    k b b

a Sk 1 is the statement: If b  0 then   b

k 1



a k 1 b k 1

(continued on next page)

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94

Chapter 7 Further Topics in Algebra

(continued) k

k 1

a k 1 b k 1 This final result is the statement for n = k + 1. It has been shown that if Sk is true, then Sk 1 is true. Therefore, the k a a a a a If b  0 , then     k     b b b b b



statement S n is true for every positive integer value of n.

Classroom Example 3 (page 736)

Let S n be the statement 4n  4n .

Step 1: Verify that S n is false for n = 1. Then show that the statement is true when n = 2: S2 is the statement 42  4  2  16  8, which is true.

Step 2: Show that S k implies Sk 1 , where Sk is the statement 4k  4k , and Sk 1 is the statement 4k 1  4  k  1 .

4k  4k  4k  4  4k  4  4k  41  16k  4k 1  16k  4k 1  4k  12k Because k > 2, we have 12k  24  12k  4 and 4k 1  4k  12k  4k  4  4k 1  4k  4  4k 1  4  k  1

It has been shown that Sk implies Sk 1 . Therefore, the statement Sn is true for every positive integer value of n greater than or equal to 2.

Section 7.6

Counting Theory

Classroom Example 1 (page 740)

Using the fundamental principle of counting, there are 6  4  3  72 possible systems.

Classroom Example 2 (page 740)

Using the fundamental principle of counting, there are 9  8  7  6  5  4  3  2  1  362,880 arrangements.

Classroom Example 3 (page 741)

Using the fundamental principle of counting, there are 9  8  7  6  5  4  60, 480 arrangements.

Classroom Example 4 (page 742) (a)

(b)

Use the formula for P(n, r) with n = 5 and r = 5.

5!

5!

P 5, 5   5  5! 0!

5  4  3  2 1   120 1 The digits can be arranged in 120 ways.

Use the formula for P(n, r) with n = 5 and r = 4.

5!

5! 5  4  3  2  1    120 P 5, 4  1 5  4! 1!

Four of the five digits can be arranged in 120 ways.

Classroom Example 5 (page 742)

The number of ways the 4 officers can be chosen from the 25 members is 25!

25!   25  24  23  22  303, 600 .

P (25, 4)  25  4! 21!

Classroom Example 6 (page 743)

The number of ways the 10 billboards can be installed in the 10 locations is 10!

10!

P 10,10    3, 628,800 . 10  10! 1!

Classroom Example 7 (page 744)

Order is not important, so use combinations.  22  22!

22!

C  22, 4       4   22  4!4! 18!4!

22  21  20  19   7315 4  3  2 1 There are 7315 ways to choose the 4 members.

Classroom Example 8 (page 744) (a)

Order is not important, so use combinations.

There are 28 total employees.  28  28!

28!

C  28, 5       5   28  5!5! 23!5!

28  27  26  25  24   98, 280 5  4  3  2 1 The committee members can be chosen in 98,280 ways.

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Section 7.7 Basics of Probability

(b)

If three surveyors and two engineers must be included on the committee, then the surveyors can be chosen in C(10, 3) ways and the engineers can be chosen in C(18, 2) ways.

10!

10! 10  C 10, 3       3  10  3!3! 7!3!

10  9  8   120 3  2 1 18!

18! 18  C 18, 2       2  18  2!2! 16!2!

18  17   153 2 1 The committee members can be chosen in 120  153  18, 360 ways.

Classroom Example 9 (page 745) (a)

Order is not important; use combinations.

(b)

Order is important; use permutations.

(c)

Order is important; use permutations.

(d)

Order is not important; use combinations.

Section 7.7

Classroom Example 3 (page 754)

Let E represent “a yellow marble being drawn.”

There are 18 marbles, so n  S   18 . Then P E 

4 5  . The odds in favor of 9 9 drawing a yellow marble are 4

P( E ) 4  9  or 4 to 5.

P( E ) 59 5

Classroom Example 4 (page 755) (a)

The events “the card is red” and “the card is a king” are not mutually exclusive because it is possible to draw a king of hearts or a king of diamonds, two outcomes satisfying both events. The probability is P  red or king   P  red   P  king   P  red and king  26 4 2 28 7      52 52 52 52 13

(b)

The events “the card is a club” and “the card is a diamond” are mutually exclusive because it is not possible to draw a card that is both a club and a diamond. The probability is P club or diamond   P club   P diamond   P club and diamond  13 13 26 1   0  52 52 52 2

For parts (a)−(d), n  S   6. (a)

E1  3, 6 , so n  E1   2. Thus, P  E1  

(b)

2 1  .

6 3

E2  1, 2, 3, 4, 5 , so n  E2   5. Thus,

P  E2  

5 .

6

Classroom Example 5 (page 756) 0  0.

6

(c)

E3  , so n  E3   0. Thus, P  E3  

(d)

E4  1, 2, 3, 4, 5, 6 , so n  E4   6. Thus, P  E4  

6  1.

6

Classroom Example 2 (page 753)

There are 52 cards in a deck and 13 of the cards are n  E  13 1   . hearts, so P  E   n  S  52 4 P E   1  P E   1 

n E  8 4   and n  S  18 9

P E   1  P E   1 

Basics of Probability

Classroom Example 1 (page 752)

95

1 3  4 4

(a)

Let A represent the event “the first die shows a 5,” and let B represent the event “the sum of the two dice is 11 or 12.” Event A has 6 elements, (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), and (5, 6), where (a, b) represents “the first die shows a and the second die shows b.”

Event B has 3 elements, (5, 6), (6, 5), and 6 3 , P B  , and (6, 6). Thus, P  A  36 36 1 . So, P  A  B  36 P  A  B   P  A  P  B   P  A  B  6 3 1 8 2      36 36 36 36 9

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96

(b)

Chapter 7 Further Topics in Algebra

The “sum of the dots showing is at least a 9” can be written as “the sum of the dots showing is 9, 10, 11, or 12.” The events represented by “9”, “10”, “11”, or “12” are mutually exclusive, so P at least 9  P 9 or 10 or 11 or 12  P 9  P 10  P 11  P 12 There are 4 ways to sum 9, (6, 3), (5, 4), 4 . There are 3 (4, 5), and (3, 6), so P 9  36 ways to sum 10, (6, 4), (5, 5), and (4, 6), so 3 P 10  . There are 2 ways to sum 11, 36 2 . There is one (5, 6) and (6, 5), so P 11  36 1 . Thus, way to sum 12, (6, 6), so P 12  36 4 3 2 1 10 5 P at least 9       36 36 36 36 36 18

Classroom Example 6 (page 758) (a)

The probability p of a 5 on one roll is 16 .

There are 12 rolls, so n = 12, and 3 successes, so r = 3. Thus, 3 12  3 12   1   1  P      1    0.197  3 6  6

(b)

The probability p of “not a 5” on one roll is 56 .

There are 12 rolls, so n = 12, and 8 successes, so r = 8. Thus, 8 12  8 12   5   5  P      1    0.089  8 6  6 Graphing calculator solution Use the binompdf( function in the DISTR menu.

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