PC Chapter 2 Solutions

8/10/2019 PC Chapter 2 Solutions

1/35

Chapter 2: Rocket Launch

Lesson 2.1.1.

2-1.

Domain: !" # x# " Range: 2 ! y ! " y-intercept! y =2

no x-intercepts

2-2.

a. Time Hours sitting Amount Earned

8PM 1 $4

9PM 2 $4*2hrs = $810PM 3 $4*3hrs = $12

11:30PM 4.5 $4*4.5hrs = $18

12:00 5 $4*5 = $20

12:30AM 5.5 $4*5+$6*!= $23

1AM 6 $4*5+$6*1 = $26

2AM 7 $4*5+$6*2 = $32

b. f(x) =4t 0 !t! 5

6(t" 5) + 20 t > 5{

2-3.a.

Forx"1 Forx> 1

x y = x2 +2 x y = 2x + 7

5 y = (!5)2 +2 =25 +2 =27 1 y = 2 !1+ 7 = 9

4 y = (!4)2 +2 = 16 +2 = 18 2 y = 2 !2 + 7 = 11

3 y = (!3)2 +2 =9 +2 = 11 3 y = 2 ! 3+ 7 = 13

2 y = (!2)2 +2 =4 + 2 =6 4 y = 2 ! 4 + 7 = 15

1 y = (!1)2 +2 = 1+2 = 3 5 y = 2 !5 + 7 = 17

0 y = (0)2 +2 =0 +2 =2 6 y = 2 !6 + 7 = 19

1 y =(1)2 +2 = 1+2 = 3 7 y = 2 ! 7 + 7 = 21

b. c. D =(!","); R = [2,") ;

Closed at (1, 3), open at (1, 9).


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2-5.

x =1

2-6.

a. D =(!","); R =(!", 5)

b. No, jump at x = 2 .

Review and Preview 2.1.1

2-7.

a. $8.00 per hour * 5 hours = $40.00Rate of pay = $8.00 per hour

b. Sample graph at right.c. 5

d. Each rectangle represents eight dollars ofKristofs earnings. ($8/hour) ! (hours) = $8

2-8.

f(x + 2) = 3(x + 2)2 + 5(x + 2) ! 2

= 3(x2 + 4x + 4)+ 5x + 10 ! 2

= 3x2 +12x + 12 + 5x + 10 ! 2

= 3x2 +17x + 20

2-9.

f(x + h) = 3(x + h)2+ 5(x + h)

!2

= 3(x2 + 2xh +h2 )+ 5x + 5h ! 2

= 3x2 + 6xh + 3h2 + 5x + 5h ! 2

= 3x2 + 6xh + 5x + 3h2 + 5h ! 2

= 3x2 +x(6h + 5) + (3h2 + 5h ! 2)

2-10.

a. (x+ 2)(x+ 2) ! 3(x+ 2)

= (x+ 2)(x+ 2 ! 3)

= (x+ 2)(x!1)

b. (x+ 2)(x+ 2)(x+ 2) ! 4(x+ 2)

= (x+ 2)(x2 + 4x+ 4! 4)

= (x+ 2)(x2 + 4x) = x(x+ 2)(x+ 4)

c. (x+ 2)(x+ 2) + 6(x+ 2)

= (x+ 2)(2(x+ 2) + 6)

= (x+ 2)(2x+ 4 + 6)

= (x+ 2)(2x+ 10)

= 2(x+ 2)(x+ 5)


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2-11.

a. 5x2 !15x = 0

5x(x! 3) = 0

Either 5x = 0 or x! 3 = 0

x = 0 or x = 3

b. 2(x! 6)2 = 18

(x! 6)2 = 9

x! 6 = 3

Either x! 6 = 3 or x! 6 =!3

x = 9 or x = 3

2-12.

a. (32 )! 3 2 = 3!3 = 133

= 127

b. (33 )2 3 =32 =9

c. 53

4 3( )

2 3

= 5

4( )3!

"#$

2 3

= 5

4( )2

= 25

16 d. 3!2( )

!1 2

= 31= 3

2-13.

a. d= 5 ! !2( )( )2+ !2 !5( )

2

= 72+ !7( )

2= 49 + 49 = 98 = 7 2

b. slope =!2 !5

5 ! !2( ) =

!7

7=!1

point-slopeform: y !5 =!(x + 2)

y +2 =!(x!5)

slope ! intercept form: y !5 =!x!2

y =!x!2 +5

y = !x + 3

2-14.

a. !

6

b. ! 3"

4

Lesson 2.1.2

2-15.

a. Shift to the left two units. b. q(x) =(x + 2)2 + (x + 2)!2

= x2+4x + 4 + x + 2 !2

= x2+5x + 4

2-16.g(x) should be shifted to the right one unit and up three units.


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2-17.

a and b. c. At x = 3

d. h(x) =(x!1)2 forx


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2-22.

See graph at right.

a. f(x) =x 3 !3x

g(x) = y =(x + 2)3!3(x + 2)

b. If f(x) = x 3 !3x

f(x + 2) =(x + 2)3 !3(x + 2)

g(x) = f(x + 2).

2-23.

f(x) =2 x!3 +1

a. f(x) = x

f(x) = x! 3 (right 3 units)

f(x) = 2 x!

3 (change slope to 2)

f(x) = 2 x! 3 +1 (up 1 unit)

b. f(x) =2 x!3 +1

f(x + 5) =2 x!3 +5 +1

f(x + 5)!3 =2 x + 2 +1!3

g(x) =2 x + 2 !2

c.

2-24.

y = x5!

2x

y = (x! 2)5 ! 2(x! 2) (right 2 units)

y = (x! 2)5 ! 2(x! 2)+3 (3 units up)

2-25.

a. The graphs are both cubic functions, but are in different places on the axes.b. The second graph is the first graph shifted to the left 2 units.

c. f(x) = x3

f(x +2) =(x +2)3 (shifted left 2 units)

2-26.

a. (2x+ 3)(2x+ 3)

=4x2 +6x+ 6x+ 9

=4x2 +12x+ 9

b. (x! a)(x! b)

= x2! ax! bx +ab

= x2! x(a + b)+ab


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7/35

2-34.

two coordinates =(1, 2.2) and (5, 3)

slope = 3!2.25!1

=0.8

4 =

810

"14 =0.2

point-slope

y !2.2 =0.2(x! 1) or

y !3 =0.2(x!5)

slope-intercept

y =0.2x!1+3

y =0.2x +2

2-35.2.2 +2.4 + 2.6 +2.8 +3.0 =

(1 !0.2 +2) + (2 !0.2 +2) + (3 !0.2 +2) + (4 !0.2 +2) + (5 !0.2 +2) = 0.2k+2

k=1

5

"

2-36.

a. 3.6 +4.0 +4.4 + 4.8 +5.2 =

3.6 + (1 !0.4 + 3.6)+ (2 !0.4 + 3.6)+ (3 !0.4 + 3.6)+ (4 !0.4 + 3.6) = 0.4k+3.6

k=0

4

"

b. Start the index at 1 and end it at 5.

2-37.

a. The difference between each term is 0.4 and the first term is 3.6.b. 0.4 ! 50 +3.6 = 20 +3.6 =23.6, so k=50 .

c. 0.4k+ 3.6k=0

50

!

d. 2.5 + (0.2 +2.5)+ (2 !0.2 +2.5)+ (3 !0.2 +2.5)+ (4 !0.2 +2.5)++ (100 !0.2 +2.5) =

0.2k+ 2.5

k=0

100

"


2-38.

y =! 2(x + 1)3 ! 5

y =!2((x + 4) +1)3 ! 5=!2(x + 5)3 ! 5 (left 4 units)

y =!2(x + 5)3 ! 5 ! 7 =! 2(x + 5)3 !12 (7 units down)

2-39.

4x + 2y = 7

2(3x ! y =5)"

6x! 2y = 10

4x + 2y = 7

10x = 17

x = 17

10 = 1.7

3(1.7)! y =5

5.1! y =5

!y =!0.1

y =0.1


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2-40.

a. x4 !81y4

=(x2 +9y2 )(x2 !9y2 )

=(x2+9y2 )(x +3y)(x !3y)

b. 8x3 + 2x7 =2 x3(4 + x4 )

2-41.

f(x) =5x ! 1"y = 5x !1

x =5y !1

x + 1=5y

x+15 =y " f!1(x) = x+1

5

2-42.

a. 12 +22 + 32 + 42 = 1+ 4 + 9 +16 = 30

b. (6(2)!22 )+ (6(3)!32 )+ (6(4)!42 )+ (6(5)!52 )+ (6(6)!62 ) =8 +9 +8 +5 +0 =30

c. 33 +34 + 35 +36 = 27 +81+243+ 729 = 1080

2-43.

12+2

2+3

2+ ...+9

2+10

2= k

2

k=1

10

!

2-44.

a. 523

= 52( )1/3

= 52/3 b. 24( )5

= 21/4( )

5

=25/4

c. c. 73 = 73( )1/2

= 73/ 2 d. 7( )3

= 71/2( )3

= 73/2

e. 7 ! 7 = 71 ! 71/2 = 73/2 f. 9 ! 33 = 32 ! 31/3 = 37/ 3

2-45.

a. x2 + x2 = 5 2

2x2= 25

x2=

25

2

x= 25

2 =

5

2=

5 2

2

sin(30!) = x8

0.5 = x8

4 = x

cos(30!) =y

8

3

2 =

y

8

8 3 =2y

4 3 = y


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10/35


2-56.

12

22 +

22

32 +

32

42 + ... =

i2

(i + 1)2i=

1

10

!

2-57.

(k! 2)2 =

k=2

6

" (2 !2)2 + (3 !2)2 + (4 !2)2 + (5 !2)2 + (6 !2)2

=02 +12 +22 +32 +42 =0 +1+4 + 9 +16 =30

2-58.

4!1 =3 (length of interval) 35 pieces

=0.6 (length of each piece)

x0 = 1

x1 = 1+ .6 = 1.6

x2 = 1.6 +0.6 =2.2x3 =2.2 +0.6 =2.8 x4 =2.8 +0.6 =3.4

2-59.

a. These summations are the same, just written in different ways.

j2

j=3

7

! = 33 +42 +52 +62 +72; k2k=3

7

! =33 +42 +52 +62 +72

b. These summations are the same, just written in different ways.

j2

j=3

7

! = 33 +42 +52 +62 +72

(j" 1)2

j=4

8

! =(4" 1)2 + (5 "1)2 + (6 "1)2 + (7 "1)2 + (8 "1)2 = 33 +42 +52 +62 +72

2-60.

Change Y1 to (k+1)2 , the starting index to B = 2 (or 2"X), and the end index to E = 7

(or X"7).

2-61.

a

x+2

4 +

x!3

5 =

5(x+2)

5(4) +

4( x!3)

4(5) =

(5 x+10)+(4 x!12)

20 =

9x!2

20

b. 42! x

+ x

5 =

5(4)

5(2! x)+

(2! x)(x)

(2! x)(5)=

20+(2 x! x2 )

10!5x=

! x2+2x+20

10!5x

c. 6x+2

!

4x!2

= 6(x!2)

(x+2)(x!2)!

4( x+2)

(x!2)(x+2) =

(6 x!12)!(4 x+8)

x2!4

= 2x!20

x2!4


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2-62.

a. 2x!56 +

x+18 =

4(2x!5)

4(6) +

3(x+1)

3(8) =

(8 x!20)+(3x+3)

24 =

11x!1724

b. xx2+2x!8

+ 2

x+4=

x

(x+4)(x!2)+

2x+4

CD = (x+ 4)(x!2)

x

(x+4)(x!2)+

2( x!2)

(x+4)( x!2)=

x+(2 x!4)

(x+4)( x!2)=

3x!4(x+4)( x!2)

Lesson 2.3.1

2-63.

a. g(x) b. It looks like steps.

c. 10+3+32 = 45

2-64.

a. and b. See graph at right.d. Answer will likely be between 7.75 and 10.75.

2-65.

Answers should range from 4.4 to 4.9.

2-66.

Students should add the results of problems 2 and 3.


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2-67.

a. See graph at right.

b. mileshr

hr = miles

c. 25(0.5)+25(0.25)+ 12! 50 ! (0.25)+75(1.25)

=

12.5+

6.25+

6.25+

93.75= 118.75 miles

d. The distance Erin traveled from 30 minutes into hertrip until 15 minutes later.

2-68.

a. slope = 75!250.75!0.5

= 500.25

=200

y !25 =200(x!0.5)

y !25 =200x!100

y =200x!75

b. E(x) =

25 for 0 < x! 0.5

200x" 75 for 0.5 < x! 0.75

75 for 0.75 < x! 2

#

$%

&%


2-69.

y =2.73

= 19.683

2-70.

slope = 3!14!0

= 24 =

12

(y !1)= 12(x!0)

y !1= 12 x

y = 12 x + 1

2-71.

The equation needs to be entered and the starting and ending values need to be adjusted.

SUM = 1520

2-72.

a. 4(1+2 +3 +4 + 5) = 4kk=1

5

! =60 b. 20 +21 +22 +23 +24 +25 = 2kk=0

5

! = 63

c. 2(1+2 +3 +4 + ...+100) = 2kk=1

100

! = 10,100

It is linear because Erin has a

constant rate of change between

30 minutes and 45 minutes.


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2-73.

a. See graph at right.b. The domain is time elapsed, and Lew walks for one hour.

The range is 0 to 4 miles.

c. The units are in miles since mihr

ihr = miles .

d. Length of rectangle (1 hr 0 hrs) = 1 hrHeight of rectangle (4 mph 0 mph) = 4 mphArea of rectangle = 1hr * 4mph = 4 miles

2-74.

a. (2x!1)(2x!1+3) =(2 x!1)(2x+ 2) =2(2x!1)(x+ 1)

b. (2x!1)2(2x!1!4) =(2 x!1)2(2x!5)

c. (2x!1)3 (2x!1)2 +4 x( ) =(2x!1)3(4 x2 !4 x+ 1+4 x)

=(2x!1)3(4 x2 +1)

2-75.

x(x2 ! 3x!10) = 0

x(x+ 2)(x! 5) = 0

Either x= 0, x+ 2 = 0, or x! 5 = 0

x= 0, !2, 5

2-76.

a. b. g(x) =x2 +2 for x


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Lesson 2.3.2

2-78.

a. (1, 1), (1.5, 2.25), (2, 4), (2.5, 6.25)b. The height .

c. x0 = 1x

1 = 1.5

x2 = 2

x3 = 2.5

x4 = 3

Width of each rectangle is 0.5.

d. The height of the rectangle is the function value at the left endpoint of each interval.e. 0.5 !1+0.5 !2.25 +0.5 ! 4 +0.5 !6.25 =0.5 +1.125 +2 +3.125 =6.75

This area is smaller than the true area.

2-79.

a. The height.b. They start from the right endpoint of the interval.c. 0.5 !2.25 +0.5 ! 4 + 0.5 !6.25 +0.5 !9 = 1.125 +2 +3.125 +4.5 = 10.75

This area is larger then the true area.

2-80.

Some possibilities are averaging the two values or using more rectangles.

2-81.

a. 0.25

b. height rectangle 1 = 12 =1

height rectangle 2 =1.252 = 1.5625



height rectangle 5 =22 = 4




c. They are they-values of the left endpoints.d. The width of each rectangle.e. The 3rdand the 6th.

2-82.

xk= 1 + 0.5 k


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12cm

2-83.

a. 0.5 1+0.5 k( )2

k=1

4

!

b. Right-hand endpoints: x1 = 1, x8 =8

0.25 0.25k+1( )2

k=1

8!

Left-hand endpoints: x0 = 0,x7 = 7

0.25 0.25k+1( )2

k=0

7!

2-84.a. See graph at right.b. height rectangle 1 = 0.25

height rectangle 2 = 1

height rectangle 3 = 2.25

height rectangle 4 = 4

c. 0.5(0.52+

12+

1.52+

22

) d. 0.5(0.52 +12 +1.52 +22 ) =0.5(7.5) = 3.75


2-85.

2 +2.5 +3 +3.5 + ...+9.5 = 4

2+

5

2+

6

2++

17

2 +

18

2 +

19

2 = !

k=4

19k

2

Other answers are acceptable as well.

2-86.

62+h

2= 12

2

36 +h2= 144

h2= 108

h = 108 = 3 ! 36 = 6 3

A = 1

2! b !h =

1

2!12 ! 6 3 = 36 3


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2-87.

a. 5 !2 =3 (length of interval)

34 peices

=0.75 (length of each piece)

x0 =2

x1 =2 +0.75 =2.75

x2 =2.75 +0.75 =3.5

x3 =3.5 +0.75 =4.25

x4 =4.25 +0.75 =5

b. f(x) = x2

f(x2 ) = f(3.5) =(3.5)2= 12.25

f(x4 ) = f(5) =52=25

2-88.

y = !x2

+6

x2!2(!x2 +6) = !3

x2

+2x2 !12 = !3

3x2 =9 " x2 =3 " x = 3

y = !( 3)2 +6 " y = !3 +6 = 3

2-89.

a. x+ 2 ! 0 " x ! #2

b. 3 ! x" 0 # !x " !3 # x $ 3

c. x2 !1"0 # (x+ 1)(x! 1)"0

# x+ 1"0, x! 1"0 # x" 1, !1

d. All reals.

2-90.

a. b. f(x)+ 2 =g(x) =4!x2 ifx " 1

x + 2 ifx > 1

#$%&

2-91.

a. 1003/2 = 1001/2( )3=(10)3 = 1000 b. 27!2/3 = 271/3( )

!2=(3)!2 = 1

32 =

19

c. 12527( )

2/ 3

= 125

27( )1/3( )

2

= 5

3( )2

= 25

9 d. a

6

27( )!2/3

= a

6

27( )1/3"

#$ %

&'!2

= a

2

3( )!2

= 3

a2( )

2

= 9

a4


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2-92.

a. xx!1

+ x

x+1=

x(x+1)

(x!1)(x+1)+

x(x!1)

(x+1)(x!1)

= (x2 + x)+(x2 ! x)

x2+ x! x!1

= 2x2

x2!1

b. 2x! 4x+2

= 2x(x+2)

(x+2) !

4x+2

= 2x2 +4 x!4

x+2

Lesson 2.3.3

2-93.

See graph at right.

a. 4!25

= 2

5 = 0.4

b. x0 = 2.0

x1 = 2.0 +0.4 = 2.4

x2 = 2.0 +2 !0.4 = 2.8

x3 = 2.0 +3 !0.4 = 3.2

x4 = 2.0 +4 !0.4 = 3.6

x5 = 2.0 +5 !0.4 = 4.0

d. 0.4 2 2+0.4k

k=0

4

!

e. 0.4(22 +22.4 +22.8 +23.2 +23.6 ) =0.4(37.558) = 15.023 grams

f. Lower because the rectangles lie below the curve.

2-94.

See graph at right.

a. 0.4 20.4k+2

k=1

5

! Change the indices.

b.0.4(22.4 +22.8 +2 3.2 +23.6 +24.0 ) =0.4(49.558)

= 19.823 grams

c. Students should realize that the indices shift so that we use the same number of rectangles,but of different heights.

2-95.

a. 60 mileshour !

2hours = 120 miles

40miles

hour !1hours =40 miles

120 +40= 160 miles

b. See graph at right.

c. Area under the curve = the distance traveled; i.e. hr! mihr

=mi


18/35

2-96.

a. Gain of about 40 million.b. 20 +25 + (!20)+15 + (!5)+5 =65 !25 =40

c. They should be counted as negative.

2-97.See graph at right.

a. 2 + (0.5) = 1.5b. f(x) is positive or negative based on the equation.

In this casef(x) is negative when x> 3 .c. 1. 2.5 + 1.5 + 0.5 = 4.5

2. 4.5 + (2) = 2.53. 2.5 + (6) = 3.5

d. There is more area below thex-axis than above.


2-98.


24 rectangles

=0.5 (length of each sub-interval)

x0 =3

x1 =3+0.5 =3.5

x2 =3.5 +0.5 =4

x3 =4 +0.5 =4.5

x4 =4.5 +0.5 =5

L =0.5 32 +2 ++ 4.52 +2( ) =0.5 11 + 14.25 + 18 + 22.25( )

R =0.5 3.52 +2 ++ 52 +2( ) =0.5 11 + 14.25 + 18 + 22.25 + 27( )

b. left endpoints = 0.5 3+0.5k( )2+2

k=0

3

! right endpoints = 0.5 3 +0.5k( )2+2

k=1

4

!

c. The left endpoint indices go fromk =0 tok = 1 whereas the right endpoint indicesgofrom k = 1 tok =4.

2-99.

a. 1x!

1

y=

1y

xy!

1x

yx=

y!x

xy b. 1

x!

1y( )(x + y) =

y!x

xy( )(x+y)

1

= xy+y

2!x

2!xy

xy=

y2!x

2

xy


19/35

f(x) f(x+2)1 12

f(x)

2-100.

slope = 5!3.54!1

= 1.53 =0.5

g(x) !5( ) =0.5(x!4)

g(x) =0.5x!2 +5

g(x) =0.5x + 3

a. 3 +0.5ii=1

4

! or 0.5i + 3i=1

4

!

b. 3.5 +4 + 4.5 +5 = 17

2-101.


25 rectangles

=0.4 (length of each sub !interval)

b. x0 = 4 x1 = 4 + 0.4 = 4.4

x2 = 4.4 + 0.4 = 4.8 x3 = 4.8 + 0.4 = 5.2

x4 = 5.2 + 0.4 = 5.6 x5 = 5.6 + 0.4 = 6

c. lower =0.4 24 +

24.4

+ 24.8

+ 25.2

+ 25.6( ) =0.0845

upper =0.4 24.4

+ 24.8

+ 25.2

+ 25.6

+ 26( ) =0.779

d. Use more rectangles .

2-102.

f(x) = 1x!4

f(x + 2) = 1(x+2)!4

= 1x!2

f(x + 2)!1=g(x) = 1x!2! 1

2-103.

a. b. c.

2-104.

Third angle = 180! ! (102! +26!) =52!

Area =62 sin(102!) sin(52!)

sin(26!)=31.65cm2


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Lesson 2.3.4

2-105.

Students should recall, lower 15.023 and upper 19.823.a. 17.312 +2 !1= 17.312 +2 = 19.312

b. The shift added a 2 by 1 rectangle with area 2.c. A(h, 2 ! x !4) = 10 +17.312 = 37.312

The shift added a 2 by 10 rectangle with area 20.d. 2a+ 17.312, the width is always 2 and the height added is a.

2-106.

a. f(x) b. g(x) Shifts the graph up two units.

A(f, 0 !x !3) =0.5(3)(6) =9

c. A(g, 0 !x !3) =9 +6 = 15 d. h(x) (See graph below.)

Area added is a 2 by 3 rectangle of area 6. A(h, 0 ! x ! 3) = 0.5(2)(4)" 0.5(1)(2)

= 4 " 1= 3

2-107.

a. No changeb. Adding a 2 by 2 rectangle, which adds an area of 4 units.c. Nothing as long as the interval shifts also.d. Add a 2 by krectangle.

2-108.

Adds a 5 by 3 rectangle with an area of 15 units.

A(j, 0 ! x !5) =W +15

2-109.

Subtracts a 2 by 2 rectangle with an area of 4 units.A(n,1! x !3) = 11.190 "4 =7.190

2-110.

See graph at right.A(g, 0 !x !3) =23.66622 "10.09887 = 13.567


21/35


2-111.

a. Width of the rectangles3 (length of interval)

6 (rectangles) = 0.5

b. 3!

2=

6, 6+

7.579=

13.579 c. 3 !1000 = 3000, 3000 +7.579 = 3007.579

2-112.

a. f(x) = x2+5

b. 6 !1= 5 (length of interval)

55 rectangles

= 1 (length of each sub ! interval)

x0 = 1 x1 = 2 x2 = 3

x3 = 4 x4 = 5 x5 = 6

L = 1 12+5 ++ 5

2+5( ) = 1 6 + 9 + 14 + 21 + 30( ) = 19.25

R = 1 22 +5 ++ 62 +5( ) = 1 9 + 14 + 21 + 30 + 41( ) = 23.2

2-113.CJ 20m Amy

60m d

John

a.!60 +40t b.

!20 +30t c. d2 =(

!60 +40t)2 + (

!20 +30t)2

d = (40t!60)2 + (30t!20)2

d. t = 1.2 hours or at 1:12p.m; The distance between them is 20 miles.


22/35

2-114.

a. 4x2

!8x

x2

!9" x2

! x!12

2x3!8x2 =

4x(x!2)

(x+3)(x!3)"

( x!4)(x+3)

2x2 (x!4)=

2(x!2)

x(x!3)

b. (x2 ! y2 ) 4x2 +4xy

x2+2xy+y2

=(x + y)(x! y) 4x(x+y)

(x+y)(x+y) =4x(x! y)

2-115.

a. 1+ xy( )

4y2

x2 !y2

"#$

%&' =

y+x

y( ) 4y2

(x+y)(x!y)"#$

%&' =

4y

(x!y)

b. x2

!6x+8

x2

!16

x2

!4x+4x+4

= (x!4)( x!2)

(x+4)(x!4)"

(x+4)

(x!2)( x!2) =

1x!2

2-116.

a. b. j(x) =x2 ! 3 for x" !2

3x + 7 for x


23/35

Lesson 2.3.5

2-120.

a. 1 ! (2 +5 +10) = 17

b. 1 ! (5 +10 +17) = 32

c. 0.5!1

!(2 +5)+0.5

!1

!(5 +10) +0.5

!1

!(10+17) =3.5 +7.5 +13.5 =24.5

d. c, the sum in c is the average of the sums in a and b.e. It is too high.

2-121.

b. Students answers should be between 126-130.

2-122.

a. See graph at right.b. Increasing.c. (20)(2) + 0.5(40)(2) = 40 + 40 = 80

d. The units are miles.

2-123.

a. See graph at right.b. Increasing.c. Not as far, at there is less area underneath the curve.d. Example of area using trapezoidal rule with four

rectangles with length 0.5.0.5 !0.5 ! (20 +2(22.5)+2(30)+2(42.5) +60)=67.5

True area #66.7 miles

2-124.

See graph at right.a. 0.5 !0.5(5 +2(4.75)+2(4)+2(2.75)+1)

=0.25(29) =7.25

b. It is too low.

2-125.

a. 0.5(5 +4.75 +4 +2.75) =8.25 square units

b. 0.5(4.75 +4 +2.75 +1)=6.25 square units

c. 8.25+6.252

= 7.25 They are the same.

2-126.

The area underneath the curve is 80.a. Half of 80 miles, which is 40 miles.b. Vertical distances are halved.c. Draw the picture again or use the fact that the first car always travels twice as fast, so goes

twice as far.


24/35


2-127.

4!1= 3 (length of interval)

3

6 rectangles

= 0.5 (length of each sub ! interval)

x0 = 1 x1 = 1.5 x2 = 2 x3 = 2.5

x4 = 3 x5 = 3.5 x6 = 4

a. left-hand approximation = 0.5 21 + 21.5 + 22 + 22.5 + 23 + 23.5( ) b. right-hand approximation = 0.5 21.5 + 22 + 22.5 + 23 + 23.5 + 24( ) c. trapezoids

1

2 0 .5 2

1+2

1.5+2

2+2

2.5+2

3+2

3.5( )( )+ 12 0 .5 21.5 +2 2 +2 2.5 +2 3 +2 3.5 +2 4( )( )

= 0.5 21+21.5

2 +

21.5+22

2 +

22 +22.5

2 +

22.5+23

2 +

23+23.5

2 +

23.5+24

2( )

2-128.

a. See graph at right.

b. 3 30+602( ) = 3 45( ) = 135 miles

c. It is the units.

2-129.


520 rectangles


b. 10!

1 =9 (length of interval)

925 rectangles =0.36 (length of each sub-interval)

c. 5 ! (!3) =8 (length of interval) 8100 rectangles


d. E! B = E! B (length of interval) E!BN rectangles

= E!B

N (length of each sub-interval)

2-130.

10 !1 =9 (length of interval) 925 rectangles


a. x0 = 1! height of rectangle = f(1)

b. x1 = 1+0.36 = 1.36! height of rectangle= f(1.36)

c. x25 =

10!

height of rectangle =

f(10) d. x24 = 10 !0.36 =9.64 "height of rectangle = f(9.64)

e. x0 = B !height of rectangle = f(B)

x1 =B +W!height of rectangle = f(B +W)

xlast =E!height of rectangle = f(E)

xone before last =E"W!height of rectangle = f(E"W)


25/35

2-131.

a. 1cos!

= 1

c /b = 1"

b

c =

b

c b. 1

sin!=

1

a/b = 1 "

b

a =

b

a

c. 1tan!

= 1

a/c= 1 "

c

a=

c

a

2-132.2 !1= 1 (length of interval)

14 rectangles

= 0.25 (length of each sub-interval)

x0 = 1 x1 =1.25 x2 =1.50

x3 =1.75 x4 = 2.0

a. Upper =0.25 ( !12 +6) + (!1.252 +6) +( (!1.52 +6) + (!1.752 +6) ) =4.031

Lower =0.25 (!1.252 +6) +( (!1.52 +6) + (!1.752 +6) + (!22 +6) ) =3.281

b. 0.25 !(0.25k+1)2 +6k=1

4

"

2-133.

a. The Hypotenuse Leg Postulate

AD ! AB, AF! AE

b. DC = BC(def square)

DF = BE(corr. Parts of ! " ' s are ! )

FC = CE(subtraction)

c. side of equilateral triangle = 3cos15!

=3.106

Area =3.106( )2 3

4 =4.177cm2

2-134.

a. 3x+3

!

5

x"5 =

3!5

x2

"5x+3x"15=

15

x2

"2x"15

b.x+y

x!

x!y

y=

y(x+y)

xy!

x(x!y)

xy

=

xy+y2!x

2+xy

xy=

y2+2xy!x2

xy

c. xy! y

x( ) x

2y2

x2y!xy2

"#$

%&' =

x2 !y2

xy

"#$

%&'

x2y2

xy(x!y)"#$

%&' =

x2 !y2

x!y = (x+y)(x!y)

x!y = x + y

2-135.

xy = 12! y = 12

x

3x "2 12

x( ) = "6

3x " 24

x= "6

3x2" 24 = "6x

3x2 +6x " 24 = 0

x2 +2x!8 =0

(x + 4 )(x!2) =0

Either x =!4 or x =2

!4y = 12 2y = 12

y =!3 y = 6

Solution : (!4, !3), (2, 6)


26/35

Lesson 2.3.6

2-136.

a. 20 rectangles b. 6!120

= 5

20 =0.25

c. Left = 0.25(0.25k+ 1)2

k=0

19

! Right = 0.25(0.25k+1)2k=1

20

!

2-137.

a. Except for the first left-hand rectangle and the last right-hand rectangle, the otherrectangles are in both areas.

b. 0.25( 0.25k+1)2

k=1

19

!

2-138.

W = E!BN

2-139.

Left-hand area = 67.34375 Right-hand area = 76.09375

2-140.

Add: (R + L)/2 "#and Disp $TRAP=$T

2-141.

The trapezoidal area should be best. Students may mention that there are fewer gaps

between the trapezoids and the curve, or that the trapezoidal area is the average of areasthat are too high and too low.To get a better estimate, increase the number of partitions.Note: Remind students of calculator limitations for rounding.

2-142.

If students use n = 20,L= 3.22846,R= 3.02846, T= 3.128.a. Using more partitions will improve your estimate.

b. y = 4 ! x2

y2= 4 ! x2

x2+ y

2= 4, y " 0

c. Students should see this as one fourth of a circle with radius 2, or 4!4

= ! .


27/35


2-143.

a. See graph at right.b. Too low, as rectangles will be under the curve.

c. Too high, as rectangles will be above the curve.d. Left = 8.360, Right = 8.960

2-144.

a. Gallons of water that passed through the pipe in the interval from 20 minutes to 50minutes.

b. Left-hand approx. ! 1.5 "10 +1 "10 +1.5 "10 = 15 +10 +15 =40 gallons

Other answers are acceptable; Between 40 and 50 gallons.

2-145.10+12+18

2 =20

Area = 20 ! (20"10)(20"12)(20"18)( ) = 20 !10 !8 !2( ) = 3200 =56.57cm2

2-146.

x2+1 =!2ax+ 7 for x= 1

12+ a =!2a "1+7

1+ a =!2a +7

3a = 6

a = 2

2-147.1x!

3y

=5 "

y !3x =5xy (for x, y # 0)

y !5xy =3x

y(1!5xy) =3x

y = 3x1!5x

2-148.

1

y!

xy( )

1

x +1( ) = 1!

xy( )

1+x

x( ) = 1+

x!

x!

x

2

xy = 1!

x

2

xy


28/35

2-149.

a. Flipped overx-axis then up 2. b. Left 2 and compressed vertically.

2-150.

0.5 0.5k+2

k=0

20

!

2-151.

Lesson 2.3.7

2-152.

The area under a velocity curve represents distance traveled.a. Students should understand that the area is found by multiplying miles/hr times hours to

get miles.b. (30)(2) + (20)(1) = 60 + 20 = 80 milesc. The second car had half the velocity of the first and the height of the rectangles has

changed.

2-153.

Area represents 80 miles.a. The height of the trapezoid is t. b. 20 and 20t+ 20

c. Area of the trapezoid:

D(t) = 0.5(20 + (20t+ 20))(t)

0.5t(20t+ 40) = 10t2 + 20t

d. D(2) = 10 ! 22 +20 ! 2

= 10 ! 4 +40 =80 miles

e. D(3) = 10 ! 32 +20 ! 3 = 10 !9 +60 = 150

V(3) =20 ! 3+20 =60 +20 = 80 mph

f(x)+2

1

2f(x+2)


29/35

2-154.

a. She used the midpoints of the sections. One might refer to them as midpoint rectangles.c. Each rectangle would have width 1.d. Rectangles are still one unit wide, change the 1 to 1.5.

e. Left = 1(21 +22 +23 +24 ) =(2 +4 + 8 +16) =30

Right = 1(22 +23 +24 +25 ) =(4 + 8 +16 +32) =60

Trap =0.5 !1(21 +2 !22 +2 !23 +2 !24 +25 ) =0.5(90) =45

Connie = 1(21.5 +22.5 +23.5 +24.5 ) =42.426

2-155.

a. 42.426


2-156.

For the first six hours, Jack makes $5/hr, meaning at the end of six hours, he has earned$30. For anytime after six hours (where he has already earned $30) he adds to the $30 bymaking $8 times the number of hours worked over six hours.

2-157.

a. J(t) =5t for 0! t! 6

30 +10(t" 6) for t> 6#$%

b. C(t) =6t for 0! t! 8

48 +12(t" 8) for t> 8#$

%

c. 30 +10(t!6) =46

30 +10t!60 = 46

!30 +10t=46

10t=76

t= 7610

= 7.6 hours

d. When is 30 +10(t!6) >48 +12(t!8)

30 +10t!60>48 +12t!96

!30 +10t>!48 +12t

18 +10t>12t

18 >2t

t


30/35

2-158.

3!1 = 2 (length of interval)

210 rectangles


a. L =0.2 3(1)3 +1( )+0.2 3(1.2)3 +1( ) ++0.2 3(2.6)3 +1( )+0.2 3(2.8)3 +1( )

= 0.2 3(0.2k+1)3 +1!" #$k=0

9

%

b. R =0.2 3(1.2)3 +1( )+0.2 3(1.4)3 +1( )++0.2 3(2.8)3 +1( )+0.2 3(3.0)3 +1( )

= 0.2 3(0.2k+ 1)3 +1!" #$k=1

10

%

c. M = 0.2 3(0.2k+1.1)3 +1!" #$k=0

9

%

2-159.

163/2 = 161.5 = 64 They are the same because 1.5 = 32

163/2 = 161/2( )3

= 43 = 64

2-160.

a. 22(x+1) = 2x( )1/2

22x+2 =2 x/2

! 2x+ 2 = x2

4x+ 4 = x

4 = "3x

x= " 43

b. 31(32x) = 33(x+1)

31+2x = 33x+3

!1 +2x= 3x+ 3

"2 = x

c. 5x( )1/3

= 5!2( )

x!2

5x/3 = 5!2x+4

" x

3 = !2x+ 4

x= !6x+ 12

7x= 12

x= 12

7

2-161.

q( )2=512 +34 2 !2 "51" 34 cos 25!

q( )2=2601+1156 +3468 " (0.906)

q( )2=614.99

q =24.824.8

sin 25=

34

sin#P$58.68 =

34

sin#P

$ sin#P =34

58.68=0.579

#P =35.4! $

#R = 180! !25! !35.4! = 119.6!


31/35

2-162.

a. 2!3

b. ! 5"6

2-163.

a. It shows that the number of customers increases approaching lunchtime and decreases after

lunchtime.b. The total number of customers served from 11 a.m. to 1 p.m.c. From the graph, it can be seen that during each hour, more than eight people are served, so

the total must be more than 16.

2-164.

a. If k(x) is continuous, !

2x2 "4 = "3x + 10 for x =2

2 #22 "4 = "3 #2 +10

8 "4 = "6 +10

4 =4

So, k(x) is continuous at x =2.

b. j(x) =2(x + 1)2 ! 4 for x


32/35

2.3.8 Review and Preview

2-167.

#3.12 to 3.16 depending on left or right rectangles.

2-169.a. g(x) =2(x3 +2) =2x3 +4

b. See graph at right.

c. The area under g(x) is twice the area underf(x).

2-170.

The interval is 8 units long and you add a height of 4 units. Therefore the new area will be9 + (8)(4) = 41.

2-171.

!M = 180!"63

!"50

!=67

!

7

sin 67!=

o

sin 50!=

n

sin 63!

7.605 =o

0.766and 7.605 =

n

0.891

o =5.8 and n =6.8

2-172.

a. g(3) = 11!32 = 11!9 =2 b. g(!2) =2(!2) =!4

c. g(!1)=2(!1)=!2 d. g(4) = 11!42 = 11!16 =!5

2-173.a. At around 50 minutesthe number of fish entering the hatchery decreases greatly.b. The total number of salmon entering the hatchery during the 60-minute interval.

2-174.

a. log416 =2

log4 42

=2

42 = 16

b. log6 36!1( ) =!2

log66!2=!2

6!2 = 136( )

c. log1000 =3

log10103

=3

103 = 1000

2-175.a. log7 49 =log7 7

2=2 b. log216 =log22

4=4 c. log33

8=8


33/35

Closure Problems

2-176.

a. R(x) =1 x! 3

2x"3 x >3

#$%

b. She started taking the potion at (3,1). The area represents the total length of Rapunzelshair after a certain period. We are multiplying inches per month by months.

c. Trapezoidal approximation with height 1 unit.

0.5(1! 20 +2 ! 21 +2 ! 22 +23) =0.5(1+4 +8 +8) = 10.5 inches

True value #10.1 inches.d. From 3 to 9 months, hair growth is 90.9 inches so total is about 93.9 inches (dont forget

the original 3 inches).e. About 8.34 months.

2-177.

a. A(3x ! 7, 2 " x "7)

b. 2x2dx

8

8

!

c. Answers will vary but, 1.6875 < A


34/35

CL 2-179.

a.

b. litersminute

minute liters! = c. 0.2 1.2( )t

0

10

! dt

d. Yes, only about 41 liters have leaked.

CL 2-180.

a.3!1

0.4=

2

0.4=5 5 sub-intervals of width 0.4 b. 1! x! 31

c. S= 0.4 20.4k+1

k=0

4

! d. left-hand

CL 2-181.

C(t) =150 for 0! t! 3

150 + 40(t" 3) for t >t3

#$%

CL 2-182.

a. (3 !12 +5)+ (3 !22 +5)+ (3 ! 32 +5) + (3 ! 42 +5) =8 +17 +32 +53

b. 0.2( 43 +43+0.2!1 +43+0.2!2 +43+0.2!3 +43+0.2!4 ) =0.2 40.2k+3

k=0

4

"

c. A(4x , 3 ! x !4 )

CL 2-183.

a. See graph at right.b. Yes, it is continuousc. See graph in part (a).

d. h(x) = (x!1)2 !1 if x" 3

9! 2x if x > 3#$%&

CL 2-184.

f(x + 2) =2x!1+2 +4(x + 2)2 =2x+1 +4(x + 2)2


35/35

CL 2-185.

a.b.

Left =1

2

4 + 2( ) + 4 + 2.5( ) + 4 + 3( )+ 4 + 3.5( ) +6 + 4 + 4.5( )

!

"

##

$

%

&&

'17.360

Right =1

2

4 + 2.5( ) + 4 + 3( ) + 4 + 3.5( )+6 + 4 + 4.5( ) + 4+ 5( )

!

"

##

$

%

&&

'17.771

CL 2-186.

a. f(x! 1)shifts right 1. b. !f(x) is a reflection over thex-axis.

CL 2-187.

a. x+1x

! x

x!1=

(x+1)(x!1)

x(x!1) !

x"x

x(x!1)

= x2 + x

!x!1!

x2x(x!1)

=!

1x(x!1)

b. 1! xy( )

x2y2

y3x!x3y

"#$

%&' =

y!xy( )

x2y2

xy(y2 !x2 )

"#$

%&'

= y!x

y( ) x

2y2

xy(y!x)(y+x)

"#$

%&' =

x

y+x

CL-2-188.

a.

b.! =

15ft

10ft = 1.5 radians

1.5 "180

!

# =

270!

# =85.94

!

c. 15 radians is about 2.4 times around the circle. The central angle is then(15 !4") #2.34radians #139.44 .

10 ft

15 ft

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PC Chapter 2 Solutions