2.a. F(z) = √2 z+3

2z + 3 ≥ 0 ; z ≥ -32

Domain :{ z ∈ R : z ≥ -32

}

b. g(v) = 1

4Q−1

4Q -1 =0; v=14

Domain { V ∈ R:v≠14

}

c. w(x) = √ x2−9

4. a. f ( x )= ⟦−2 x ⟧ n≤−2 x<n+1

n2≤−x<( n+1

2 )−n2≥ x>−( n+1

2 )−( n+1

2 )<x≤−n2−n+1

2=−1

Untuk n=1 , −1<x ≤−12

n=0 ,−12< x≤0

n=−1 ,0<x≤ 12

n=−2 ,12< x≤1

f ( x )= ⟦−2 x ⟧={−2 ,12<x≤1

−1 ,0<x≤ 12

0 ,12<x ≤ 1

21 ,−1<x≤1

b. f ( x )= ⟦2x ⟧ , n≤2x<n+1

n2≤ x< n+1

2

n2=−1 , n=2

Untuk

n=−2 ,−1≤x<−12

n=−1 ,−12≤ x<0

n=0 ,0≤x< 12

,

n=1 ,12≤ x<1

f ( x )= ⟦2x ⟧=¿

SOAL III

2. f ( x )=sinx x+sin x

Misalkan P ( x )= sin2 x => Dp =R

g ( x )=sin x=¿D g=¿R ¿

Jadi Df :Dp∩Dg=R

y=f ( x )sin2 x + sin x

= (sin x+ 12 )2

−1≤sin x ≤1

−12

≤ sin x+ 12≤

32

0≤(sin x+ 12)2 ≤

94

−14≤(sin x+1

2)2 -

14≤2

−14≤ y≤2

Jadi Df = {−14,2}

3. a. limx→2

7 x3+x2−5 x−403 x2+x−10

¿7 (2 )3+ (2 )2–5 (2 )−40

3 (2 )2+2−10

¿ 56+4−10−4012+2−10

=104

=52

B. limy→2

( 4 y2

y+4 )1/3

= [ limy→2( 4 y2+8 yy+4 )]1/3

= [ limy→2

(4 y2)+ limy→2

8 y

limy→

( y+4) ]= [ 4 lim

y→2y2+ lim

y→28 y

2+4 ]= ¿

= [ 4(2)2+166 ]

= [ 16+166 ]

= [ 326 ]

4. limX→2 √ 10−x2

2x

= limx→1

√10−x2

limx→1

2 x

= √10−12

2(1)

= √92

= 32

6. f ( x )=x−⟦x ⟧

Untuk

n=−1 ,−1≤x<0

n=0 ,0≤x<1

n=1 ,1≤x<2

n=2 ,2≤ x<3

n=3 , x=3

f ( x )=f ( x )={x+1 ,−1≤∧x<0x ,0≤∧x<1x−1 ,1≤x<2x−2 ,2≤x<3x−3 , x=3

7.a. |(2 x−4 )+8<ε|

↔|2x+4|<ε

↔|2(x+2)|<ε

↔2|x+2|<ε

|x+2|< ε2

|(2 x−4 )+8| = |2 x+4|

|2 ( x+2 )|=2|x+2|<2 ε

b. O<|x−1|<δ →|x2+5 x−6x−1

−7|<ε→|(x−1 )(x+6)

x−1−7|<ε

→|( x+6 )−7|<ε

→|x−1|<ε

δ=ε