Nama : Ayu Lusi Natallia

NIM : DBD 115 022

Dosen : Murniati, ST. MT.

Tugas Kalkulus 1

I. Ketaksamaan

1. 2x + 16 < x + 22

= 2x + 16 – 16 < x + 22 – 16 ◄ )

= 2x < x + 6 9

= 2x – x < x – x + 6

= x < 6

Hp: { x: x < 6 } atau ( -∞, 6 )

2. 10x + 6 > 8x + 2

= 10x + 6 – 6 > 8x + 2 – 6 ( ►

= 10x > 8x - 4 - 2

= 10x – 8x > 8x – 8x - 4

= 2x > - 4

= x > - 2

Hp: { x: x > -2 } atau (-2, ∞)

3. -2 < 2 – 5x ≤ 3

= -2 – 2 < 2 – 2 – 5x ≤ 3 – 2 ( ]

= -4 < -5x ≤ 1 4/5 -1/5

= -4 (1/5) < -5x (1/5) ≤ 1 (1/5)

= -4/5 < -x ≤ 1/5

Dikalikan dengan (-1)

= 4/5 > x ≥ -1/5

Hp: { x: 4/5 > x ≥ -1/5 } atau ( 4/5, -1/5 ]

4. 2 + 3x < 5x + 2 < 16

= 2 + 3x < 5x + 2 dan 5x + 2 < 16

= 2 – 2 + 3x < 5x + 2 – 2 = 5x + 2 – 2 < 16 – 2

= 3x < 5x = 5x < 14

= 3x – 3x < 5x – 3x = 5x (1/5) < 14 ( 1/5)

= 0 < 2x = x < 14/5

= 0 (½) < 2x (½)

= 0 < x

Hp: { x: 0 < x < 14/5} atau ( 0, 14/5) ( )

0 14/5

5. ( x + 2 ) ( 2x – 1 )2 ( x – 2 ) ≤ 0

i. x + 2 = 0

x + 2 – 2 = 0 – 2

x = -2

ii. 2x – 1 = 0

2x – 1 + 1 = 0 + 1

2x = 1

2x (½) = 1 (½)

x = ½

karena kuadrat maka menghasilkan nilai x = ½ sebanyak 2 buah

iii. x – 2 = 0

x – 2 + 2 = 0 + 2

x = 2

+ (0) - (0) - (0) +

-2 ½ 2

Titik penguji :

x = 3, ( 3 + 2 ) ( 2(3) – 1 )2 ( 3 – 2 ) = +

x = 1, ( 1 + 2 ) ( 2(1) – 1 )2 ( 1 – 2 ) = -

x = 0, ( 0 + 2 ) ( 2(0) – 1 )2 ( 0 – 2 ) = -

x = -3, ( -3 + 2 ) ( 2(-3) – 1)2 ( -3 – 2 ) = +

Diminta soal bertanda ≤ yang berarti area negatif sehingga,

Hp: { x: -2 < x ≤ ½ U ½ < x ≤ 2 }

6. 3 ≤ 2

1 – x

= 3 - 2 ≤ 0

1 – x

= 3 - 2 ( 1 – x ) ≤ 0

1 – x 1 – x

= ( 3 – 2 + 2x ) ≤ 0

1 – x

= 1 + 2x ≤ 0

1 – x

i. 1 + 2x = 0 ii. 1 – x = 0

1 – 1 + 2x = 0 – 1 1 – 1 – x = 0 – 1

2x = -1 -x = -1

2x (½) = - 1 (½) -x (-1) = -1 (-1)

x = -½ x = 1

Hp: { x: -½ ≤ x < 1 } atau [ -½, 1 )

- ( 0 ) + ( 0 ) -

-½ 1

II. Nilai Mutlak, Akar Kuadarat dan Kuadrat

1. | x – 2 | < 2

= -2 < x – 2 < 2

= -2 + 2 < x – 2 + 2 < 2 + 2

= 0 < x < 4

Hp: { x: 0 < x < 4 } atau ( 0, 4 )

2. 3 – 2x < 2 2 + x

-2 < 3 – 2x < 2 2 + x

-2 ( 2 + x ) < 3 – 2x ( 2 + x ) < 2 ( 2 + x ) 2 + x

-4 – 2x < 3 – 2x < 4 + 2x

-4 + 4 – 2x < 3 + 4 – 2x < 4 + 4 + 2x

-2x < 7 – 2x < 8 + 2x

-2x + 2x < 7 – 2x + 2x < 8 + 2x + 2x

0 < 7 < 8 + 4x

0 - 8 < 7 – 8 < 8 – 8 + 4x

-8 < -1 < 4x

-8 (¼) < -1 (¼) < 4x (¼)

-2 < -¼ < x

Hp: { x: -2 < -¼ < x } atau ( -2, -¼ )

3. | 2x – 2 | > 3

= -3 > 2x – 2 > 3

= -3 + 2 > 2x – 2 + 2 > 3 + 2

= -1 > 2x > 5

= -1 (½) > 2x (½) > 3 (5/2)

= -½ > x > 5/2

Hp: { x: -½ > x > 5/2 } atau ( -½, 5/2 )

4. | 4x + 2 | ≥ 12

i. 4x + 2 ≥ 12 ii. 4x + 2 ≤ - 12

4x + 2 – 2 ≥ 12 -2 4x + 2 – 2 ≤ -12 - 2

4x ≥ 10 4x ≤ -14

4x (¼) ≥ 10 (¼) 4x (¼) ≤ -14 (¼)

x ≥ 10/4 x ≤ - 14/4

Hpi = [ 10/4, ∞ ) Hpii = ( -∞, -14/4 ]

Hp, Hpi U Hpii = [ 10/4, ∞ ) U ( -∞, -14/4 ]

5. 2 | 2x – 3 | < | x + 12 |

= | 4x – 6 | < x + 12 |

= ( 4x – 6 ) 2 < ( x + 12 )2

= 16x2 – 48x + 36 < x2 + 24x + 144

= 15x2 – 72x – 108 < 0

X1,2 −(−72 )±√ (−72 )2−4 (15 )(−108)

2(15)

72±√5284+648030

72 + √11664 72 - √11664

30 30

72 + 108 72 - 108

30 30 6 -6/5

Hp = ( -6, -6/5)

6. 4x2 – 5x – 2 < 0

X1,2 −(−5 )±√(5 )2−4 ( 4 )(−2)

2 (4)

5±√25∓328

5 + √77 5 - √77

8 8Hp = ( 5 + √77, 5 - √77)

8 8

II. Sistem Koordinat Cartesius1. P ( -1, 5 ) dan Q ( 2, 2 )

d ( P, Q ) = √ (2+1 )2+(2−5) ² = √9+9 = √18 = 3√2Gambar :

y ( -1, 5 )

( 2, 2 ) 6

5

4

3

2

1

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 x

-1

-2

2. Pusat ( 4, 3 ) melalui titik ( 2, 2 )r = √ (4−2 )2+(2−3 ) ² = √4+1 = √5( x – 4 )² + ( y – 3 )² = 5

= x2 – 8x + 16 + y2 - 6y + 9 = 5

= x2 + y2 – 8x – 6y – 20 = 0

3. x² + y² - 8x + 2y = 0pusat : ( -½ A, -½ B )

= ( -½ ( - 8 ), -½ ( 2 ) ) = ( 4, -1 )

Jari-jari ( r ) : √¼ A2+¼B2−C = √¼ (8 )2+¼ (2 )2−0 = √¼ (64 )+¼(4) = √16+1 = √17

4. melalui ( 2, 2 ) dan ( 4, 2 )

00

kemiringan ( m ) = y2 – y1

x2 – x1

= 2 – 2 4 – 2 = 0 2

Persamaan garis = y2 – y1 = m ( x – x1 ) y – 2 = 0 ( x – 2 )

2 2 ( y – 2 ) = 0 ( x – 2 )

2y – 4 = 0 2y = 4

y = 4/2y = 2

5. kemiringan dan perpotongan dengan sumbu y untuk garis 2y = 5x + 22y = 5x + 2 y = 5x + 2 2 y = 5x + 1

2 Kemiringan ( m ) = 5

2 Perpotongan dengan sumbu y, x = 0 y = 5x + 1

2 = 5 (0) + 1

2 = 1 Perpotongan di y = 1

6. melalui ( 3, -3) tegak lurus 2x + 3y = 22x + 3y = 2 3y = 2 – 2x y = 2 – 2x 3Sehingga m = -2

3

Tegak lurus m = -1 m2

= -1 -2/3

= 3 2

persamaan : ( y + 3 ) = 3/2 ( x – 3 ) y + 3 = 3/2x – 9/2 2 ( y + 3 ) = 3x - 9

2y + 6 = 3x – 9 2y = 3x – 9 – 6 2y = 3x – 15 2y – 3x + 15 = 0

7. melalui ( 3, -3 ) sejajar 2x + 3y = 23y = 2 – 2x y = 2 – 2x

3 kemiringan (m) = -2

3

persamaan : ( y + 3 ) = -2 ( x – 3 ) 3

y + 3 = -2x + 6 3 3

3 ( y + 3 ) = -2x + 6

3y + 9 = -2x + 6

3y = -2x + 6 – 9

3y = -2x - 3

3y + 2x + 3 = 0

8. y = -2x + 2 dan y = -x2 – x + 3 y1 = y2

-2x + 2 = -x2- x + 3 x2 – x – 1 = 0

x1,2 = −(−1)±√ (−1 ) ²−4 (1 )(−1)

2(1)

x1 = 1±√1+42

x2 = 1±√1+42

= 1 + √5 = 1 - √5 2 2y = -2 (1 + √5) + 2 y = -2 (1 - √5) + 2 2 2 = 1 - √5 = 1 + √5

titik potong (1 + √5, 1 - √5), (1 - √5, 1 + √5) 2 2Pada persamaan y = -2x + 2

tipot sumbu y, x = 0 tipot sumbu x, y = 0 y = -2 (0) + 2 0 = -2x + 2

= 2 2x = 2 x = 1

( 0, 2 ) ( 1, 0 )Pada persamaan y = -x2 – x + 3

tipot sumbu y, x = 0 tipot sumbu x, y = 0 y = 3 0 = -x2 – x + 3

( 0, 3 ) X1,2 = −(−1)±√ (−1 )2−4 (−1 )(3)2(−1)

= 1±√13−2

( 1 + √13, 0 ) ( 1 - √13, 0 )-2 -2

titik puncak koordinat Xmax = ½ (x1 + x2 ) = 1/2 (( 1 + √13 + (1 - √13 )) -2 -2 = ½ (2)

-2 = ½ (-1)

= - ½

Ymax = (- ½ )2 – ½ + 3

= ¼ - ½ +3

= 11/4

titik koordinat ( - ½ , 11/4 )

y = -x2 – x + 3

(1 + √5, 1 - √5) 2 y

6

5

4

3

2

1

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 x

-1 (1 - √5, 1 + √5) 2

-2

-3

-4 y = -2x + 2

-5

00