Nama : Gusliando

NIM : DBD 115 045

Dosen : Murniati, ST. MT.

Tugas Kalkulus 1

I. Ketaksamaan

1. 2x + 14 < x + 25

= 2x + 14 – 14 < x + 25 – 14 ◄ )

= 2x < x +11 11

= 2x – x < x – x+11

= x < 11

Hp: { x: x < 11 } atau ( -∞, 11 )

2. 10x + 4 > 8x + 5

= 10x + 4 – 4 > 8x + 5 – 4 ( ►

= 10x > 8x+1 ½

= 10x – 8x > 8x – 8x +1

= 2x > 1

= 2x (½) > 1 (½)

= x > ½

Hp: { x: x > ½ } atau (½, ∞)

3. -2 < 5 – 5x ≤ 3

= -2 – 5 < 5 – 5 – 5x ≤ 3 – 5 ( ]

= -7 < -5x ≤ -2 7/5 2/5

= -7 (1/5) < -5x (1/5) ≤ -2 (1/5)

= -7/5 < -x ≤ -2/5

Dikalikan dengan (-1)

= 7/5 > x ≥ 2/5

Hp: { x: 7/5 > x ≥ 2/5 } atau ( 7/5, 2/5 ]

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

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

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

= -3+3x < 5x = 5x < 11

= -3+3x – 3x < 5x – 3x = 5x (1/5) < 11 ( 1/5)

= -3 < 2x = x < 11/5

= -3 (½) < 2x (½)

= -3 < x

Hp: { x: -3 < x < 11/5} atau ( -3, 11/5) ( )

-3 11/5

5. ( x + 4 ) ( 2x – 1 )2 ( x – 5 ) ≤ 0

i. x + 4 = 0

x + 4 – 4 = 0 – 4

x = -4

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 – 5 = 0

x – 5 + 5 = 0 + 5

x = 5

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

-4 ½ 5

Titik penguji :

x = 6, ( 6 + 4 ) ( 2(6) – 1 )2 ( 6 – 5 ) = +x = 1, ( 1 + 4 ) ( 2(1) – 1 )2 ( 1 – 5 ) = -x = 0, ( 0 + 4 ) ( 2(0) – 1 )2 ( 0 – 5 ) = -x = -5, ( -5 + 4 ) ( 2(-5) – 1)2 ( -5 – 5 ) = +

Hp = [ -4, 5 ]

6. 3 ≤ 5

1– x

= 3 - 5 ≤ 0

1 – x

= 3 - 5 ( 1 – x ) ≤ 0

1 – x 1 – x

= ( 3 – 5 + 5x ) ≤ 0

1 – x

= -2 + 5x ≤ 0

1 – x

Titik pemecah adalah 2/5 dan 1,tetapi tidak terdefenisi di 1(u)

i. -2+ 5x = 0 ii. 1 – x = 0

5x = 2 1 = x

x = 2/5

Hp: { x: 2/5 ≤ x < 1 } atau [ 2/5, 1 )

- ( 0 ) + ( u ) -

2/5 1

Titik penguji

x = 2 -2+ 5 ( 2 ) = ( - ) 1 - 2

x = ½ -2 + 5 ( ½ ) = ( + ) 1 - ½

x = 0 -2 + 5 ( 0 ) = ( - ) 1 + 0

II. Nilai Mutlak, Akar Kuadarat dan Kuadrat

1. | x – 4 | < 5

= -5 < x – 4 < 5

= -5 + 4 < x – 4 + 4 < 5+ 4

= -1 < x < 9

Hp: { x: -1 < x < 9 } atau ( -1, 9)

2. ¿ 3−2 x2+x

∨¿< 5

= -5¿3−2 x2+x

∨¿ < 5

= -5 ( 2 + x ) < 3−2 x2+x ( 2 + x ) < 5 ( 2 + x )

= -10 – 5x <3 – 2< 10 + 5x

i. -10 + 5x < 3 – 2x ii. 3 -2x < 10 + 5x= -10 – 10 – 5x < 3 + 10 – 2x = 3 – 3 – 2x < 10 – 3 + 5x= -5x < 13 – 2x = -2x < 7 + 5x= -5x + 2x < 13 – 2x + 2x = -2x + 2x < 7 + 5x + 2x= -3x <13 = 0 < 7 + 7x

= -3x (13) < 13 (

13¿ = 0 – 7 < 7 – 7 + 7x

= -x < 133 = -7 < 7x

Dikalikan ( -1 ) = -7 (17) < 7x (

17)

= x > 133 = - 1 < x

HPi = (133,∞ ) HPii = ( -1,∞)

HP,HPi , U HPii = (133,∞ ) U ( -1,∞)

3. | 2x – 4 | > 3

= -3 > 2x – 4 > 3

= -3 + 4 > 2x – 4 + 4 > 3 +4

= 1 > 2x > 7

= 1 (½) > 2x (½) > 7 (½)

= ½ > x > 7/2

Hp: { x: ½ > x > 7/2 } atau ( ½, 7/2 )

4. | 4x + 4 | ≥ 15

i. 4x + 4 ≥ 15 ii. 4x + 4 ≤ - 15

4x ≥ 11 4x ≤ -19

x ≥ 11/4 x ≤ - 19/4

Hpi = [ 11/4, ∞ ) Hpii = ( -∞, -19/4 ]

Hp, Hpi U Hpii = [ 11/4, ∞ ) U ( -∞, -19/4 ]

5. 2 | 2x – 3 | < | x + 15 |

= | 4x – 6 | < x + 15 |

= ( 4x – 6 ) 2 < ( x + 15 )2

= 16x2 – 48x + 36 < x2 + 30x + 225

= 15x2 – 78x – 189 < 0

X1,2 −(−78 )±√(−78 )2−4 (15 )(−189)

2(15)

78±√6084+1134030

78 + √17424 78 - √17424

30 30

78 + 132 78 - 132

30 30 7 -9/5

Hp = ( −95 ,7)

( + ) + ( + )

−95 7

6. 4x2 – 5x – 5 < 0

X1,2 −(−5 )±√(5 )2−4 ( 4 )(−5)

2(4)

5±√25∓808

5 + √105 5 - √105

8 8Hp = ( 5 + √105, 5 - √105)

8 8

II. Sistem Koordinat Cartesius1. P ( -1, 5 ) dan Q ( 4, 5 )

d ( P, Q ) = √ (4+1 )2+(5−5)² = √ (5 )2+(0 )2

= √25 = 5Gambar : ( -4, 5 )

y ( -1, 5 )

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 ( 5, 2 )r = √ (5−4 )2+(2−3 ) ² = √ (1 )2+ (−1 )2

= √2( x – 4 )² + ( y – 3 )² =(√2 )2

= x2 – 8x + 16 + y2 - 6y + 9 = 2

= x2 + y2 – 8x – 6y – 25 = 2

=x2+y2–8x–6y+23=0

3. x² + y² - 8x + 5y = 0

00

pusat : ( -½ A, -½ B ) = ( -½ ( - 8 ), -½ ( 5 ) ) = ( 4, - 5/2 )

Jari-jari ( r ) : √¼ A2+¼B2−C = √¼ (−8 )2+¼ (5 )2−0 = √¼ (64 )+¼(25)

= √ 644

+254

= √ 894

4. melalui ( 2, 4 ) dan ( 4, 5 )kemiringan ( m ) = y2 – y1

x2 – x1

= 5– 4 4 –2

= 12

Persamaan garis ( y –4 ) =12 ( x – 2 )

2(y– 4) = x -22y–8 = x-22y–x–6 = 0

5. kemiringan dan perpotongan dengan sumbu y untuk garis 2y = 5x + 52y = 5x + 5 y = 5x + 5 2

Kemiringan ( m ) = 5 2

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

2 = 5 (0) + 2

2

= 52

Perpotongan di y = 52

6. melalui ( 3, -3) tegak lurus 2x + 3y = 5kemiringan(m) 2x + 3y = 5 3y = 5 – 2x y = 2 – 2x 3Sehingga m2 = -2

3

Tegak lurus m = -1 m2

= - 1 -2/3

= 3 2

persamaan : ( y + 3 ) = 32 ( x – 3 )

2 ( y + 3 ) = 3(x– 3) 2y + 6 = 3x – 9 2y – 3x – 9 – 6 = 0

2y – 3x – 15 = 0

7. melalui ( 3, -3 ) sejajar 2x + 3y = 52x + 3y = 2 – 2x y = 2 – 2x

3 kemiringan (m) = -2

3

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

3 (y + 3) = -2( x – 3 )

3y + 9 = -2x + 6

3y + 2x + 9 – 6 = 0

3y + 2x + 3 = 0

8. y = -4x+5 tipot sumbu y, x = 0 tipot sumbu x, y = 0

y = -4 (0) + 5 0 = - 4x + 5 y = 5 -5 = - 4x

x = 45

( 0, 5 ) ( 45 , 0 )

Pada persamaan y = -x2 – x + 3

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

koordinat perpotongany1 = y2

-4x + 5 = -x2 – x + 3x2 – 3x + 2 = 0(x – 2)(x – 1) = 0x – 2 = 0 x – 1 = 0 x = 2 = 1

y = -4x + 5 y = -4x + 5 = -4(2) + 5 = -4(1) + 5 = -3 = 1(2 ,– 3) (1 , 1)

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

= −1+2+12

4

= 134

titik koordinat ( - ½ , 134 )

y = -4x + 5

6

5

4

3

2

1

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

-1

-2

-3

-4

-5

y = -x2 – x + 3

00