jawaban uts kalkulus wa



1.a  (x + 4)(2x – 1)²(x - 3) ≤ 0
(x + 4)(2x - 1)(2x -1)(x - 3) ≤ 0
x + 4    0              atau                       2x – 1 ≤ 0                             atau       x – 3 ≤ 0
x + 4 – 4 ≤ 0 – 4                                  2x – 1 + 1 ≤ 0 + 1                               x – 3 + 3 ≤ 0 + 3
x    -4                                                   2x ≤ 1                                                    x ≤ 3
                                                                   x ≤
Hp {| , x
1.b (  -   ≤ 0 ×
       ≤ 0
       (  ) ≤ 0( )
      2x – x + 1 ≤ 0
      x – 1 ≤ 0
      x – 1 + 1 ≤ 0 + 1
      x ≤ 1
Hp {| , x

2.a  → 2 +  > 7                         atau                      2 +  < -7
                                (2 + ) > 7()                                     (2 + )<-7 ()
                                   2 > 7                                                      2 + 5 < -7
                                2 > 7                                              2 < -7 
                                   -5 > 0                                                      9 < 0
                                   -5> 0 – 5                                         9 < 0 – 5
                                   -1(-5 >( -5)-1                                          9 < -5
                                   5 < 5                                                                  <
                                    <
                                 < 1
Hp {| <    , x

2.b  2|2 – 3|< | + 10|→ 2 ( 2- 3 )² < (+ 10)²
                                                      (4-6)² < + 20+100
                                                16- 48+36 <+ 20+100
                                                16- -48+36 < -+ 20+100
                                                15-48-20+36 < 20-20+100
                                                15-68+36-100 < 100-100
                                                15- 68- 64 < 0
                                                (15 - 80)( 15 + 12) < 0
                                                15 - 80 < 0                         atau                       15 + 12 < 0
                                                15 – 80 + 80 < 80                              15 + 12 – 12 < -12
                                            15 < 80                                                   15 < -12
                                             <                                                                    <
                                             < 5.3                                                                 < -0.8
Hp {| < -0.8  , x


3. Persamaan garis yang melalui (0,-4) tegak lurus pada garis + 2 = -( - 1)
 - = m₁ ( - ₁)
m₁ = -
m₂ =              m₂ = 2
 - = m₂( - ₂)
 – (-4) = 2 ( - 0)
 + 4 = 2
 + 4 - 4 = 2– 4
 = 2 – 4

4.  +  = 36
  =
    +  = 6
 = 6 - 
 
f()= 6 -  
( ,  )
-3
6 - (-3) = 9
-3,9
-2
6 - (-2) = 8
-2,8
-1
6 - (-1) = 7
-1,7
0
6 -  0    = 6
0,6
1
6 -  1    = 5
1,5
2
6 -  2    = 4
2,4
3
6 -  3    = 3
3,3

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