Page 26 - TUTORIAL_MAT_IPS_K11-2

TURUNAN FUNGSI ALJABAR

′
2
30. ( ) = − 6 + 5
′
→ ( ) = 0
2
→ 0 = − 6 + 5 → ( − 5)( − 1) = 0 → = 1 , = 5
+++++++++ ----------- ++++++++
1 5
HP = { |1 < < 5, ∈ }

′
2
31. ( ) = 1. ( − 4 + 1) + ( − 2)(2 − 4) =
2
2
3 − 12 + 9 = − 4 + 3 = ( − 3)( − 1)
′
→ ( ) = 0
→ ( − 3)( − 1) = 0 → = 1 , = 3
+++++++++ ----------- ++++++++
1 3
HP = { |1 < ⋁ > 3, ∈ }

′
2
32. ( ) = 9 + 2 − 1
2 1 2 2
1
1
′
(− ) = 9 (− ) + 2 . (− ) − 1 = − − 1 = 0 → − = 1 … ( )
3 3 3 3 3
′
2
(1) = 9 (1) + 2 . (1) − 1 = 9 + 2 − 1 = 0 → 9 + 2 = 1 … ( )
1 1 2
Eliminasi (i) dan (ii) → = −1, = → + = − 1 = −
3 3 3

2
33. ′( ) = + 2 + .
Agar fungsi ( ) selalu naik untuk sembarang nilai x maka haruslah a > 0 dan diskriminan
′
( ) < 0.
Di sini syarat pertama nilai a (=1) > 0 sudah terpenuhi.
2
Syarat kedua : < 0 → − 4 < 0
2
2
→ (2 ) − 4.1. < 0 → 4 − 4 < 0 → 4 ( − 1) < 0
+++++++++ ----------- ++++++++
0 1
HP = { |0 < < 1, ∈ }

2
′
2
34. ( ) = −3 + − − 3 = −3 + ( − 1) − 3.
Agar fungsi ( ) selalu turun untuk sembarang nilai x maka harsulah a < 0 dan diskriminan
′
( ) < 0 → < 0.
Di sini syarat pertama nilai a (= -3) < 0 sudah terpenuhi.
2
Syarat kedua : < 0 → − 4 < 0
2
2
→ ( − 1) − 4. (−3). (−3) = − 2 − 35 < 0 → ( − 7)( + 5) < 0
+++++++++ ----------- ++++++++
-5 7
HP = { |−5 < < 7, ∈ }

‘LEARNING IS FUN’ 25