Fungsi f(x) = (ax³+bx²+cx+d)(1/d)
–> turun pada interval penyelesaian dari |2x|÷|3| > |2x+1|
–> melewati titik (0, 1)
–> memiliki nilai ordinat titik stasioner maksimum yaitu -⅛
Nilai d – a + b(c) = ………..
Quiz (+50):
complications
•
|2x| / |3| > |2x + 1|
|2x| > |6x + 3||
(8x + 3)(-4x – 3) > 0
-3/4 < x < -3/8
•
f(x) = (ax³ + bx² + cx + d)(1/d)
interval TURUN
f'(x) < 0
1/d (3ax² + 2bx + c) < 0
3ax² + 2bx + c < 0
-3/4 < x < -3/8
akar dr 3ax² + 2bx + c = 0 :
-3/4 dan -3/8
(8x + 3)(-4x – 3) = 0
-32x² – 36x – 9 = 0
32x² + 36x + 9 = 0
kesamaan letak
3a = 32
a = 32/3
2b = 36
b = 18
c = 9
•
f(x) = ∫(32x² + 36x + 9)
f(x) = 32/3 x³ + 18x² + 9x + C
melalui titik (0,1)
f(0) = 1
C = 1
f(x) = 32/3 x³ + 18x² + 9x + 1
f(x) = 1/3 (32x³ + 54x² + 27x + 3)
•
f(x) = (ax³ + bx² + cx + d)(1/d)
f(x) = 1/3 (32x³ + 54x² + 27x + 3)
d – a + bc
= 3 – 32 + (54 × 27)
= 1429
