mohon bantuan jawabannnyaa . bab turunan
Matematika
raniihr
Pertanyaan
mohon bantuan jawabannnyaa . bab turunan
1 Jawaban
-
1. Jawaban whongaliem
1) fungsi naik untuk f' (x) > 0 dan turun f' (x) < 0
f (x) = x³ + 5x² - 8x - 5
f'(x) 3x² + 10x - 8
interval naik
f' (x) > 0
3x² + 10x - 8 > 0
(3x - 2)(x + 4) > 0
<======o-----------o====>
- 4 2/3
fungsi naik : x < - 4 atau x > 2/3
fungsi turun :
3x² + 10x - 8 < 0
(3x - 2) (x + 4) < 0
--------o==========o-------
- 4 2/3
interval turun : - 4 < x < 2/3
b) f (x) = x (x - 1)²
= x (x² - 2x + 1)
= x³ - 2x² + x
f' (x) = 3x² - 4x + 1
fungsi naik ..
3x² - 4x + 1 > 0
(3x - 1) (x - 1) > 0
<======o---------o=====>
1/3 1
interval naik : x < 1/3 atau x > 1
fungsi turun ..
3x² - 4x + 1 < 0
(3x - 1) (x - 1) < 0
-------o========o-------
1/3 1
interval turun : 1/3 < x < 1
2) titik stasioner terjadi saat f' (x) = 0
f (x) = x² + 12x - 28
f' (x) = 2x + 12
0 = 2x + 12
2x = - 12
x = - 12/2
x = - 6
y = ( - 6)² + 12 ( - 6) - 28
= 36 - 72 - 28
= - 36 - 28
= - 64
titik stasioner ( - 6 , - 64 ) dengan jenis titik balik minimum
[tex]f (x) = \frac{1}{4} X^{4} - 2 X^{3} + \frac{9}{2} X^{2} - 8[/tex]
f' (x) = x³ - 6x² + 9x ⇒ f" = 3x² - 12x
0 = x (x² - 6x + 9)
0 = x (x - 3)²
x = 0 atau (x - 3)² = 0
x - 3 = 0
x = 3
f' (x) = 3x² - 12x
f' (3) = 3 (3)² - 12 (3)
= 27 - 36
= - 9
f (3) < 0 ⇒ memberikan nilai maksimum
y = (1/4) 3^4 - 2 (3)³ + (9/2) (3)² - 8
= 81/4 - 54 + 81/2 - 8
= 81/4 + 162/4 - 248/4
= - 5/4 titik stasioner (3 , - 5/4) jenis titik balik maksimum