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Menentukan Titik Stationer
Pada maximum Pada minimum
+
dy
dx =0
-
dy
> 0
dx
dy
< 0
dx
-
dy
dx =0
Contoh (continued)
f(x)= x 3 - 12x + 1
“… dan gambarkan grafiknya”
Tititk Stationer
x = -2
+ve -ve
Maximum
y = (-2) 3 -12(-2) +1
= -8 + 24 + 1
= 17
Maximum at (-2,17)
x = +2
-ve
f’(x)= 3x 2 - 12
+ve
Minimum
y = (2) 3 -12(2) +1
= 8 - 24 + 1
= -15
Minimum at (2,-15)
+
x=0
[y-axis]
y = 0 3 -12x0 +1
= 1
(0,1)
Contoh
f(x)= x3 - 12x + 1
“dapatkan titik stationer dan tentukan tipenya”
f’(x)= 3x 2 - 12
Titik stationer
terjadi ketika
gradient
[turunan] adl 0
3x 2 - 12 = 0
3x 2 = 12
x 2 = 4
x = 2 or x = -2
x = -2
Pilih titik dikiri
(x = -2.1)
f`(x) = 3x(-2.1) 2 -12
= 1.23 [+ve]
Pilih titik dikanan
(x = -1.9)
f`(x) = 3x(-1.9) 2 -12
= -1.17 [-ve]
+ve -ve
Maximum
x = +2
Contoh (continued)
f(x)= x 3 - 12x + 1
Fungsi kubik
Maximum pd (-2,17)
-2
20
-20
Y
Pilih titik dikiri
(x = 1.9)
f`(x) = 3x(1.9) 2 -12
= -1.17 [-ve]
Pilih titik dikanan
(x = 2.1)
f`(x) = 3x(2.1) 2 -12
= 1.23 [+ve]
-ve +ve
Minimum
memotong y-axis pd (0,1)
2
X
Minimum pd (2,-15)
2