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f(x) = ax + b
f(2) = 3
f(2) = a.2 +b = 3
f(2) = 2a + b = 3 (p1)
f(-1) = -3
f(-1) = a.-1 + b = -3
f(-1) = -a + b = -3 (p2)
((p1) - (p2))
(metode eliminasi)
2a + b = 3
-a + b = -3
___________-
3a = 6
a = 6 : 3
a = 2 <= diketahui a
2a + b = 3
2.2 + b = 3
4 + b = 3
b = 3 - 4
b = -1 <= diketahui b
jadi.. jawaban no1 a= 2 b= -1
no2
g(x) = 5x - b
g(3) = 19
g(3) = 5.3 - b = 19
g(3) = 15 - b = 19
(karna koefisien (15) tidak boleh dikiri maka dipindahkan kekanan)
g(3) = -b = 19 -15
g(3) = -b = 4
g(3) = b = -4
jadi...jawaban no2 adalah... b = -4
no3
f(a) = 2a + 3
f(2x + 1) = ?
f(2x + 1) = 2.2x + 1 + 3
f(2x + 1 = 4x + 4
jadi... jawaban no3 adalah f(2x + 1) = 4x + 4
no4
f(x) = ax + b
f(-1) = 3
f(-1) = a.-1 + b = 3
f(-1) = -a + b = 3 (p1)
f(2) = 18
f(2) = a.2 + b = 18
f(2) = 2a + b = 18 (p2)
((p1 - p2))
(metode eliminasi)
-a + b = 3
2a + b = 18
__________-
-3a = -15
a = -15 + 3
a = -12 <= diketahui a
2a + b = 3
2.-12 + b = 3
-24 + b = 3
b = 3 + 24
b = 27 <= diketahui b
f(x) = ax + b
f(x) = -12x + 27 (fungsi f sekarang)
f(-5) = -12.-5 + 27
= 60 + 27
= 87
semoga bermanfaat, kalau ada yg tidak ngerti tanya yaa..