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g(5) = P(5) + B g(7) = P(7) + B
5P + B = 0 ..... (pers 1) = 7P + B = 4.........(pers 2)
subtitusi
5p+b =0 5p+b=0 nilai g(6)= p(6)+b
7p+b =4 - 5(2)+b=0 = 2(6) + (-10)
-2p = -4 b= -10 = 12 -10 = 2
p = 2
2. f(x) = 5x -p
f(3) = 5(3)-p
9 = 15 -p
p=15-9
=6
3. f(x) = 1/3(x²-4)
7 = 1/3x² - 4/3
7 + 4/3 = 1/3x²
25/3 = 1/3x²
25/3 x 3/1 = x²
25 = x²
x=
= 5
g(7)=px+b=4
⇒ g(5)=p(5)+b=0
g(7)=p(7)+b=4
⇒g(5)=5p+b=0
g(7)=7p+b=4
_________________-
-2p=-4
p=-4/-2
p=2
Subtitusikan p=2 pada g(5)= p(5)+b=0
g(x)=px+b
g(5)=(2 x 5)+b = 0
g(5)=10+b = 0
b = 0-10
b = -10
Bentuk Fungsi = g(x)=2x-10
g(x)=2x-10
g(6)=2(6)-10
g(6)=12-10
g(6)=2
2. f(x) = 5x-p
f(3) = -9
f(3)= 5(3) - p =9
= 15 -p = 9
= -p = 9-15
= -p = -6
= p = 6
3.f(x)=1/3(x²-4)
f(x)= 1/3 (x² -4) = 7
f(x)= x²-4 = 7 x 3
x² -4 = 21
x² = 21+4
x² = 25
x = √25
x = 5
Maka daerah asal = 5