diketahui f(x)=2x+5 dan g (x-1)=x²+2x+3 tentukan: a. rumus komposisi fungsi (g•f) (x) b. nilai a jika (g•f) (a) =27

g(x-1) = x²+2x+3
g(x) = (x+1)²+2(x+1)+3
g(x) = x²+2x+1+2x+2+3
g(x) = x²+4x+6

(g•f)(x)
= g(f(x))
= (2x+5)²+4(2x+5)+6
= 4x²+20x+25+8x+20+6
= 4x²+28x+51

(g•f)(a) = 27
4a²+28a+51 = 27
4a²+28a+24 = 0
a²+7a+6 = 0
(a+6)(a+1) = 0
a = -6 atau a = -1

Mapel : Matematika

Kelas : 10

Bab : Fungsi Komposisi
__________________

g(x - 1) = x² + 2x + 3

g(x) = (x + 1)² + 2(x + 1) + 3
g(x) = x² + 2x + 1 + 2x + 2 + 3
g(x) = x² + 4x + 6

a] Fungsi komposisi (g o f) (x)

(g o f) (x) = g(2x + 5)
(g o f) (x) = (2x + 5)² + 4(2x + 5) + 6
(g o f) (x) = 4x² + 20x + 25 + 8x + 26
(g o f) (x) = 4x² + 28x + 51

b] Nilai a jika (g o f) (a) = 27

4x² + 28x + 51 = 27
4x² + 28x + 51 - 27 = 0
4x² + 28x + 24 = 0 → Bagi 4

x² + 7x + 6 =
(x + 6)(x + 1) = 0

→ (x + 6) = 0

x1 = -6

→ (x + 1) = 0

x2 = -1

Jadi, nilai a yang memenuhi :

a = -6 atau a = -1