Page 21 - Modul Ajar Matematika Fungsi Komposisi Kelas X Genap

2
g(x) = 2x + 3x + 1
(hogof )(x) = h(g( f (x)))
7 E
= h(g(2 + x))
= h((2 + x) − 1)
2
= h(x + 4x + 3)
2
= 2(x + 4x + 3)
2
2
= 2x + 8x + 6
8 ( fog )(x) = 101 A
f (g(x)) = 101
f (2x − 1) = 101
2
3(2x − 1) − 4(2x − 1) + 6 = 101
2
12 x − 12 x + 3 − 8x + 4 + 6 = 101
2
12 x − 20 x − 88 = 0
2
3x − 5x − 22 = 0
(3x − 11)(x + 2) = 0
2
3 atau −2
3
2
Jadi, nilai x yang memenuhi adalah 3 dan −2.
3
9. (g o f)(x) = 4x – 24x + 32 E
2
g (2x – 4) = 4x – 24x + 32
2
2
= (2x – 4) – 8x + 16
2
= (2x – 4) – 4(2x – 4)
g (x) = x – 4x
2
Alternatif 2:
2
(g o f)(x) = 4x – 24x + 32
2
g(2x – 4) = 4x – 24x + 32
1
Misalnya 2x – 4 = w, maka x = (w + 4) , sehingga
2
1  2 1 
g(w) = 4 (w + 4) − 24 (w + 4) + 32
 2   2 
= w + 8w + 16 – 12w – 48 + 32
2
= w – 4w
2
g (x) = x – 4x
2
Jadi, rumus fungsi g adalah g (x) = x – 4x
2
2
10 (gof )(x) = 2x + 4x − 3 E
2
g( f (x)) = 2x + 4x − 3
2
g(x + 3) = 2x + 4x − 3
2
g(x) = 2(x − 3) + 4(x − 3) − 3
2
g(x) = 2x − 12 x + 18 + 4x − 12 − 3
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