fungsi-komposisi(1)
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Fungsi Komposisi
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Komposisi FungsiPenggabungan operasi dua fungsi
secara berurutan akan
menghasilkan sebuah fungsi baru.Penggabungan tersebut disebut
komposisi fungsi dan hasilnya
disebut fungsi komposisi.
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x A dipetakan oleh f ke y B
ditulis f : x y atau y = f(x)
y B dipetakan oleh g ke z C
ditulis g : y z atau z = g(y)
atau z = g(f(x))
A
x
C
z
B
yf g
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maka fungsi yang memetakan
x A ke z C
adalah komposisi fungsi f dan g
ditulis (g o f )(x) = g(f(x))
A B C
x zyf g
g o f
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contoh 1
f : A B dan g: B C
didefinisikan seperti pada gambar
Tentukan (g o f )(a) dan (g o f )(b)
A B C
a
b
p
q
1
2
3
f g
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Jawab:
A B C
a
b
p
q
12
3
f g
f(a) = 1 dan g(1) = q
Jadi (g o f )(a) = g(f(a)) = g(1) q
(g o f )(a) = ?
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A B C
a
b
p
q
1
23
f g
f(b) = 3 dan g(3) = p
Jadi (g o f ) = g(f(b)) = g(3) = p
(g o f )(b) = ?
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contoh 2
Ditentukan g(f(x)) = f(g(x)).Jika f(x) = 2x + p dan
g(x) = 3x + 120maka nilai p = « .
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Jawab:
f(x) = 2x + p dan g(x) = 3x + 120
g(f(x)) = f(g(x))
g(2x+ p) = f(3x + 120)3(2x + p) + 120 = 2(3x + 120) + p
6x + 3p + 120 = 6x + 360 + p
3p ± p = 360 ± 120
2p = 240 p p = 120
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Sifat Komposisi Fungsi1.Tidak komutatif:
f o g g o f
2. Bersifat assosiatif:
f o (g o h) = (f o g) o h = f o g o h
3. Memiliki fungsi identitas: I(x) = xf o I = I o f = f
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contoh 1f : R R dan g : R R
f(x) = 3x ± 1 dan g(x) = 2x2 + 5
Tentukan: a. (g o f )(x)
b. (f o g)(x)
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Jawab:f(x) = 3x ± 1 dan g(x) = 2x2 + 5
a. (g o f )(x) = g[f(x)] = g(3x ± 1)
= 2(3x ± 1)2 + 5
= 2(9x2 ± 6x + 1) + 5
= 18x2 ± 12x + 2 + 5
= 18x2 ± 12x + 7
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f(x) = 3x ± 1 dan g(x) = 2x2 + 5
b. (f o g)(x) = f[g(x)] = f(2x2 + 5)
= 3(2x2 + 5) ± 1
= 6x2 + 15 ± 1
(f o g)(x) = 6x2 + 14 (g o f )(x) = 18x2 ± 12x + 7
(g o f )(x) (f o g )(x)
tidak bersifat komutatif
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contoh 2f(x) = x ± 1, g(x) = x2 ± 1 dan
h(x) = 1/x
Tentukan: a. (f o g) o h
b. f o (g o h)
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Jawab:f(x) = x ± 1, g(x) = x2 ± 1
dan h(x) = 1/x
((f o g) o h)(x) = (f o g)(h(x))(f o g)(x) = (x2 ± 1) ± 1
= x2 ± 2
(f o g(h(x))) = (f o g)(1/x) = (1/x)2 ± 2
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f(x) = x ± 1, g(x) = x2 ± 1
dan h(x) = 1/x
(f o (g o h))(x) = (f(g oh)(x))
(g o h)(x)= g(1/x)
= (1/x)2 ± 1
= 1/x2 - 1
f(g o h)(x)= f(1/x2
± 1)= (1/x2 ± 1) ± 1
=(1/x)2 ± 2
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contoh 3I(x) = x, f(x) = x2 dan g(x) = x + 1
Tentukan:
a.(f o I)(x) dan (g o I)
b.(I o f ) dan (I o g)
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Jawab:I(x) = x, f(x) = x2 dan g(x) = x + 1
(f o I)(x) = x2
(g o I)(x) = x + 1
(I o f )(x) = x2
(I o g)(x) = x + 1
(I o f )(x) = (f o I) = f
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MenentukanSuatu Fungsi
Jika Fungsi Komposisidan
Fungsi Yang L
ain Dik
e
tahui
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Contoh 1
D
iketahui f(x) = 3x ± 1dan (f o g)(x) = x2 + 5
Tentukan g(x).
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Jawabf(x) = 3x ± 1dan (f o g)(x) = x2 + 5
f ?g(x)] = x2 + 5
3.g(x) ± 1 = x2 + 5
3.g(x) = x2 + 5 + 1 = x2 + 6
Jadi g(x) = (x2 + 6)
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contoh 2
Diketahui g(x) = x +
9dan
(f o g)(x) = x2 ± 6
maka f(x) = « .
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Jawab:g(x) = x + 9
(f o g)(x) = f(g(x)) = x2 ± 6
f(x + 9) = x2 ± 6
Misal: x + 9 = y p x = y ± 9
f(y) = (y ± 9)2 ± 6
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f(y) = (y ± 9)2 ± 6
= (y2 ± 18y + 81) ± 6
= y2 ± 6y + 27 ± 6
Jadi f(x) = x2 ± 6x + 21
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contoh 3
Diketahui f(x) = x ± 3 dan(g of )(x) = x2 + 6x + 9
maka g(x ± 1) = « .
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Jawab:f(x) = x ± 3;
(g o f )(x) = g (f(x)) = x2 + 6x + 9
g(x ± 3) = x2 + 6x + 9
Misal: x ± 3 = y p x = y + 3
g(y) = (y + 3)2 + 6(y + 3) + 9
= y2 + 6y + 9 + 6y + 18 + 9
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g(y) = y2 + 6y + 9 + 6y + 18 + 9
= y2 + 12y + 36
g(x ± 1) = (x ± 1)2 + 12(x ± 1) + 36
= x2 ± 2x + 1 + 12x ± 12 + 36
= x2 + 10x + 25
Jadi g(x ± 1) = x2 + 10x + 25
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Contoh 4
Diketahui f(x) = 2x + 1
dan (f o g)(x + 1)= -2x2 ± 4x + 1
Nilai g(-2) =«.
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Jawaban:f(g(x + 1))= -2x2 ± 4x + 1
f(x) = 2x + 1 f(g(x))= 2g(x) + 1
f(g(x + 1)) = 2g (x + 1) + 1
2g(x + 1) + 1 = -2x2 ± 4x ± 1
2g(x + 1) = -2x2 ± 4x ± 2
g(x + 1) = -x2 ± 2x ± 1
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g(x + 1) = -x2 ± 2x ± 1
g(x) = -(x ± 1)2 ± 2(x ± 1) ± 1
g(2) = -(2 ± 1)2
± 2(2 ± 1 ) ± 1= -1 ± 2 ± 1 = -4
Jadi g(2) = - 4
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