FUNGSI
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MEMAHAMI
NOTASI, DOMAIN, RANGE,
DAN
GRAFIK SUATU FUNGSI
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Contoh
Daerah
Asal
Semua nilai x ≥ –2
memenuhi, sehingga daerah
asalnya adalah {x : x ≥ –2, x ϵ R }
atau x ϵ (–2, ∞).
Daerah
Hasil
Semua nilai y ≥ –6
memenuhi, sehingga daerah hasilnya
adalah {y : y ≥ –6, y ϵ R }
atau yϵ(–6, ∞).
Daerah
Asal
Semua nilai x, sehingga
daerah asalnya adalah {x |xϵR }
atau x ϵ R.
Daerah
Hasil
Nilai y yang memenuhi
adalah y ≤ 1 atau dengan
kata lain, y tidak mungkin bernilai
lebih dari satu, sehingga
daerah hasilnya adalah {y | y ≤ 1, y ϵ R } atau y ϵ (–∞, 1).
Daerah
Asal
Semua nilai x memenuhi kecuali x = 2, sehingga daerah asalnya
adalah {x | x ≠ 2, x ϵ R }.
Daerah
Hasil
Semua nilai y memenuhi
kecuali y = 1, sehingga daerah asalnya adalah {y | y ≠ 1, y ϵ R }.
Latihan 1
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Tentukanlah daerah asal dan daerah
hasil fungsi yang disajikan pada grafik
berikut.
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Menentukan
Domain dan range dari sebuah Fungsi
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Contoh 1
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Tentukan domain dan range dari x(x) =
2x
Jawab :
Bentuk 2x akan
terdefinisi pada semua nilai x bilangan real, jadi Df
= {x|x ϵ R}
Dan bentuk 2x akan
bernilai positif untuk semua x bilangan real,
Jadi Rf = {y|y > 0, y ϵ R}
Untuk lebih jelasnya, perhatikan
sketsa grafik dari f(x)=2x
Contoh 2
--------------------------------------------------------------------------
Tentukan Domain dan range untuk f(x)
= x2 − 4x + 3
Jawab
Semua fungsi polinom terdefinisi di
bilangan real, termasuk fungsi kuadrat.
jadi Df = {x|x ϵ R}
Sedangkan untuk mencari rangenya
dari fungsi kuadrat, dengan cara
(i) koefisien x2 adalah 1 > 0 ,
grafiknya terbuka ke atas, sehingga mempunyai nilai minimum
(ii)
Jadi puncaknya (2,−1)
Jadi range dari fungsi f adalah Rf
= {y|y ≥ −1, y ϵ R}
Untuk lebih jelasnya, perhatikan
sketsa grafik dari
f(x) = x2 − 4x + 3
Contoh 3
--------------------------------------------------------------------------
Tentukan Domain dan range untuk g(x)
=
Jawab :
Fungsi rasional (bentuk akar) akan
terdefinisi jika dalam akar bernilai non negatif,
Jadi
x2 − 4x + 3 ≥ 0
(x − 3)(x − 1) ≥ 0
x = 3 atau x = 1
Bandingkan dengan gambar di atas,
grafik x2 − 4x + 3 ≥ 0 untuk interval 1 ≤ x ≤ 3 (contoh 2)
Dg = {x|x ≤ 1 atau x ≥ 3,
x ∈ R}
Pada gambar di atas, x2 −
4x + 3 ≥ −1
maka
≥ 0
Rg = {y|y ≥ 0, y ∈ R}
Latihan 2
---------------------------------------------------------------------------
Tentukanlah daerah asal dan daerah hasil
fungsi berikut.
a. f(x) = 2x + 3
b. f(x) = x2 –
2x – 8
c. f(x) = x2
–1 2 ≤ x ≤ 6
d. f(x) =
e. f(x) =
f. h(x) =
g. h(x) =
h. h(x) =
i. h(x) =
j. h(x) =
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Operasi
Aljabar pada Fungsi
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Jika
f suatu fungsi dengan daerah
asal Df dan g suatu fungsi dengan daerah asal Dg,
maka pada operasi aljabar penjumlahan,
pengurangan, perkalian, dan pembagian
dinyatakan sebagai berikut.
1. Jumlah f dan g ditulis
f + g didefinisikan sebagai
(f
+ g)(x) = f(x) + g(x)
dengan
daerah asal
Df
+ g = Df
∩Dg.
2. Selisih
f dan g ditulis f – g didefinisikan sebagai
(f – g)(x) = f(x)
– g(x)
dengan daerah asal
Df – g = Df ∩Dg.
3. Perkalian
f dan g ditulis f × g didefinisikan sebagai
(f × g)(x) = f(x) × g(x)
dengan daerah
asal Df × g = Df ∩Dg.
4. Pembagian
f dan g ditulis didefinisikan
sebagai
dengan daerah asal
.
Contoh
--------------------------------------------------------------------------
Diketahui
fungsi f(x) = x + 3 dan g(x)= x2
– 9. Tentukanlah fungsi-fungsi berikut dan
tentukan pula daerah asalnya.
a) (f
+ g)(x)
b) (f
– g) (x)
c) (f
× g) (x)
d)
(x)
Jawab :
Daerah asal
fungsi f(x)
= x + 3 adalah Df = {x | x ∈ R }
dan daerah
asal
fungsi g(x) = x2
– 9 adalah Dg = {x | x ∈ R }.
a) (f
+ g)(x) = f(x) + g(x)
= (x + 3 ) + (x2 – 9)
= x2 + x –
6
Daerah asal
fungsi (f + g)(x) adalah
Df
+ g
= Df
∩ Dg
= {x
| x ∈ R } ∩ {x
| x ∈ R }
= {x
| x ∈ R }
b)
(f – g)(x) = f(x)
– g(x)
= (x + 3) – (x2–
9)
= –x2
+ x + 12
Daerah asal fungsi (f – g)(x)
adalah
Df
- g
= Df
∩ Dg
= {x
| x ∈ R } ∩ {x
| x ∈ R }
= {x
| x ∈ R }
c)
(f × g)(x) = f(x) × g(x)
= (x + 3) × (x2
– 9)
= x3 + 3x2
– 9x – 27
Daerah asal
fungsi (f × g)(x) adalah
Df
x g
= Df
∩ Dg
= {x
| x ∈ R } ∩ {x
| x ∈ R }
= {x
| x ∈ R }
d)
= Df
∩ Dg dan g(x) ≠ 0
= {x
| x ∈ R} ∩ {x
| x ∈ R } dan x2 – 9 ≠ 0}
= {x
| x ∈ R }
dan (x + 3) (x
– 3) ≠ 0}
= {x
| x ∈ R } dan x
≠ –3, x
≠ 3}
= {x
| x ∈ R, x
≠ –3, x
≠ 3}
Latihan 3
---------------------------------------------------------------------------
1. Diketahui
fungsi f(x) = dan g(x)= . Tentukanlah fungsi-fungsi berikut dan tentukan pula daerah
asalnya.
a) (f
+ g)(x)
b) (f
– g) (x)
c) (f
× g) (x)
d) (x)
2. Diketahui
f(x) = x2 – 3x + 2 dan g(x)
= x – 1. Tentukan:
a. (f + g)(x) c. (f . g)(x)
b. (f – g)(x) d. (x)
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Fungsi Komposisi
---------------------------------------------------------
Jika
f dan g fungsi serta Rf ∩Dg
≠
O, maka terdapat suatu
fungsi h dari himpunan bagian Df ke
himpunan bagian Rg yang disebut fungsi komposisi f dan g (ditulis gof) yang ditentukan dengan
h(x) = (gof)(x) = g(f(x))
daerah
asal fungsi komposisi f dan g adalah
Dgof = {x ∈ Df
| f(x) ∈ Dg }, dengan
Df
= daerah asal (domain) fungsi f;
Dg
= daerah asal (domain) fungsi g;
Rf
= daerah hasil (range) fungsi f;
Rg
= daerah hasil (range) fungsi g.
Contoh
--------------------------------------------------------------------------
Diketahui fungsi f: R → R dengan f(x) = 2x +
1 dan fungsi g: R → R dengan g(x) = x2
– 1.
(1) Apakah fungsi
komposisi (gof)(x)dan (fog)(x) terdefinisi?
(2) Tentukanlah
rumus fungsi komposisi (gof)(x) dan (fog)(x).
Jawab
f(x) = 2x
+ 1; g(x) = x2 – 1
Df
={x
| x ∈ R } = ; Rf
= {y
| y ∈ R } = R
Dg
={x
| x ∈ R } = ; Rg
= {y
| y ∈ R } = R
(1) Untuk
menentukan fungsi komposisi (gof)(x) dan (fog)(x) terdefinisi, maka dapat diketahui berdasarkan
i. Jika
Rf ∩Dg
≠ O, maka (gof)(x) terdefinisi.
{y| y∈ } ∩ {x| x∈ } = ∩ =
≠ O karena Rf
∩Dg
≠ O, maka (gof)(x) terdefinisi.
ii. Jika
Rg∩Df ≠ 0, maka (fg)(x) terdefinisi.
{y| y∈ } ∩ {x
| x∈ } = ∩ =
≠ O karena Rg∩Df ≠ O,
maka (fog)(x) terdefinisi.
(2) Rumus
fungsi komposisi (gof)(x)dan (fog)(x) ditentukan dengan
i. (gof)(x) = g(f(x))
= g(2x + 1)
= (2x + 1)2 –1
= (4x2 + 4x +
1) – 1
= 4x2 + 4x
ii. (fog)(x) = f(g(x))
= f(x2 –
1)
= 2(x2 – 1) + 1
= 2x2 – 2 + 1
= 2x2 –1
Dengan
demikian diperoleh
(gof)(x) = 4x2
+ 4x dan
(fog)(x) = 2x2
– 1.
Latihan 4
---------------------------------------------------------------------------
1. Diketahui
fungsi f: R → R dengan
f(x) = x2 – 4x + 2 dan fungsi g: R → R dengan g(x)
= 3x – 7. Tentukanlah
a) (gof)(x)
b) (fog)(x)
c) (gof)(5)
d) (fog) (10)
2. Diketahui f : R → R;
g : R → R dengan f(x) = 2x2
+ 1 dan g(x) = x + 2. Tentukan:
a. (gof)(x) c. (gof)(1)
b. (fog)(x) d. (fog)(–2)
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Contoh
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Diketahui fungsi komposisi (gof) (x) = 18x2
+ 24x + 2 dan fungsi g(x) = 2x2 – 6.
Tentukanlah rumus untuk fungsi
berikut.
a) Fungsi f(x)
b)
Fungsi komposisi (fog)(x)
Jawab
(gof) (x) = 18x2
+ 24x + 2; g(x) = 2x2 – 6
a) Menentukan
fungsi f(x)
(gof) (x) = g(f(x))
= 18x2 + 24x + 2
↔ 2
× f(x)2 – 6 = 18x2
+ 24x + 2
↔ 2
× f(x)2 = 18x2
+ 24x + 2 + 6
↔ 2
× f(x)2 = 18x2
+ 24x + 8
↔ f(x)2 =
↔
f(x)2
= 9x2 + 12x +
4
↔
f(x)
= ±
↔
f(x)
= ±
(3x + 2)
Jadi,
ada dua fungsi f yang mungkin,
yaitu
f(x) = 3x + 2 dan f(x) = –3x – 2.
b) Menentukan
fungsi komposisi (fog)(x)
i. Untuk
f(x) = 3x + 2
(fog)(x) = f(g(x))
= 3 × g(x) + 2, karena f(x)
= 3x + 2
= 3 × (2x2 –
6) + 2
= 6x2 – 18 + 2
= 6x2 – 16
Jadi, fungsi
komposisi (fog)(x) = 6x2
– 16
ii. f(x) = –3x – 2
(fog)(x) = f(g(x))
= –3 × g(x) – 2, karena f(x)
= –3x – 2
= –3 × (2x2
– 6) – 2
= –6x2 + 18 – 2
= –6x2 + 16
Jadi, fungsi
komposisi (fog)(x) = -6x2
+ 16
Latihan 5
---------------------------------------------------------------------------
1. Jika f(x)
= 2x – 3 dan g ° f(x) = 2x + 1 maka g(x)
= ....
2. Diketahui
(gof)(x) = 4x2
+ 4x dan g(x) = x2 - 1. Tentukanlah nilai f(x –
2).
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Sifat-Sifat
Operasi Fungsi Komposisi
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Diketahui
f, g, dan h suatu fungsi. Jika Rh∩Dg
≠
O; Rgoh∩Df ≠ O;
Rg∩Df
≠
O; Rh∩Dfog ≠ O, maka pada operasi komposisi fungsi berlaku
sifat asosiatif, yaitu
fo(goh) = (fog)oh
---------------------------------------------------------------------------Diketahui f suatu fungsi
dan I merupakan fungsi identitas. Jika RI∩Df
≠ O, maka terdapat
sebuah fungsi identitas, yaitu I(x) = x, sehingga berlaku
sifat identitas,
yaitu
foI = Iof = f
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Contoh
---------------------
Diketahui fungsi f: R → R dengan f(x)
= 2x – 1, fungsi g: R → R dengan g(x) = 4x + 5, dan
fungsi h: R → R dengan
h(x) = 2x – 3.
a) Tentukanlah rumus fungsi komposisi
go(foh) dan (gof)o h.
b) Apakah
go(foh) = (gof)oh. Coba selidiki.
Jawab
--------------------
a) Rumus fungsi komposisi (go(foh))(x) dan ((gof)oh)(x)
i) Misalkan
k(x) = (foh)(x)
k(x) =
f(h(x)) = 2h(x) – 1
= 2(2x – 3) – 1
= 4x – 6 – 1
= 4x – 7
(go(foh)(x)) =
(gok)(x)
= g(k(x))
= 4(k(x)) + 5
= 4(4x – 7) + 5
= 16x – 28 +5
= 16x – 23
Jadi, fungsi komposisi (go(foh)(x)) = 16x – 23
ii) Misalkan
l(x) = (gof)(x)
l(x) =
g(f(x)) = 4(f(x)) + 5
= 4(2x –
1) + 5
= 8x –
4 + 5
= 8x +
1
((gof)oh)(x) =
(loh)(x)
= l(h(x))
= 8(h(x))
+ 1
= 8(2x –
3) + 1
= 16x –
24 + 1
= 16x –
23
Jadi, rumus fungsi komposisi
((gof)oh)(x) = 16x – 23.
b) Dari
butir (a), diperoleh nilai
(go(foh)(x)) = 16x – 23 dan
((gof)oh)(x) = 16x – 23
Berdasarkan
nilai-nilai ini disimpulkan bahwa
(go(foh)(x)) = ((gof)oh)(x) = 16x – 23
Latihan 6
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1. Suatu pabrik kertas berbahan dasar kayu memproduksi kertas melalui dua tahap. Tahap pertama menggunakan mesin I yang
menghasilkan bahan kertas setengah jadi, dan tahap kedua
menggunakan mesin II yang menghasilkan bahan kertas. Dalam produksinya
mesin I menghasilkan bahan setengah jadi dengan mengikuti fungsi
f(x) = 6x
– 10 dan mesin II mengikuti
fungsi
g(x) = x2
+ 12, x merupakan
banyak bahan dasar kayu dalam satuan ton.
a)
Jika
bahan dasar kayu yang tersedia untuk suatu produksi sebesar 50 ton, berapakah kertas yang dihasilkan? (Kertas dalam
satuan ton).
b)
Jika
bahan setengah jadi untuk kertas yang dihasilkan
oleh mesin I sebesar 110 ton, berapa tonkah kayu yang
sudah terpakai? Berapa banyak kertas yang dihasilkan?
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2. Diketahui
fungsi f(x) = , x
≠
0 dan g(x) = Tentukan rumus fungsi berikut apabila terdefinisi dan tentukan daerah
asal dan daerah hasilnya.
a) (f
+ g)(x)
b) (f
– g) (x)
c) (f
× g) (x)
d) (x)
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3. Misalkan f
fungsi yang memenuhi
= 2x
untuk setiap x
≠
0.
Tentukanlah
nilai f(2).
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5. Jika f(xy)
= f(x + y)
dan f(7)
= 7. Tentukanlah nilai f(49).
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6. Diketahui
fungsi f dan g dinyatakan dalam pasangan terurut
f = {(1,5), (2,6), (3,–1), (4,8)}
g = {(2,–1), (1,2), (5,3), (6,7)}
Tentukanlah
a) gof
b) fog
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7. Jika f fungsi yang memenuhi persamaan f(1) = 4 dan f(x+1)
= 2 f(x).
Tentukanlah f(2014).
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8. Jika f(x)
= dan x2
≠
1, buktikanlah bahwa
f(–x)
=
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