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BAB 5
FUNGSI, KOMPOSISI FUNGSI, DAN FUNGSI INVERS
A. Fungsi 1. Definisi Relasi khusus yang memasangkan setiap anggota daerah asal(domain) dengan tepat satu anggota daerah kawan (ko domain). Jika f merupakan suatu fungsi yang memetakan setiap ang gota himpunan A ke tepat satu anggota himpunan B, maka dapat dinotasikan dengan: f:A→B 2. Domain dan Range Syarat domain agar fungsi terdefinisi: •
f (x) = p (x) Syarat: p (x) ≥ 0
•
f (x) =
p (x) q (x)
Syarat: q (x) ≠ 0 •
f (x) =
q( x )
log p (x)
Syarat: q (x) > 0, q (x) ≠ 1, p (x) > 0
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B. Komposisi Fungsi Komposisi merupakan penggabungan operasi dua buah fungsi secara berurutan yang akan menghasilkan fungsi yang baru. Fungsi f komposisi g dapat dinotasikan dengan (f g)(x).
(f g)(x) = f (g (x)) Sifat-sifat komposisi fungsi: •
(f g)(x) ≠ (g f )(x)
•
((f g) h)(x) = (f (g h))(x)
C. Fungsi Invers Sifat-sifat fungsi invers: •
( f g ) ( x ) = ( g −1 f −1 ) ( x )
•
( g f ) ( x ) = ( f −1 g −1 ) ( x )
•
(f ) ( x ) = f ( x )
•
( f f )( x ) = I ( x ) = x
−1
−1
−1 −1
−1
Invers berbagai bentuk fungsi antara lain: Bentuk Fungsi f (x) = ax + b f (x) =
110
ax + b cx + d
Invers f −1 ( x ) = f −1 ( x ) =
x−b a
−dx + b cx − a
f (x) = ax 2 + bx + c f (x) = apx ±q f (x) = a log (px ± q)
f −1 ( x ) =
−b ± 4ax + D 2a
f −1 ( x ) =
(
a
f −1 ( x ) =
log x) ± q p ax ± q p
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LATIHAN SOAL UM UGM 2019 1. SOAL 1. Diberikan fungsi f (x) =
2x − 1 . Nilai x+1
adalah .... A. 1
C. 3
B. 2
D. 4
(f
−1
1 f −1 ) 2
E. 5
UM UGM 2019 2. SOAL 1. Diberikan f (x) = x 2 + 1 dan g (x) = ax + 2 , dengan a ≠ 0 . Jika (f g−1 )(1) = 5 , maka 4a2 − 3 = .... A. –3 B. –2
C. –1 D. 1
E. 2
SIMAK UI 2019 3. SOAL 1. Jika (g−1 f −1 )(x) = 3x − 1 dan f (x) =
x−2 untuk x+1
x ≠ 1 , maka g (a − 2) = .... A. B.
−a + 9 a−4 −( a + 8 ) a−1
C.
D.
−( a + 5 ) a−4
E.
− ( a + 6) a−3
SIMAK UI 2019 4. SOAL 1. Jika f (x) = 2x 2 − 3x + 1 , g (x) = ax + b dan
(g f )(x − 1) = 4x 2 − 14x + 11 , maka ....
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−a + 5 a−3
1) a = 2
3) (f g)(1) = 0
2) b = –1
4)
f (x) g (x)
= x+1
UTBK 2019 5. SOAL 1. 1 x Jika f = dan f −1 (a) = −1 , maka a = …. x 2 + 3x A. –2 B. –1
C. 1 D. 2
E. 3
UTBK 2019 6. SOAL 1. Jika f (x) = ax + 3 dan (f f )(x) = 4x − 3 maka nilai f (a) = .... A. 1 B. 3
C. 5 D. 7
E. 9
UM UGM 2018 7. SOAL 1. 2x − 1 Jika f −1 adalah invers fungsi f dengan f −1 (1 − x) = , 1−x maka A. 2 B. 1
f(x − 2) − f −1 (x) .... = ... 2
C.
1 +2 x
E.
1 −2 x
D. –2
UM UGM 2017 8. SOAL 1. Jika f (x) = b x , b konstanta positif, maka
f ( x 2 − 1) f (1 − x 2 )
= ....
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A. f (1 − x 2 ) ⋅ f (1 − x 2 ) B. f (1 − x 2 ) ⋅ f (x 2 − 1) C. f (x 2 − 1) ⋅ f (x 2 − 1) D. f (1 − x 2 ) + f (1 − x 2 ) E. f (x 2 − 1) + f (x 2 − 1) SBMPTN 2017 9. SOAL 1. Jika f(x) = x 2 − 1 dan g(x) =
x−2 , maka daerah x+1
fungsi f ⋅ g adalah .… A. {x | −∞ < x < ∞}
D. {x | x < −1}
B. {x | x ≠ −1} C.
E.
{x | x ≥ 2}
{x | x ≠ 2}
SBMPTN 2016 10. SOAL 1. Jika tabel berikut menyatakan hasil fungsi f dan g. X
0
1
2
3
f(x)
1
3
1
-1
g(x)
2
0
1
2
Maka (fogof)(1) + (gofog)(2) = .... A. –1 C. 2 E. 5 B. 1 D. 3 SIMAK UI 2016 11. SOAL 1. 1 Jika f(x) = x − 3 dan g(x) = x 3 , maka .... 8
114
(1) (fog)(x) =
1 3 x −3 8
(2) (fog)(2) = −2
(x) (3) (fog) −1=
3
8x + 24
(4) (fog) −1 (−2) = 2 UM UGM 2016 12. SOAL 1.
= 2x − 6 dan g −1 (x) = Jika f(x) fog(2) = .... A. 20 B. 16
x−5 , maka nilai 4
C. 15 D. 10
E. –2
UM UGM 2016 13. SOAL 1. Diberikan fungsi f dan g dengan f(x − 2) = 3x 2 − 16x + 26 dan g(x) = ax − 1 . Jika (f g)(3) = 61 , maka nilai a yang memenuhi ada lah .… A. –2 B.
8 9
C.
9 8
E. 4
D. 2
SIMAK UI 2015 14. SOAL 1. Misalkan g(x)= 4 − x 2 dan = f(g(x))
2 − x2 ,x ≠ 0 4x 2
maka ....
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1 1 1 1 −1 1 (1) f ⋅ f = (3) f − f = 4 2 80 4 2 105 1 1 −47 (2) f + f = 4 2 210
1 f 2 45 (4) = 1 49 f 4
STANDAR UTBK 2019 15. SOAL 1. 7 ax + 1 , g(x)= x − 2 , dan (g −1 of −1 )(2) = f(x) = 2 3x − 1 maka a = .... A. –4 B. –3
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C. 2 D. 3
E. 4
PEMBAHASAN CERDIK: 1. PEMBAHASAN 1. 2x − 1 f (x) = x+1 −x − 1 f −1 ( x ) = x−2
(f
−1
1 1 f −1 ) = f −1 f −1 2 2 1 − − 1 −1 2 = f 1 − 2 2 3 − = f 2 3 − 2 = f −1 ( 1 ) −1
−1 − 1 1−2 −2 = =2 −1 =
Jawaban: B CERDIK: 2. PEMBAHASAN 1. g (x) = ax + 2 g −1 ( x ) =
x−2 a
117
( f g ) ( 1) = 5 f (g (1)) = 5 −1 −1
1 − 2 f =5 a 1 f − = 5 a 2
1 − + 1 = 5 a 1 +1 = 5 a2 1 =4 a2 1 a2 = 4 1 Jadi, 4a2 − 3 = 4 − 3 = 1 − 3 = −2 . 4 Jawaban: B CERDIK: 3. PEMBAHASAN 1.
(g
−1
f −1 )(x) = 3x − 1 −1
(f g) (x) = 3x − 1 x+1 3 x+1 (f g)(x) = 3 x+1 f (g (x)) = 3 g (x) − 2 x + 1 = g (x) + 1 3
((f g)
−1
(x)) = −1
3 ( g ( x ) − 2 ) = ( x + 1) ( g ( x ) + 1 )
118
3g (x) − 6 = xg (x) + x + g (x) + 1 3g (x) − xg (x) − g (x) = x + 1 + 6
(
)
3 g (x) − 2 x + 1 = g (x) + 1 3 3 ( g ( x ) − 2 ) = ( x + 1) ( g ( x ) + 1 ) 3g (x) − 6 = xg (x) + x + g (x) + 1 3g (x) − xg (x) − g (x) = x + 1 + 6 2g (x) − xg (x) = x + 7 g (x)(2 − x) = x + 7 g ( x) = g (a − 2) = = =
x+7 2−x
( a − 2) + 7 2 − ( a − 2) a+5 4−a −( a + 5 ) a−4 Jawaban: C
CERDIK: 4. PEMBAHASAN 1. Diketahui f (x) = 2x 2 − 3x + 1 , g (x) = ax + b , dan
(g f )(x − 1) = 4x 2 − 14x + 11 (g f )(x − 1) = 4x 2 − 14x + 11 2
(g f )(x − 1) = 4 (x − 1) − 6 (x − 1) + 1 (g f )(a) = 4a2 − 6a + 1 (g f )(x) = 4x 2 − 6x + 1 g (f (x)) = 4x 2 − 6x + 1 a (2x 2 − 3x + 1) + b = 4x 2 − 6x + 1 2ax 2 − 3ax + a + b = 4x 2 − 6x + 1 Diperoleh: 2a = 4 ⇒ a = 2
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a+b = 1 2+b = 1 b = −1 Maka, g (x) = 2x − 1 Dengan demikian:
(f g)(1) = f (g (1)) = f ( 2 ( 1) − 1 ) = f ( 1) 2
= 2 ( 1) − 3 ( 1) + 1 = 2−3 + 1 =0 f (x) g (x)
=
2x 2 − 3x + 1 (2x − 1)(x − 1) = = x−1 2x − 1 2x − 1
Pernyataan 1), 2), dan 3) benar. Jawaban: A CERDIK: 5. PEMBAHASAN 1. 1 x Diketahui: f = dan f −1 (a) = −1 , x 2 + 3x Misal:
1 1 =y⇒x= x y
Diperoleh: 1 y
1 y y 1 1 f (y) = = = ⋅ = 1 2y + 3 y 2y + 3 2y +3 2 + 3 y y Maka: f (x) =
120
1 2x + 3
Ingat-ingat! ax + b −dx + b f (x) = ⇒ f −1 ( x ) = cx + d cx − a f (x) =
1 −3x + 1 ⇒ f −1 ( x ) = 2x + 3 2x
Sehingga: −3a + 1 2a −3a + 1 −1 = 2a −2a = −3a + 1 a=1
f −1 ( a ) =
Jawaban: C CERDIK: 6. PEMBAHASAN 1.
(f f )(x) = 4x − 3 f (f (x)) = 4x − 3 f (ax + 3) = 4x − 3 Misal: k = ax + 3 k −3 =x a Sehingga: k − 3 f (k ) = 4 −3 a 4k − 12 f (k ) = −3 a 4x − 12 f (x) = −3 a 4x − 12 ax + 3 = −3 a 4x 12 ax + 3 = − −3 a a
121
4k − 12 −3 a 4x − 12 f (x) = −3 a 4x − 12 ax + 3 = −3 a 4x 12 ax + 3 = − −3 a a f (k ) =
Berdasarkan persamaan di atas dapat disimpulkan: 4x 12 • ax = • 3 = − −3 a a −12 − 3a a2 x = 4x 3= a a2 = 4 3a = − 12 − 3a a = ±2 6a = −12 a = −2 Maka, nilai a yang memenuhi adalah a = –2. Sehingga, f (x) = −2x + 3 . Jadi nilai f(a) adalah f (−2) = −2 (−2) + 3 = 4+3 =7 Jawaban: D
CERDIK: 7. PEMBAHASAN 1. 2x − 1 f −1 (1 − x) = 1−x Misal: y = 1 − x maka x = 1 − y f −1 ( y ) =
2 (1 − y ) − 1 1 − (1 − y )
2 − 2y − 1 1−1+ y −2y + 1 = y =
122
f −1 (x) =
−2x + 1 1 maka f(x) = x x+2
Selanjutnya akan dihitung nilai
f ( x − 2) − f
−1
(x)
=
2
f ( x − 2 ) − f −1 ( x ) 2
(−2x + 1) 1 − x ( x − 2) + 2 2
1 − x =
(−2x + 1) 2
x
2x = x 2 =1 Jawaban: B CERDIK: 8. PEMBAHASAN 1. f ( x 2 − 1) f (1 − x 2 )
=
bx
2
−1
b 1− x
=b
(
x 2 −1 − 1 − x 2 2
−2
2
−1
= b2x
(
2
= bx
)
)
2
= f ( x 2 − 1) ⋅ f ( x 2 − 1)
Jawaban: C
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CERDIK: 9. PEMBAHASAN 1. x − 2 f ⋅ g = (x 2 − 1) x + 1 x − 2 = (x − 1)(x + 1) x + 1 = (x − 1)(x − 2) Domain dari f ⋅ g = {x | −∞ < x < ∞} . Jawaban: A CERDIK: 10. PEMBAHASAN 1. Diketahui: x
0
1
2
3
f(x)
1
3
1
–1
g(x)
2
0
1
2
•
(fogof)(1) = f(g(f(1)))
•
= f(g(3)) = f(2) =1 (gofog)(2) = g(f(g(2))) = g(f(1)) = g(3)
= 2 (fogof)(1) + (gofog)(2) =1 + 2 =3 Jawaban: D
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CERDIK: 11. PEMBAHASAN 1. (1) (f g)(x) = f (g (x)) =
1 3 x −3 8
(2) (f g)(2) =
1 3 ⋅ 2 − 3 = −2 8
(3) (f g)(x) =
1 3 x −3 8
11 3 yy = = xx 3 − −3 3 8 8 1 +3 3= = 1 xx 33 yy + 8 8 3 8y + 24 = x 8y + 24 = x 3 = 33 8y + 24 24 xx = 8y +
−1
⇒ ( f g) −1
(4) (f g)
(x) = 3 8x + 24
(−2) = 3 8 (−2) + 24 = 3 8 = 2
Diperoleh (1), (2), (3), dan (4) benar semua. Jawaban: E CERDIK: 12. PEMBAHASAN 1.
f(x) = 2x − 6 dan g −1 (x) =
x−5 4
x−5 4 x−5 Misal: y = 4 •
g −1 (x) =
4y= x − 5 4y + 5 = x Maka: g(x) = 4x + 5
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•
g (2) = 4 ⋅ 2 + 5 = 13
•
f (13) = 2 (13) − 6 = 20
•
(f g)(2) = f (g (2)) = f (13) = 20 Jawaban: A
CERDIK: 13. PEMBAHASAN 1. • f(x − 2) = 3x 2 − 16x + 26 Misal: y= x − 2
y+2 = x f(y) = 3(y + 2)2 − 16(y + 2) + 26 f(y)= 3(y 2 + 4y + 4) − 16y − 32 + 26 f(y) = 3y 2 − 4y + 6
f(x) = 3x 2 − 4x + 6
= ax − 1 • g(x)
g(3) = 3a − 1
•
(f g)(3) = 61 f (g (3)) = 61 2
3 (3a − 1) − 4 (3a − 1) + 6 = 61 3 (9a2 − 6a + 1) − 12a + 4 + 6 = 61 27a2 − 30a − 48 = 0 9a2 − 10a − 16 = 0
(9a + 8)(a − 2) = 0
a= −
8 atau a = 2 9 Jawaban: D
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CERDIK: 14. PEMBAHASAN 1. g(x)= 4 − x 2 dan= f(g(x))
2 − x2 ,x ≠ 0 4x 2
2 − x2 f(4 − x 2 ) = 2 4x
Misal: y= 4 − x 2 maka= x f(y) =
2− 4
(
(
4−y 4−y
)
)
4 − y diperoleh
2
sehingga f(x) =
2
2− 4
(
(
4−x 4−x
)
)
2
2
2
1 f = 2
1 2 − 4 − 2
7 3 − 3 2 = = 2 =− 7 14 28 4 2 2−
2
1 4 4 − 2
2
1 f = 4
1 2 − 4 − 4 2
1 4 4 − 4
15 7 − 4 = 4 =− 7 = 15 15 60 4 4 2−
Maka: 1 1 7 3 1 1) f ⋅ f = − ⋅ − = 4 2 60 28 80 1 1 7 3 2) f + f = − + − 4 2 60 28 −49 − 45 = 420 −94 −47 = = 420 210
127
1 1 7 3 3) f − f = − − − 4 2 60 28 −49 + 45 420 −4 −1 = = 420 105 =
1 3 f − 28 4) 2 = 1 7 − f 4 60 3 60 45 =− ⋅ − = 28 7 49 Pernyataan (1), (2), (3), dan (4) benar (semua benar). Jawaban: E CERDIK: 15. PEMBAHASAN 1. ax + 1 • f(x) = , g(x)= x − 2 3x − 1 −1 −1 −1 • (g of )(x) = (fog) (x) (fog)(x) = f(g(x))
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=
a(x − 2) + 1 3(x − 2) − 1
=
ax − 2a + 1 3x − 7
7 • (g −1 of −1 )(2) = 2 Trik Praktis:
f(a) = b
f −1 (b) = a 7 2
(fog) −1 (2) =
7 (fog) = 2 2
7 a − 2a + 1 2 =2 7 3 − 7 2
3 a+1 2 =2 7 2
3 a+1 = 7 2
3 a=6 2
a=4 Jawaban: E
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