Latihan Soal Fungsi Komposisi 1) Fungsi-fungsi f dan g : R ke R ditentukan oleh f(x) = 1 + 1/x dan g(x) = 2x - 4. Tentukan nilai (f⁻¹ o g⁻¹)(6) 2) Diketahui (go

(gof)(x) = 2(1 + 1/x) - 4
= 2/x - 2
(gof)x atau y
y = 2/x - 2
y+2=2/x
x= 2/y+2
(gof)^-1(x) = 2/x+2
(gof)^-1(6)=2/6 + 2 = 2 1/3
(gof)^-1 (x) = (f^-1 o g^-1)(x)
oh ya... bisa
2.(goh)^-1(x) =2x - 7; h(x) = 3x +2 ------ (soalnya ya)
h^-1(x) = (x-2)/3
(goh)^-1(x) = 2x - 7
(h^-1of^-1)(x) = 2x -7 (anggap f^-1(x) = p), maka
(p - 2)/3 = 2x - 7
(p - 2) = 6x - 21
p = 6x - 19
nilai (g^-1 o h^-1)(x) =
6((x-2)/3) - 19
=2x - 4 - 19
= 2x - 23
(mudah2an membantu)

Bab Fungsi
Matematika SMA Kelas XI

1] (f^-1 o g^-1) (x) = (gof)^-1 (x)
(gof) (x) = 2 (1 + 1/x) - 4
(gof) (x) = 2 + 2/x - 4
(gof) (x) = 2/x - 2
(gof) (x) = (2 - 2x)/x

✩ h (x) = (ax + b)/(cx + d)
h^-1 (x) = (-dx + b)/(cx + d) ✩

(gof) (x) = (2 - 2x)/(x + 0)
(gof)^-1 (x) = (-2x)/(x - 2)
(gof)^-1 (6) = (-2 . 6)/(6 - 2)
(gof)^-1 (6) = -12/4
(gof)^-1 (6) = -3

2] (g^-1 o h^-1) (x) = (hog)^-1 (x)

((goh)^-1)^-1 (x) = (goh) (x)

(goh)^-1 (x) = 2x - 7
y = 2x - 7
2x = y + 7
x = (y + 7)/2
(goh) (x) = (x + 7)/2

(goh) (x) = g (3x + 2) = (x + 7)/2
g(3 . x/3 + 2) = (x/3 + 7)/2
g(x + 2) = (x + 21)/(3 . 2)
g(x - 2 + 2) = (x - 2 + 21)/6
g(x) = (x + 19)/6

(hog) (x) = h((x + 19)/6)
(hog) (x) = 3 ((x + 19)/6) + 2
(hog) (x) = (x + 19)/2 + 4/2
(hog) (x) = (x + 19 + 4)/2
(hog) (x) = (x + 23)/(0x + 2)
(hog)^-1 (x) = (-2x + 23)/-1
(hog)^-1 (x) = 2x - 23