Penjelasan dengan langkah-langkah:

Invers Fungsi menggunakan rumus cepat untuk invers tunggal / SPLSV

f(x) = ax + b → f-¹(x) = (x – b)

a

g(x) = kx – m → g-¹(x) = (x + m)

k

h(x) = a.x + b → h-¹(x) = m.x + b

kx – m kx – a

Maka,

f(x) = 3x + 1

a) f-¹(x) = ⅓(x – 1)

g(x) = 2x – 5

b) g-¹(x) = ½(x + 5)

h(x) = 3x + 1

2x – 5

c) h-¹(x) = 5x + 1

2x – 3

d) (f-¹ o g-¹)(x) = ⅓[½(x+5) – 1]

(f-¹ o g-¹)(x) = ⅓[½x+5/2 – 2/2]

(f-¹ o g-¹)(x) = ⅓[½x + 3/2)

(f-¹ o g-¹)(x) = ⅓[½(x + 3)]

(f-¹ o g-¹)(x) = [x + 3]

6

e) (g o f)(x) = 2(3x + 1) – 5

(g o f)(x) = 6x + 2 – 5

(g o f)(x) = 6x – 3

(g o f)-¹(x) = x + 3

6

kesimpulan :

(f-¹ o g-¹)(x) = (g o f)-¹(x)