1. Jika fungsi f : R->R dan fungsi g: R->R ditentukan f(x) = x³ dan g(2x-3)=6x-1 maka nilai (g-¹o f-¹)(27) adalah…

Jawaban : (g¯¹of¯¹)(27) = 13/3

Perhatikan penjelasan berikut ya.

Ingat kembali:
1. (fog)(x) = f(g(x))
2. (gof)(x) = g(f(x))
3. f(x) = y → f¯¹(x) = x

Diketahui: f(x) = x³ dan g(2x-3) = 6x – 1

Ditanya : (g¯¹of¯¹)(27) = … ?

Maka:
(i) menentukan f¯¹(27)
f(x) = x³
y = x³
x³ = y
x = ³√y
f¯¹(x) = ³√x
f¯¹(27) = ³√27
f¯¹(27) = 3

(i) menentukan g¯¹(x)
g(2x-3) = 6x – 1

misalkan:
2x – 3 = A
2x = A – 3
x = (A – 3) / 2

g(2x-3) = 6x – 1
g(A) = 6[(A – 3) / 2] – 1
g(A) = 3(A – 3) – 1
g(A) = 3A – 9 – 1
g(A) = 3A – 10
g(x) = 3x – 10
y = 3x – 10
3x – 10 = y
3x = y + 10
x = (y + 10) / 3
g¯¹(x) = (x + 10) / 3

Sehingga:
(g¯¹of¯¹)(27)
= g¯¹(f¯¹(27))
= g¯¹(3)
= (3 + 10) / 3
= 13/3

Jadi, (g¯¹of¯¹)(27) = 13/3.

Semoga membantu.