Fungsi f: R -> R dan g: R -> R ditentukan oleh f(x)=x+6 dan

(f o g)^-1(a+1) = (a - 6)^(1/3) dan (g^-1 o f^-1)(a+1) = (a - 6)^(1/3)

Pembahasan

Untuk menghitung (f o g)^-1(a+1) dan (g^-1 o f^-1)(a+1), kita perlu mencari invers dari masing-masing fungsi terlebih dahulu.

Fungsi f(x) = x + 6
Misalkan y = x + 6
x = y - 6
Maka, f^-1(x) = x - 6

Fungsi g(x) = x^3 + 1
Misalkan y = x^3 + 1
y - 1 = x^3
x = (y - 1)^(1/3)
Maka, g^-1(x) = (x - 1)^(1/3)

a. (f o g)(x) = f(g(x)) = f(x^3 + 1) = (x^3 + 1) + 6 = x^3 + 7
Untuk mencari inversnya, misalkan y = x^3 + 7
y - 7 = x^3
x = (y - 7)^(1/3)
Maka, (f o g)^-1(x) = (x - 7)^(1/3)
Jadi, (f o g)^-1(a+1) = ((a+1) - 7)^(1/3) = (a - 6)^(1/3).

b. (g^-1 o f^-1)(x) = g^-1(f^-1(x)) = g^-1(x - 6) = ((x - 6) - 1)^(1/3) = (x - 7)^(1/3)
Jadi, (g^-1 o f^-1)(a+1) = ((a+1) - 7)^(1/3) = (a - 6)^(1/3).

Kesimpulan: 
a. (f o g)^-1(a+1) = (a - 6)^(1/3)
b. (g^-1 o f^-1)(a+1) = (a - 6)^(1/3)