Fungsi f:R->R dan g:R->R ditentukan oleh rumus f(x)=3x+1

(gof)(2) = 33, (fog)(2) = 40

Pembahasan

Untuk menentukan (gof)(2) dan (fog)(2) dengan f(x)=3x+1 dan g(x)=4x+5:
a. (gof)(x) = g(f(x)) = g(3x+1) = 4(3x+1)+5 = 12x+4+5 = 12x+9. Maka, (gof)(2) = 12(2)+9 = 24+9 = 33.
b. (fog)(x) = f(g(x)) = f(4x+5) = 3(4x+5)+1 = 12x+15+1 = 12x+16. Maka, (fog)(2) = 12(2)+16 = 24+16 = 40.