Tentukan (fog)(2) & (gof)(-1) untuk fungsi-fungsi berikut ! a. F(x) = 4x + 1 & g(x) = 3x – 5 b. F(x) = 2x² & g(x) = x + 1

a. Dik : f(x) = 4x + 1
g(x) = 3x - 5
Dit : (fog)(2) & (gof)(-1)

Jawab :
(fog)(x) = 4(3x - 5) + 1
= 12x - 20 + 1
= 12x - 19
(fog)(2) = 12x - 19
= 12(2) - 19
= 24 - 19
= 5
(gof)(x) = 3(4x + 1) - 5
= 12x + 3 - 5
= 12x - 2
(gof)(-1) = 12x - 2
= 12(-1) - 2
= - 12 - 2
= - 14

b. Dik : f(x) = 2x^2
g(x) = x + 1
Dit : (fog)(2) & (gof)(-1)

Jawab :

(fog)(x) = 2(x + 1)^2
= 2(x^2 + 2x + 1)
= 2x^2 + 4x + 2
(fog)(2) = 2x^2 + 4x + 2
= 2(2)^2 + 4(2) + 2
= 8 + 8 + 2
= 18
(gof)(x) = (2x^2) + 1
= 2x^2 + 1
(gof)(-1) = 2(-1)^2 + 1
= 2 + 1
= 3

a.
(fog)(x) = f(g(x))
=f(3x-5)
=4(3x-5)+1
=12x-20+1
=12x-19
(fog)(1)= 12(1)-19 = -7
(gof)(x) = g(4x+1)
=3(4x+1)-5
=12x+3-5
=12x-2
(gof)(-1) = 12(-1)-2 = -14

b.
(fog)(x) = f(x+1)
= 2(x+1)^2
=2(x^2+2x+1)
=2x^2+4x+2
(fog)(1) = 2(1)^2+4(1)+2 = 2+4+2 = 8
(gof)(x) = g(2x^2)
=2x^2+1
(gof)(-1) = 2(-1)^2+1 = 2+1 = 3

maaf kalo hitungannya ada yg keliru..