Diketahui g(x)=2x-3 , (fog)(x)=12xkuadrat-38x+32. Rumus fungsi f(x)?

Mapel : Matematika
Kelas : X SMA
Bab : Fungsi Komposisi dan Fungsi Invers

Pembahasan :
Fog(x) = f(g(x))
12x² - 38x + 32 = f(2x - 3)

Misal 2x - 3 = a
x = (a + 3)/2

f(a) = 12[(a + 3)/2]² - 38[(a + 3)/2] + 32
f(a) = 3(a² + 6a + 9) - 19(a + 3) + 32
f(a) = 3a² + 18a + 27 - 19a - 57 + 32
f(a) = 3a² - a + 2
f(x) = 3x² - x + 2

g(x)= 2x-3, fog(x)= 12x^2-38x+32
     fog(x)= 12x^2-38x+32
   f(g(x)) = 12x^2-38x+32
   f(2x-3)= 12x^2-38x+32
             = 12(x+3/2)^2 -38(x+3/2)+32
             = 12(x^2+6x+9/4)-19(x+3)+32
             = 3(x^2+6x+9) - 19x-57+32
             = 3x^2 +18x+27-19x-57+32
             = 3x^2-x+2