mohon bantuan guys yang ini

Mapel : Matematika
Kelas : X SMA
Bab : Fungsi

Pembahasan :
Gof(2x + 5) = 5x + 7

Misal 2x + 5 = a, sehingga x = (a - 5)/2

Gof(a) = 5(a - 5)/2 + 7
Gof(a) = (5a - 25)/2 + 7
Gof(a) = (5a - 11)/2
G(f(x)) = (5x - 11)/2
f(x) - 1 = (5x - 11)/2
2f(x) - 2 = 5x - 11
2f(x) = 5x - 9
f(x) = (5x - 9)/2
y = (5x - 9)/2
2y = 5x - 9
2y + 9 = 5x
x = (2y + 9)/5
f¯¹(x) = (2x + 9)/5
f¯¹(-12) = (2.-12 + 9)/5
f¯¹(-12) = (-24 + 9)/5
f¯¹(-12) = -15/5 = -3

(fog)(2x + 5) = 5x + 7
g(x) = x - 1

misal :
2x + 5 = a
x = (a-5)/2

(fog)(a) = 5(a-5 / 2) + 7
(fog)(a) = (5a - 25 + 14)/2
(fog)(a) = (5a - 11)/2

(fog)(x) = f{g(x)}
(5x - 11)/2 = f(x - 1)

misal :
x - 1 = a
x = a + 1

soal menjadi →
f(a) = (5(a+1) - 11) / 2
f(a) = (5a + 5 - 11) / 2

f(x) = (5x - 6) / 2
y = (5x - 6) / 2
2y = 5x - 6
5x = 2y + 6
x = (2y + 6) / 5

f`¹(x) = (2x + 6) / 5
f`¹(-12) = (2 (-12) + 6) / 5
f`¹(-12) = (-24 + 6) / 5
f`¹(-12) = (-15) / 5
f`¹(-12) = -3

Opsi B

semoga berguna +_+