Materi Fungsi Komposisi dan Fungsi Invers Soal berupa lampiran Jawab dengan penjelasan/langkah

Matematika (Fungsi komposisi dan invers)

f(x) = (2x - 3) / (x + 2)

y (x + 2) = 2x - 3
2x - xy = 2y + 3
x = (2y + 3) / (2 - y)
f`¹(x) = (2x + 3) / (2 - x)

g(x) = 2x - 1

y = 2x - 1
x = (y + 1) / 2
gˉ¹(x) = (x + 1) / 2

Nomor 1
(fog)(x) = f{g(x)}
y = (2g(x) - 3) / (g(x) + 2)
y = (2(2x - 1) - 3) / (2x - 1 + 2)
y (2x + 1) = 4x - 5
4x - 2xy = y + 5
x = (y + 5) / (4 - 2y)

(fog)ˉ¹(x) = (x + 5) / (4 - 2x)

Nomor 2
(gof)(x) = g{f(x)}
y = 2f(x) - 1
y = 2((2x - 3) / (x + 2)) - 1
y = (4x - 6 - (x + 2)) / (x + 2)
y(x + 2) = 3x - 8
3x - xy = 2y + 8
x = (2y + 8) / (3 - y)

(gof)ˉ¹(x) = (2x + 8) / (3 - x)

Nomor 3
(gˉ¹of)(x) = gˉ¹{f(x)}

= (f(x) + 1) / 2
= ((2x - 3) / (x + 2) + 1) / 2
= (2x - 3 + x + 2) / 2(x + 2)
= (3x - 1) / (2x + 4)

Nomor 4
(fogˉ¹)(x) = f{gˉ¹(x)]
y = (2gˉ¹(x) - 3) / (gˉ¹(x) + 2)
y = (2(x+1)/2 - 3) / ((x+1)/2 + 2)
y = (x - 2) / (x + 5)/2
y(x + 5) = 2(x - 2)
xy + 5y = 2x - 4
2x - xy = 5y + 4
x = (5y + 4) / (2 - y)

(fogˉ¹)ˉ¹(x) = (5x + 4) / (2 - x)

Nomor 5
(gofˉ¹)(x) = g{fˉ¹(x)}

= 2fˉ¹(x) - 1
= 2((2x + 3) / (2 - x)) - 1
= (4x + 6 - (2 - x) / (2 - x)
= (5x + 4) / (2 - x)

Nomor 6
(fˉ¹og)(x) = fˉ¹{g(x)]
y = (2g(x) + 3) / (2 - g(x))
y = (2(2x - 1) + 3) / (2 - (2x - 1))
y = (4x + 1) / (3 - 2x)
y(3 - 2x) = 4x + 1
3y - 2xy = 4x + 1
4x + 2xy = 3y + 1
x = (3y + 1) / (4 + 2y)

(fˉ¹og)ˉ¹(x) = (3x + 1) / (4 + 2x)

Nomor 7
(fˉ¹)ˉ¹(x) = f(x)
= (2x - 3) / (x + 2)

Nomor 8
(g`¹)`¹(x) = g(x)
= 2x - 1