Question

Poin 50 . Tolong Selesain Soal Dari 6 sampai 10 kaka master . Boleh di tulis tangan dan Di foto kaka. terimakasih

Answer 1

8) f(x) = 3x - 5, g(x) = (x - 1)/(2 - x)
(f o g)(x) = f(g(x)) = 3 g(x) - 5
= 3 (x - 1)/(2 - x) - 5
= (3(x - 1) - 5(2 - x)) / (2 - x)
= (3x - 3 - 10 + 5x) / (2 - x)
= (8x - 13)/(2 - x)

7) f(x) = (x + 1)/(x - 3) => f(2) = (2 + 1)/(2 - 3) = 3/(-1) = -3
g(x) = x^2 + x + 1
(g o f)(2) = g(f(2)) = g(-3) = (-3)^2 + (-3) + 1 = 9 - 3 + 1 = 7

9) f(x) = x^2 - 4, g(x) = 2x - 6
(f o g)(x) = -4
=> f(g(x)) = -4
=> f(2x - 6) = -4
=> (2x - 6)^2 - 4 = -4
=> (2x - 6)^2 = 0
=> 2x - 6 = 0
=> 2x = 6
=> x = 3

10) f(x) = x - 2, g(x) = x^2 + 4x - 3
(g o f)(x) = 2
=> g(f(x)) = 2
=> g(x - 2) = 2
=> (x - 2)^2 + 4(x - 2) - 3 = 2
=> x^2 - 4x + 4 + 4x - 8 - 3 = 2
=> x^2 - 9 = 0
=> (x + 3)(x - 3) = 0
=> x = -3 atau x = 3

6) f(x) = 1/x bukan fungsi karena jika x = 0 nilai f(x) tidak terdefinisi => f(0) = 1/0 = ??
f(x) = √x bukan fungsi karena jika x = -1, -4, -9 dst tdk ada nilainya
f(-1) = √-1 = ???
f(-4) = √-4 = ???