Question

p........................​

Answer 1

[tex] \begin{align} (g\circ f)(x) &= -4x^2-12x-8 \\ g(f(x)) &= -4x^2-12x-8 \\ -(f(x))^2+1 &= -4x^2-12x-8 \\ -(f(x))^2 &= -4x^2-12x-9 \\ (f(x))^2 &= 4x^2+12x+9 \:\:\green{, a^2 =b \Leftrightarrow a = \pm \sqrt{b} } \\ f(x) &= \pm \sqrt{4x^2+12x+9} \\ f(x) &= \pm \sqrt{(2x+3)^2} \:\:\green{,\sqrt{a^2}=|a|} \\ f(x) &= \pm\: |2x+3| \end{align} [/tex]

Answer 2

Verified answer

Jawab:
Terdapat dua penyelesaian
f(x) = 2x+3,
f(x) = -2x-3

Penjelasan:
(7.) Tentukan f(x) jika

(b.) (gof)(x) = -4x²-12x-8
g(x) = -x²+1
---------------------------------
g(f(x)) = -4x²-12x-8
Anggap f(x) = u
g(u) = -4x²-12x-8
-u²+1 = -4x²-12x-8
u² = 4x²+12x+8+1
u² = 4x²+12x+9
u² = (2x)²+2x(3)+3²
∵ a²+2ab+b² = (a+b)² ∴
u² = (2x+3)²
|u| = √(2x+3)²
|u| = 2x+3
u = 2x+3,    u = -(2x+3)
u = 2x+3,    u = -2x-3
f(x) = 2x+3, f(x) = -2x-3

(xcvi)