Hitunglah nilai komposisi dari : B. (Gof) = xpangkat2 + x - 12 dan f(x) = x - 3 maka g(x) = ..? C. F(x) = akar x+1 , g(x) = xpangkat2 - 2 dan h (x) = akar 1 - 2

B
.
(gof) (x)= x²+x-12
g(x-3)= x²+x-12
g(x)=(x+3)²+(x+3)-12
= x²+6x+9+x-9
=x²+7x

C.
f(x)= √(x+1)
g(x)= x²-2
h(x)= √(1-2x)

(fogoh)(x) =5

(fog)oh) (x)
(fog) (x) = √(g(x) + 1)
=√(x²-2+1)
=√(x²-1)

maka
(fog)oh) (x)
√(h(x) ²-1)
√(√(1-2x)²-1)
√(1-2x)-1)
=√-2x

(fogoh) (x) = 5
√(-2x) =5
-2x=5^½
x= -5^½ / 2

Langsung aja y

Bagian B
(gof)(x) = g{f(x)}
x² + x - 12 = g(x - 3)

misal :
x - 3 = a
x = a+3

soal menjadi →
g(a) = (a+3)² + (a+3) - 12
g(a) = a² + 6a + 9 + a + 3 - 12
g(a) = a² + 7a

g(x) = x² + 7x

Bagian C

misal
(fog)(x) = s(x)

= f{g(x)}
= √g(x) + 1
= √x² - 2 + 1
= √x²-1

(soh)(x) = s{h(x)}

= √h(x) - 1
= √1-2x - 1
= √-2x

(fogoh)(x) = √-2x
5 = √-2x
5² = -2x
-2x = 25
x = 25 : -2
x = -12,5

Semoga berguna +_+