2. Nilai x real yang memenuhi (f o f)(x) = 0, jika f (x) = x^2 – 4 adalah . . .
3. F (x) = x – 1 / x dan g (x) = x + 3. Rumus (g(f (x)))^-1 = . . .
4. Jika f (x) = 1 / x + 1 dan g (x) = 2 / 3 – x, maka (f o g)^-1 (x) = . . .
5. Jika g (x) = x + 2, (f o g)(x) = 3x – 1 / 5 – 2x dan f^-1 (a) = 2, maka a = . . .
6. Jika (f o g)(x) = x^2 + 8x – 3 dan g (x) = x + 4, maka f^-1 (x) = . . .
1. Diketahui f (x) = x^2 – nx dan g (x) = 3x + 14. Nilai n yang memenuhi (f o g)(-4) + 10 = (g o f)(2) adalah . . .
Fungsi Komposisi dan Fungsi Invers
1.
fog(-4) + 10 = (gof)(2)
f{g(-4)} + 10 = g{f(2)}
f(3(-4)+14) + 10 = g(2² – 2n)
f(-2) + 10 = g(4- 2n)
(-2)² – n(-2) + 10 = 4 – 2n + 14
4 + 2n + 10 = 4 – 2n + 14
2n + 14 = -2n + 18
2n+ 2n = 18 -14
4n = 4
n = 1
2) f(x)= x² – 4
fof(x)= 0
f{ f(x)}= 0
f(x² -4)=0
(x²- 4)² – 4 = 0
(x²-4)² = 4
x² – 4= 2 atau x² – 4 = – 2
x² = 6 atau x² = 2
x = √6 atau x = √2
3) f(x)= (x-1)/x , g(x) = x+ 3
{g(f(x))}⁻¹ (x) = (gof)⁻¹
gof(x) = g {f(x)}= (x -1)/x + 3 = { x – 1 + 3x }/(x)
gof(x) = (4x -1)/ (x)
(gof)⁻¹(x) = 1./(-x +4) atau (gof)⁻¹ (x) = – 1/ (x – 4)
4) f(x) = 1/ (x + 1), g(x) = 2/(3-x)
fog(x) = f (g(x)) = f (2/(3-x)
fog(x) = 1/ {(2/(3-x) + 1)}
fog(x) = 1/ {2 + 3 – x}/(3-x)
fog(x) = (3-x) /(5- x)
fog(x)= (-x + 3)/(-x +5)
(fog)⁻¹(x) = (5x – 3)/(x – 1)
5) g(x) = x+2,
fog(x) =(3x-1)/(-2x + 5)
f{g(x)} = (3x -1)/ (-2x + 5)
f(x+2) = (3x -1)/ (-2x + 5)
f⁻¹(a) = 2
f(2)= a
x + 2 = 2
x= 0
a= (3x -1) /(-2x + 5) untuk x= 0
a = {3(0) -1} / (-2(0) + 5}
a = (0 – 1)/(0 + 5)
a = (-1)/(5)
a = – 1/5
6) fog(x)= x² + 8x – 3, g(x)= x+ 4
f {g(x)} = x² +8x – 3
f(x+4) = x² +8x – 3
f(x+4)= (x + 4)² – 19
f(x)= x² – 19
x² = y + 19
x = √(y+19)
f⁻¹ (x) = √(x + 19)