Question

Tolong bantuin ya
Trimkasih

Answer 1

Milih soal :
1a)
f(x) = - 1/5 x + 3
y = - 1/5 x + 3 .....kalikan 5
5y = - x + 15
x = -5y + 15
f⁻¹(x) = - 5x + 15
f⁻¹(3) = - 5(3) + 15 = -15 +15 --> f⁻¹()= 0

1c)
f(x) = 2/3 x (2x - 1)
 y = 2/3 x (2x-1)....kalikan 3
3 y = 2(2x - 1)
3y = 4x - 2
4x = 3y +2
x = 1/4 (3y + 2)
f⁻¹(x) = ¹/₄ (3x + 2)
f⁻¹(3) = ¹/₄ {3(3) +2)}
f⁻¹(3) = ¹/₄(11) = ¹¹/4

2b)
g(x) = √(2x+1)
daerah asal 2x+1 ≥ 0
2x ≥ - 1
x ≥ - ¹/₂

daerah hasil y ≥ 0

invers
y= √(2x+1) ...kuadratkan kedua ruas
y² = (√(2x+1))²
y² = 2x+ 1
2x = y² -1
x = ¹/₂ (y² -1)
g⁻¹(x)= ¹/₂(x² -1)