Kak tolong bantu yaa jwab soalnya

komposisi fungsi dan fungsi invers

14. f(x) = (2x + 1) /(3x)
f⁻¹(a) = 1/2 --> f(1/2) = a

a= {2(1/2)  +1}/{3(1/2)}
a = (1+1) /(3/2)
a = 2/(3/2) = 4/3

15. karena (fog)⁻¹ (x ) = (g⁻¹ o f⁻¹)(x)
maka (fog)⁻¹(6) = (g⁻¹ o f⁻¹ )(6)

g⁻¹ {f⁻¹(6)} = g⁻¹ {(6-1)/5} = g⁻¹(5/5) = g⁻¹(1)
g⁻¹(1) = (3 -1)/(2) = 2/2 = 1
(fog)⁻¹(6) = 1

16. f⁻¹og⁻¹(x) = (gof)⁻¹
gof(x) = g {f(x)}= g {x²+2x + 1} = 5(x² +2x +1) +1
gof(x) = 5x² + 10x + 5 + 1
(gof)(x) = 5x² +10x + 6
5x²+10x = y - 6
x² + 2 x = 1/5 (y - 6)
(x +1)² = 1/5 y - 6/5 + 1 = 1/5 y - 1/5
x+ 1 = √(1/5 y - 1/5)
x = - 1 +-√(1/5y  - 1/5)
(gof)⁻¹(x) = - 1 +- √{(x-1)/5} dengan x ≥1