Question

Sertakan cara pembahasan.

Answer 1

F(x)= (2x+1) / (x-3)

y = (2x+1) / (x-3)
y(x-3) = 2x + 1
xy - 3y = 2x + 1
xy - 2x = 3y + 1
x(y-2) = 3y + 1
x = (3y + 1). (y-2)
f⁻¹ (x)= (3x + 1) /(x -2)

f⁻¹ (x+1) = {3(x+1) + 1} / ( x+1-2)
f⁻¹ (x+1) = (3x + 4)/(x - 1)

f⁻¹ o f(x+1) = f⁻¹ { f(x) } = f⁻¹ {(2x+1)/(x-3)}
= { 3 (2x+1)/(x-3)  + 4 }/ { (2x+1)/(x-3) - 1 }
= { (6x +3)/(x-3) + 4)} / ((2x+1)/(x-3) - 1}
= { 6x + 3 + 4(x-3) )/ (x-3) } / {(2x+1- x + 3)/(x-3)
= {(10x  -9) /(x-3) } / {(x +4)/(x-3)
= ( 10x - 9)/ (x+ 4)

b) f(x+1) =  { 2(x+1)+1 / (x+1) -3}
f(x+1) = (2x+3)/(x -2)

f⁻¹o f (x+1) = f⁻¹  { (2x+3)/(x-2)} = 
= { 3 (2x+3)/(x-2)   -1}/ {(2x+3)/(x-2) - 2}--> kalikan (x-2)/(x-2)
= [ 3(2x+3) - x + 2 ] \ [ (2x+3) - (2x - 4) }
= (5x + 11 )/ (7)