Find g(x) = f o f o … o f(x) = f¹⁹⁹⁴ (x), where f(x) = (x√3 - 1) / (x + √3)?

Answer 1

For this type of problems involving composit functions of the form, f of f of f of ..of(x), trick is to look for the repeating pattern after evaluating a first few compositions.

f²(x) = f of f(x) =f((x√3 - 1) / (x + √3)) = [√3(x√3 - 1) -(x + √3)]/[(x√3 - 1) +√3(x + √3)]

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

f³(x) = f(f²(x))=[√3(x-√3) -(√3x+1)]/[(x-√3) + √3(√3x+1)] = (-3 -1)/(x + 3x) = -1/x

f⁴(x) = f(f³(x)) = (-√3/x -1)/(-1/x + √3) =- (x+√3)/( √3 x -1)

f⁵(x) = f(f⁴(x)) = [-√3(x+√3) -(√3x -1)]/[-(x+√3)+√3(√3x -1)] = -(√3 x + 1)/(x - √3)

f⁶(x) = f(f⁵(x)) =[-√3(√3 x + 1) -(x-√3)]/[-(√3 x +1) + √3(x-√3)] = [-3x -√3 -x +√3]/[-√3x -1 +√3 x -3] =-4x/(-4) =x

f⁷(x) = f(f⁶(x)) = f(x)

i.e. after every 6 compositions the function repeats itself

ie. f(x) = f⁷(x) =f¹³(x) =f¹⁹(x) =..

x =f⁶(x) = f¹²(x) =f¹⁸(x) = ...

1998 = 333 * 6.

g(x) = f¹⁹⁹⁸(x) = x.