+876 A function $f(x)$ is defined by $f(x) = f(x) = \frac{x - \sqrt{3}}{x\sqrt{3} + 1}.$
Find $f^{2021}(x)$
+876 Solution:
$f(x) = \frac{x - \sqrt{3}}{x\sqrt{3} + 1}$
f2(x)=x−√3x√3+1−√3√3(x−√3x√3+1)+1=x−√3√3x+1−√3√3(x−√3)√3x+1+1=−2x−2√3√3x+1√3(x−√3)√3x+1+1=−2x−2√3(x√3+1)((x−√3)√3x√3+1+1)=−2x−2√32√3x−2√3x+1(√3x+1)=−2x−2√32√3x−2=−2(x+√3)2(√3x−1)=−x+√3√3x−1
f3(x)=f(−x+√3√3x−1)=−x−√3x√3+1+√3√3(x−√3x√3+1)−1=x−√3√3x+1+√3√3(x−√3)√3x+1−1=−4x√3x+1√3(x−√3)√3x+1−1=−4x(√3x+1)(√3(x−√3)√3x+1−1)=−4x(−4√3x+1)(√3x+1)=4x(x√3+1)4x√3+1=4x4=x
f4(x)=f(f3(x))=f(x)=x−√3x√3+1
$f^1(x) = \frac{x - \sqrt{3}}{x\sqrt{3} + 1}$
$f^2(x) = -\frac{x+\sqrt{3}}{\sqrt{3}x-1}$
$f^3(x) = x$
$f^4(x) = \frac{x - \sqrt{3}}{x\sqrt{3} + 1}$
$2021 \pmod 3 \equiv 2 \pmod 3$
$f^{2021}(x) = f^2(x) = \boxed{-\frac{x+\sqrt{3}}{\sqrt{3}x-1}}$
