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Question Number 169922 by cortano1 last updated on 12/May/22

  Let f(x)=((2x−7)/(x+1)) . Compute f^(1989) (x).   note f^2 (x)= f(f(x))

$$\:\:{Let}\:{f}\left({x}\right)=\frac{\mathrm{2}{x}−\mathrm{7}}{{x}+\mathrm{1}}\:.\:{Compute}\:{f}^{\mathrm{1989}} \left({x}\right). \\ $$$$\:{note}\:{f}^{\mathrm{2}} \left({x}\right)=\:{f}\left({f}\left({x}\right)\right) \\ $$

Answered by floor(10²Eta[1]) last updated on 12/May/22

f(f(x))=f(((2x−7)/(x+1)))=((2((2x−7)/(x+1))−7)/(((2x−7)/(x+1))+1))=((−3x−21)/(3x−6))=((−x−7)/(x−2))  f^3 (x)=f(((−x−7)/(x−2)))=((−2((x+7)/(x−2))−7)/(((−x−7)/(x−2))+1))=((9x)/(−9))=−x  ⇒f^6 (x)=f^3 (−x)=x  ⇒f^n (x)=−x, n=3a, a is odd  ⇒f^n (x)=x, n=3b, b is even  ⇒f^(1989) (x)=−x

$$\mathrm{f}\left(\mathrm{f}\left(\mathrm{x}\right)\right)=\mathrm{f}\left(\frac{\mathrm{2x}−\mathrm{7}}{\mathrm{x}+\mathrm{1}}\right)=\frac{\mathrm{2}\frac{\mathrm{2x}−\mathrm{7}}{\mathrm{x}+\mathrm{1}}−\mathrm{7}}{\frac{\mathrm{2x}−\mathrm{7}}{\mathrm{x}+\mathrm{1}}+\mathrm{1}}=\frac{−\mathrm{3x}−\mathrm{21}}{\mathrm{3x}−\mathrm{6}}=\frac{−\mathrm{x}−\mathrm{7}}{\mathrm{x}−\mathrm{2}} \\ $$$$\mathrm{f}^{\mathrm{3}} \left(\mathrm{x}\right)=\mathrm{f}\left(\frac{−\mathrm{x}−\mathrm{7}}{\mathrm{x}−\mathrm{2}}\right)=\frac{−\mathrm{2}\frac{\mathrm{x}+\mathrm{7}}{\mathrm{x}−\mathrm{2}}−\mathrm{7}}{\frac{−\mathrm{x}−\mathrm{7}}{\mathrm{x}−\mathrm{2}}+\mathrm{1}}=\frac{\mathrm{9x}}{−\mathrm{9}}=−\mathrm{x} \\ $$$$\Rightarrow\mathrm{f}^{\mathrm{6}} \left(\mathrm{x}\right)=\mathrm{f}^{\mathrm{3}} \left(−\mathrm{x}\right)=\mathrm{x} \\ $$$$\Rightarrow\mathrm{f}^{\mathrm{n}} \left(\mathrm{x}\right)=−\mathrm{x},\:\mathrm{n}=\mathrm{3a},\:\mathrm{a}\:\mathrm{is}\:\mathrm{odd} \\ $$$$\Rightarrow\mathrm{f}^{\mathrm{n}} \left(\mathrm{x}\right)=\mathrm{x},\:\mathrm{n}=\mathrm{3b},\:\mathrm{b}\:\mathrm{is}\:\mathrm{even} \\ $$$$\Rightarrow\mathrm{f}^{\mathrm{1989}} \left(\mathrm{x}\right)=−\mathrm{x} \\ $$

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