Question Number 206136 by cortano21 last updated on 07/Apr/24
Answered by A5T last updated on 07/Apr/24
![f(f(f(f(x))))=[f(x)]^2 −3[f(x)]+4=f(x^2 −3x+4) [f(2)]^2 −3[f(2)]+4=f(2)⇒[f(2)]^2 −4f(2)+4=0 [f(2)−2]^2 =0⇒f(2)=2](https://www.tinkutara.com/question/Q206137.png)
$${f}\left({f}\left({f}\left({f}\left({x}\right)\right)\right)\right)=\left[{f}\left({x}\right)\right]^{\mathrm{2}} −\mathrm{3}\left[{f}\left({x}\right)\right]+\mathrm{4}={f}\left({x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{4}\right) \\ $$$$\left[{f}\left(\mathrm{2}\right)\right]^{\mathrm{2}} −\mathrm{3}\left[{f}\left(\mathrm{2}\right)\right]+\mathrm{4}={f}\left(\mathrm{2}\right)\Rightarrow\left[{f}\left(\mathrm{2}\right)\right]^{\mathrm{2}} −\mathrm{4}{f}\left(\mathrm{2}\right)+\mathrm{4}=\mathrm{0} \\ $$$$\left[{f}\left(\mathrm{2}\right)−\mathrm{2}\right]^{\mathrm{2}} =\mathrm{0}\Rightarrow{f}\left(\mathrm{2}\right)=\mathrm{2} \\ $$