Question Number 204671 by mr W last updated on 25/Feb/24
$${if}\:{f}\left({f}\left({x}\right)\right)={x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{4},\:{find}\:{f}\left(\mathrm{1}\right)=? \\ $$
Answered by witcher3 last updated on 25/Feb/24
$$\mathrm{f}\left(\mathrm{f}\left(\mathrm{x}\right)\right)=\mathrm{x}^{\mathrm{2}} −\mathrm{3x}+\mathrm{4} \\ $$$$\mathrm{f}\left(\mathrm{f}\left(\mathrm{1}\right)\right)=\mathrm{2} \\ $$$$\mathrm{f}\left(\left(\mathrm{f}\left(\mathrm{2}\right)\right)=\mathrm{2}\right. \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{f}\left(\mathrm{f}\left(\mathrm{2}\right)\right)=\mathrm{f}\left(\mathrm{2}\right)=\mathrm{f}^{\mathrm{2}} \left(\mathrm{2}\right)−\mathrm{3f}\left(\mathrm{2}\right)+\mathrm{4}\Rightarrow\mathrm{f}\left(\mathrm{2}\right)=\mathrm{2}\right. \\ $$$$\mathrm{f}\left(\mathrm{f}\left(\mathrm{f}\left(\mathrm{1}\right)\right)=\mathrm{f}\left(\mathrm{2}\right)=\mathrm{2}\right. \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{1}\right)^{\mathrm{2}} −\mathrm{3f}\left(\mathrm{1}\right)+\mathrm{4}=\mathrm{2}\Rightarrow\mathrm{f}^{\mathrm{2}} \left(\mathrm{1}\right)−\mathrm{3f}\left(\mathrm{1}\right)+\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{1}\right)\in\left\{\mathrm{1},\mathrm{2}\right\} \\ $$$$\mathrm{f}\left(\mathrm{f}\left(\mathrm{1}\right)\right)=\mathrm{2}\Rightarrow\mathrm{f}\left(\mathrm{1}\right)=\mathrm{2}\:\mathrm{impossible} \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{1}\right)=\mathrm{2} \\ $$
Commented by mr W last updated on 25/Feb/24
$${great}! \\ $$
Answered by A5T last updated on 25/Feb/24
![f(x)=f(y)⇒f(f(x))=f(f(y))⇒x^2 −3x=y^2 −3y ⇒(x−y)(x+y)=3(x−y)⇒x=y or x=3−y ⇒f(x)=f(3−x)⇒f(1)=f(2) f(f(2))=2 f(f(f(2)))=f(2)=[f(2)]^2 −3f(2)+4⇒[f(2)−2]^2 =0 ⇒f(2)=2=f(1)](https://www.tinkutara.com/question/Q204676.png)
$${f}\left({x}\right)={f}\left({y}\right)\Rightarrow{f}\left({f}\left({x}\right)\right)={f}\left({f}\left({y}\right)\right)\Rightarrow{x}^{\mathrm{2}} −\mathrm{3}{x}={y}^{\mathrm{2}} −\mathrm{3}{y} \\ $$$$\Rightarrow\left({x}−{y}\right)\left({x}+{y}\right)=\mathrm{3}\left({x}−{y}\right)\Rightarrow{x}={y}\:{or}\:{x}=\mathrm{3}−{y} \\ $$$$\Rightarrow{f}\left({x}\right)={f}\left(\mathrm{3}−{x}\right)\Rightarrow{f}\left(\mathrm{1}\right)={f}\left(\mathrm{2}\right) \\ $$$${f}\left({f}\left(\mathrm{2}\right)\right)=\mathrm{2} \\ $$$${f}\left({f}\left({f}\left(\mathrm{2}\right)\right)\right)={f}\left(\mathrm{2}\right)=\left[{f}\left(\mathrm{2}\right)\right]^{\mathrm{2}} −\mathrm{3}{f}\left(\mathrm{2}\right)+\mathrm{4}\Rightarrow\left[{f}\left(\mathrm{2}\right)−\mathrm{2}\right]^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{f}\left(\mathrm{2}\right)=\mathrm{2}={f}\left(\mathrm{1}\right) \\ $$
Commented by mr W last updated on 25/Feb/24
$${great}! \\ $$
Answered by es last updated on 25/Feb/24
$${f}\left({x}\right)={f}^{−\mathrm{1}} \left({x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{4}\right)\rightarrow \\ $$$${f}\left(\mathrm{1}\right)={f}^{−\mathrm{1}} \left(\mathrm{2}\right) \\ $$$${f}^{−\mathrm{1}} \left({x}\right)=\frac{\mathrm{3}\pm\sqrt{\mathrm{9}−\mathrm{4}\left(\mathrm{4}−{x}\right)}}{\mathrm{2}} \\ $$$${f}^{−\mathrm{1}} \left(\mathrm{2}\right)=\frac{\mathrm{3}\pm\mathrm{1}}{\mathrm{2}}={f}\left(\mathrm{1}\right)=\begin{cases}{\mathrm{2}}\\{\mathrm{1}}\end{cases} \\ $$$$ \\ $$