Question Number 123267 by mohammad17 last updated on 24/Nov/20
$${if}\:{f}\left({x}\right)=\mid\mathrm{3}−{x}\mid\:{find}\:\left({f}^{\:\mathrm{2}} \right)^{'} \left(\mathrm{1}\right)\:\:? \\ $$
Commented by MJS_new last updated on 24/Nov/20
$$\mathrm{if}\:\mathrm{you}\:\mathrm{mean}\:\left(\left({f}\left({x}\right)\right)^{\mathrm{2}} \right)' \\ $$$$\left({f}\left({x}\right)\right)^{\mathrm{2}} =\left(\mathrm{3}−{x}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{9} \\ $$$$\left(\left({f}\left({x}\right)\right)^{\mathrm{2}} \right)'=\mathrm{2}{x}−\mathrm{6} \\ $$$$\mathrm{then}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{is}\:−\mathrm{4} \\ $$
Answered by TANMAY PANACEA last updated on 24/Nov/20
$${f}\left({x}\right)=\mathrm{3}−{x}\:\:{when}\:\left(\mathrm{3}−{x}\right)>\mathrm{0}\:\:\:\:\mathrm{3}>{x} \\ $$$${f}\left({x}\right)=−\mathrm{3}+{x}\:\:\:{when}\:{x}>\mathrm{3} \\ $$$${when}\:{x}<\mathrm{3}\:\:\left({when}\:{x}=\mathrm{1}\right) \\ $$$${f}\left({x}\right)=\mathrm{3}−{x} \\ $$$$\frac{{df}}{{dx}}=−\mathrm{1} \\ $$$$\frac{{d}^{\mathrm{2}} {f}}{{dx}^{\mathrm{2}} }=\mathrm{0} \\ $$$${i}\:{think}\:{so} \\ $$$$ \\ $$
Answered by mathmax by abdo last updated on 24/Nov/20
$$\mathrm{f}\left(\mathrm{x}\right)=\mid\mathrm{x}−\mathrm{3}\mid\:\Rightarrow\mathrm{f}^{\mathrm{2}} \left(\mathrm{x}\right)\:=\left(\mathrm{x}−\mathrm{3}\right)^{\mathrm{2}} \:=\mathrm{x}^{\mathrm{2}} −\mathrm{6x}+\mathrm{9}\:\Rightarrow \\ $$$$\left(\mathrm{f}^{\mathrm{2}} \right)^{'} \left(\mathrm{x}\right)=\mathrm{2x}−\mathrm{6}\:\Rightarrow\left(\mathrm{f}^{\mathrm{2}} \right)^{'} \left(\mathrm{1}\right)=−\mathrm{4} \\ $$