Question Number 146523 by mnjuly1970 last updated on 13/Jul/21
Answered by ajfour last updated on 13/Jul/21
![f(((x+y)/2))=((2+f(x)+f(y))/3) f(0)=2 f(x)=2x+2 3f(x+y)=2+f(2x)+f(2y) ⇒ 3[2(x+y)+2]=2+4x+2 +4y+2 ⇒ x+y=0 so y=−x ⇒ 3f(0)=2+f(2x)+f(−2x) but f(0)=2 ⇒ f(2x)+f(−2x)=4 2f ′(2x)−2f ′(−2x)=0 ⇒ f ′(2x)=f ′(−2x) g(x)=∣f(∣x∣)−3∣ =∣2∣x∣+2−3∣ =∣2∣x∣−1∣ g(x) isn′t differentiable at x=±(1/2) , x=0](Q146530.png)
$${f}\left(\frac{{x}+{y}}{\mathrm{2}}\right)=\frac{\mathrm{2}+{f}\left({x}\right)+{f}\left({y}\right)}{\mathrm{3}} \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{2} \\ $$$${f}\left({x}\right)=\mathrm{2}{x}+\mathrm{2} \\ $$$$\mathrm{3}{f}\left({x}+{y}\right)=\mathrm{2}+{f}\left(\mathrm{2}{x}\right)+{f}\left(\mathrm{2}{y}\right) \\ $$$$\Rightarrow \\ $$$$\mathrm{3}\left[\mathrm{2}\left({x}+{y}\right)+\mathrm{2}\right]=\mathrm{2}+\mathrm{4}{x}+\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{4}{y}+\mathrm{2} \\ $$$$\Rightarrow\:\:{x}+{y}=\mathrm{0} \\ $$$${so}\:{y}=−{x}\:\:\Rightarrow \\ $$$$\mathrm{3}{f}\left(\mathrm{0}\right)=\mathrm{2}+{f}\left(\mathrm{2}{x}\right)+{f}\left(−\mathrm{2}{x}\right) \\ $$$${but}\:\:{f}\left(\mathrm{0}\right)=\mathrm{2}\:\:\Rightarrow \\ $$$${f}\left(\mathrm{2}{x}\right)+{f}\left(−\mathrm{2}{x}\right)=\mathrm{4} \\ $$$$\mathrm{2}{f}\:'\left(\mathrm{2}{x}\right)−\mathrm{2}{f}\:'\left(−\mathrm{2}{x}\right)=\mathrm{0} \\ $$$$\Rightarrow\:{f}\:'\left(\mathrm{2}{x}\right)={f}\:'\left(−\mathrm{2}{x}\right) \\ $$$${g}\left({x}\right)=\mid{f}\left(\mid{x}\mid\right)−\mathrm{3}\mid \\ $$$$\:\:\:\:\:\:\:\:\:=\mid\mathrm{2}\mid{x}\mid+\mathrm{2}−\mathrm{3}\mid \\ $$$$\:\:\:\:\:\:\:\:\:=\mid\mathrm{2}\mid{x}\mid−\mathrm{1}\mid \\ $$$${g}\left({x}\right)\:{isn}'{t}\:{differentiable}\:{at} \\ $$$$\:\:{x}=\pm\frac{\mathrm{1}}{\mathrm{2}}\:,\:{x}=\mathrm{0} \\ $$
Commented by ajfour last updated on 13/Jul/21
Commented by mnjuly1970 last updated on 14/Jul/21
$${thank}\:{you}\:{so}\:{much}... \\ $$
Answered by mathmax by abdo last updated on 13/Jul/21
$$\mathrm{y}=\mathrm{x}\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{2}+\mathrm{2f}\left(\mathrm{x}\right)}{\mathrm{3}}\:\Rightarrow\mathrm{3f}\left(\mathrm{x}\right)=\mathrm{2}+\mathrm{2f}\left(\mathrm{x}\right)\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\mathrm{2}\:\:\mathrm{how}\:\mathrm{to}\:\mathrm{get} \\ $$$$\mathrm{f}^{'} \left(\mathrm{2}\right)=\mathrm{2}\:\:\mathrm{if}\:\mathrm{f}\:\mathrm{is}\:\mathrm{constant}?\:\:\mathrm{something}\:\mathrm{went}\:\mathrm{wrong}.... \\ $$