Integration Questions

Question Number 209353 by alcohol last updated on 08/Jul/24

Answered by Berbere last updated on 08/Jul/24

$${f}\left({x}+{y}\right)={f}\left({x}\right).{f}\left({y}\right) \\$$$${f}\left({x}\right)={f}\left(\frac{{x}}{\mathrm{2}}+\frac{{x}}{\mathrm{2}}\right)=\left({f}\left(\frac{{x}}{\mathrm{2}}\right)\right)^{\mathrm{2}} \geqslant\mathrm{0} \\$$$$\forall{x}\in\mathbb{R}\:{f}\left({x}\right)\geqslant\mathrm{0} \\$$$${f}\left({x}+{y}\right)={f}\left({x}\right){f}\left({y}\right)\:\forall{y}\in\mathbb{R}\:{fixe}\:{x}\rightarrow{f}\left({x}+{y}\right)\:{est}\:{derivable} \\$$$$\Rightarrow{f}'\left({x}+{y}\right)={f}'\left({x}\right){f}\left({y}\right)\Rightarrow{f}'\left({x}\right)={f}'\left({x}\right){f}\left(\mathrm{0}\right) \\$$$${si}\:{f}\left(\mathrm{0}\right)=\mathrm{0}\Rightarrow{f}'\left({x}\right)=\mathrm{0}\Rightarrow{f}\left({x}\right)={c}\:\Rightarrow{c}={f}\left(\mathrm{0}\right)=\mathrm{0}\Rightarrow{f}=\mathrm{0}_{\mathbb{R}} \\$$$${f}'\left({y}+{x}_{\mathrm{0}} \right)={f}\:\left({x}_{\mathrm{0}} \right){f}'\left({y}\right) \\$$$$\Rightarrow{f}'\left({x}_{\mathrm{0}} \right)={f}\left({x}_{\mathrm{0}} \right){f}'\left(\mathrm{0}\right) \\$$$$\Rightarrow{f}\:{est}\:{solution}?\begin{cases}{{y}'={f}'\left(\mathrm{0}\right){y}}\\{{y}\left(\mathrm{0}\right)=\mathrm{1}}\end{cases} \\$$$${y}={ke}^{{f}'\left(\mathrm{0}\right){x}} \\$$$${y}\left(\mathrm{0}\right)=\mathrm{1}\Rightarrow{k}=\mathrm{1} \\$$$${f}\left({x}\right)={e}^{{f}'\left(\mathrm{0}\right){x}} \Rightarrow{f}\left({x}\right)={e}^{{ax}} \\$$$$\\$$$$\\$$