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

Answered by Berbere last updated on 08/Jul/24

f(x+y)=f(x).f(y) f(x)=f((x/2)+(x/2))=(f((x/2)))^2 ≥0 ∀x∈R f(x)≥0 f(x+y)=f(x)f(y) ∀y∈R fixe x→f(x+y) est derivable ⇒f′(x+y)=f′(x)f(y)⇒f′(x)=f′(x)f(0) si f(0)=0⇒f′(x)=0⇒f(x)=c ⇒c=f(0)=0⇒f=0_R f′(y+x_0 )=f (x_0 )f′(y) ⇒f′(x_0 )=f(x_0 )f′(0) ⇒f est solution? { ((y′=f′(0)y)),((y(0)=1)) :} y=ke^(f′(0)x) y(0)=1⇒k=1 f(x)=e^(f′(0)x) ⇒f(x)=e^(ax)

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

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