Question Number 199167 by tri26112004 last updated on 28/Oct/23
![Give a function f: R→(0;+∞) continous on R and such that f(x+y) = f(x).f(y) a. Prove f(0) = 1 b. Let h(x) = ln[f(x)]. Prove that: h(x+y) = h(x) + h(y) c. Find all the function f such that problem request](https://www.tinkutara.com/question/Q199167.png)
$${Give}\:{a}\:{function}\: \\ $$$${f}:\:{R}\rightarrow\left(\mathrm{0};+\infty\right)\:{continous}\:{on}\:{R}\:{and}\:{such}\:{that} \\ $$$${f}\left({x}+{y}\right)\:=\:{f}\left({x}\right).{f}\left({y}\right) \\ $$$${a}.\:{Prove}\:{f}\left(\mathrm{0}\right)\:=\:\mathrm{1} \\ $$$${b}.\:{Let}\:{h}\left({x}\right)\:=\:{ln}\left[{f}\left({x}\right)\right].\:{Prove}\:{that}: \\ $$$$\:{h}\left({x}+{y}\right)\:=\:{h}\left({x}\right)\:+\:{h}\left({y}\right) \\ $$$${c}.\:{Find}\:{all}\:{the}\:{function}\:{f}\:{such}\:{that}\:{problem}\:{request} \\ $$$$\:\:\: \\ $$$$\: \\ $$
Answered by AST last updated on 28/Oct/23
![a.f(0)=[f(0)]^2 ⇒f(0)=0 or 1 but f(0)≠0⇒f(0)=1 b. h(x+y)=In[f(x+y)]=In[f(x).f(y)] =In[f(x)]+In[f(y)]=h(x)+h(y) c. e^(h(x)) =f(x) for continuous h(x) Or generally, a^(h(x)) for continuous h(x) where a>1](https://www.tinkutara.com/question/Q199169.png)
$${a}.{f}\left(\mathrm{0}\right)=\left[{f}\left(\mathrm{0}\right)\right]^{\mathrm{2}} \Rightarrow{f}\left(\mathrm{0}\right)=\mathrm{0}\:{or}\:\mathrm{1}\:{but}\:{f}\left(\mathrm{0}\right)\neq\mathrm{0}\Rightarrow{f}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$${b}.\:{h}\left({x}+{y}\right)={In}\left[{f}\left({x}+{y}\right)\right]={In}\left[{f}\left({x}\right).{f}\left({y}\right)\right] \\ $$$$={In}\left[{f}\left({x}\right)\right]+{In}\left[{f}\left({y}\right)\right]={h}\left({x}\right)+{h}\left({y}\right) \\ $$$${c}.\:{e}^{{h}\left({x}\right)} ={f}\left({x}\right)\:{for}\:{continuous}\:{h}\left({x}\right) \\ $$$${Or}\:{generally},\:{a}^{{h}\left({x}\right)} \:{for}\:{continuous}\:{h}\left({x}\right) \\ $$$${where}\:{a}>\mathrm{1} \\ $$
Commented by tri26112004 last updated on 28/Oct/23
$${Thank}\:{you}\:{so}\:{much}! \\ $$$$ \\ $$