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f-0-R-f-ux-x-u-f-x-f-x-




Question Number 3878 by 123456 last updated on 23/Dec/15
f:[0,+∞)→R  f(ux)=x^u f(x)  f(x)=?
$${f}:\left[\mathrm{0},+\infty\right)\rightarrow\mathbb{R} \\ $$$${f}\left({ux}\right)={x}^{{u}} {f}\left({x}\right) \\ $$$${f}\left({x}\right)=? \\ $$
Commented by Rasheed Soomro last updated on 23/Dec/15
f(ux)=x^u f(x)  Let x=0    f(0)=0^u f(0)=0  x=1  f(ux)=x^u f(x)⇒f(u)=f(1)  ⇒f(1)=f(u)⇒u=1..............................(i)  So,    f(ux)=x^u f(x)⇒x^u =1⇒x=1  For x=1 and u=1 f(x) may have  any  definition.  For example f(x)=x^2   f(ux)=x^u f(x)⇒(ux)^2 =x^u (x^2 )⇒1=1 for x=1,u=1
$${f}\left({ux}\right)={x}^{{u}} {f}\left({x}\right) \\ $$$${Let}\:{x}=\mathrm{0}\:\: \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{0}^{{u}} {f}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${x}=\mathrm{1} \\ $$$${f}\left({ux}\right)={x}^{{u}} {f}\left({x}\right)\Rightarrow{f}\left({u}\right)={f}\left(\mathrm{1}\right) \\ $$$$\Rightarrow{f}\left(\mathrm{1}\right)={f}\left({u}\right)\Rightarrow{u}=\mathrm{1}…………………………\left({i}\right) \\ $$$${So},\:\:\:\:{f}\left({ux}\right)={x}^{{u}} {f}\left({x}\right)\Rightarrow{x}^{{u}} =\mathrm{1}\Rightarrow{x}=\mathrm{1} \\ $$$$\mathcal{F}{or}\:{x}=\mathrm{1}\:{and}\:{u}=\mathrm{1}\:{f}\left({x}\right)\:{may}\:{have} \\ $$$${any}\:\:{definition}. \\ $$$${For}\:{example}\:{f}\left({x}\right)={x}^{\mathrm{2}} \\ $$$${f}\left({ux}\right)={x}^{{u}} {f}\left({x}\right)\Rightarrow\left({ux}\right)^{\mathrm{2}} ={x}^{{u}} \left({x}^{\mathrm{2}} \right)\Rightarrow\mathrm{1}=\mathrm{1}\:{for}\:{x}=\mathrm{1},{u}=\mathrm{1} \\ $$
Commented by prakash jain last updated on 23/Dec/15
f(x)=0  I think f(x)=0 is only solution.
$${f}\left({x}\right)=\mathrm{0} \\ $$$$\mathrm{I}\:\mathrm{think}\:{f}\left({x}\right)=\mathrm{0}\:\mathrm{is}\:\mathrm{only}\:\mathrm{solution}. \\ $$
Answered by RasheedSindhi last updated on 24/Dec/15
  f(ux)=x^u f(x)..............(i)  x=1⇒f(u)=f(1)⇒u=1  u=1⇒ (i) will become  f(x)=xf(x)  xf(x)−f(x)=0  f(x)(x−1)=0  f(x)=0 ∣  x=1
$$ \\ $$$${f}\left({ux}\right)={x}^{{u}} {f}\left({x}\right)…………..\left({i}\right) \\ $$$${x}=\mathrm{1}\Rightarrow{f}\left({u}\right)={f}\left(\mathrm{1}\right)\Rightarrow{u}=\mathrm{1} \\ $$$${u}=\mathrm{1}\Rightarrow\:\left({i}\right)\:{will}\:{become} \\ $$$${f}\left({x}\right)={xf}\left({x}\right) \\ $$$${xf}\left({x}\right)−{f}\left({x}\right)=\mathrm{0} \\ $$$${f}\left({x}\right)\left({x}−\mathrm{1}\right)=\mathrm{0} \\ $$$${f}\left({x}\right)=\mathrm{0}\:\mid\:\:{x}=\mathrm{1} \\ $$

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