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Question Number 56921 by rahul 19 last updated on 26/Mar/19

Given :  f(xy)=f(x).f(y)∀x,yεR and f(0)≠0  then f(x)=?

$${Given}\:: \\ $$$${f}\left({xy}\right)={f}\left({x}\right).{f}\left({y}\right)\forall{x},{y}\epsilon\mathbb{R}\:{and}\:{f}\left(\mathrm{0}\right)\neq\mathrm{0} \\ $$$${then}\:{f}\left({x}\right)=? \\ $$

Answered by mr W last updated on 26/Mar/19

f(xy)=f(x)f(y)  with y=0  f(0)=f(x)f(0)  since f(0)≠0 ⇒f(x)=1

$${f}\left({xy}\right)={f}\left({x}\right){f}\left({y}\right) \\ $$$${with}\:{y}=\mathrm{0} \\ $$$${f}\left(\mathrm{0}\right)={f}\left({x}\right){f}\left(\mathrm{0}\right) \\ $$$${since}\:{f}\left(\mathrm{0}\right)\neq\mathrm{0}\:\Rightarrow{f}\left({x}\right)=\mathrm{1} \\ $$

Commented by rahul 19 last updated on 26/Mar/19

Sir, how to find value of n if we assumed  f(x)=x^n  ?

$${Sir},\:{how}\:{to}\:{find}\:{value}\:{of}\:{n}\:{if}\:{we}\:{assumed} \\ $$$${f}\left({x}\right)={x}^{{n}} \:? \\ $$

Commented by mr W last updated on 27/Mar/19

f(x)≠x^n , since f(0)≠0.  f(x)=1 is the only one solution.

$${f}\left({x}\right)\neq{x}^{{n}} ,\:{since}\:{f}\left(\mathrm{0}\right)\neq\mathrm{0}. \\ $$$${f}\left({x}\right)=\mathrm{1}\:{is}\:{the}\:{only}\:{one}\:{solution}. \\ $$

Commented by rahul 19 last updated on 27/Mar/19

thank you sir!

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