Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

Question Number 33375 by rahul 19 last updated on 15/Apr/18

If f:R → R is an odd function such that : a) f(1+x) = 1+f(x) . b) x^2 f((1/x)) = f(x) , x≠0. Then find f(x) ?

$${If}\:{f}:{R}\:\rightarrow\:{R}\:{is}\:{an}\:\boldsymbol{{odd}}\:{function}\:{such} \\ $$$${that}\:: \\ $$$$\left.{a}\right)\:{f}\left(\mathrm{1}+{x}\right)\:=\:\mathrm{1}+{f}\left({x}\right)\:. \\ $$$$\left.{b}\right)\:{x}^{\mathrm{2}} \:{f}\left(\frac{\mathrm{1}}{{x}}\right)\:=\:{f}\left({x}\right)\:,\:{x}\neq\mathrm{0}. \\ $$$${Then}\:{find}\:\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)\:? \\ $$

Commented by prof Abdo imad last updated on 15/Apr/18

f is odd⇒f(0)=0 ⇒f(1+0)=1+f(0)⇒f(1)=1 f(2)=f(1+1)=1+f(1) =2 let suppose f(n)=n ⇒f(n+1)=1+f(n)=1+n⇒ ∀n∈N f(n)=n if n∈Z^− n=−m f(n)=f(−m)=−f(m)=−m =n ⇒ ∀n∈Z f(n)=n f(1)=f((n/n))=nf((1/n))=1⇒f((1/n))=(1/n) f((p/n))=f((1/n) +..+(1/n))(p×) =pf((1/n))=(p/n) ⇒ ∀x∈Q f(x)=x but Q is dense inside R ⇒ ∀x∈R f(x)=x we have for x≠o x^2 f((1/x))=x^2 .(1/x) =x =f(x) cond.b is verified.

$${f}\:{is}\:{odd}\Rightarrow{f}\left(\mathrm{0}\right)=\mathrm{0}\:\Rightarrow{f}\left(\mathrm{1}+\mathrm{0}\right)=\mathrm{1}+{f}\left(\mathrm{0}\right)\Rightarrow{f}\left(\mathrm{1}\right)=\mathrm{1} \\ $$$${f}\left(\mathrm{2}\right)={f}\left(\mathrm{1}+\mathrm{1}\right)=\mathrm{1}+{f}\left(\mathrm{1}\right)\:=\mathrm{2} \\ $$$${let}\:{suppose}\:{f}\left({n}\right)={n}\:\Rightarrow{f}\left({n}+\mathrm{1}\right)=\mathrm{1}+{f}\left({n}\right)=\mathrm{1}+{n}\Rightarrow \\ $$$$\forall{n}\in{N}\:{f}\left({n}\right)={n}\:{if}\:{n}\in{Z}^{−} \:{n}=−{m} \\ $$$${f}\left({n}\right)={f}\left(−{m}\right)=−{f}\left({m}\right)=−{m}\:={n}\:\Rightarrow \\ $$$$\forall{n}\in{Z}\:{f}\left({n}\right)={n}\: \\ $$$${f}\left(\mathrm{1}\right)={f}\left(\frac{{n}}{{n}}\right)={nf}\left(\frac{\mathrm{1}}{{n}}\right)=\mathrm{1}\Rightarrow{f}\left(\frac{\mathrm{1}}{{n}}\right)=\frac{\mathrm{1}}{{n}} \\ $$$${f}\left(\frac{{p}}{{n}}\right)={f}\left(\frac{\mathrm{1}}{{n}}\:+..+\frac{\mathrm{1}}{{n}}\right)\left({p}×\right)\:={pf}\left(\frac{\mathrm{1}}{{n}}\right)=\frac{{p}}{{n}} \\ $$$$\Rightarrow\:\forall{x}\in{Q}\:{f}\left({x}\right)={x}\:\:\:{but}\:{Q}\:{is}\:{dense}\:{inside}\:{R}\:\Rightarrow \\ $$$$\forall{x}\in{R}\:{f}\left({x}\right)={x}\:\:{we}\:{have}\:{for}\:{x}\neq{o} \\ $$$${x}^{\mathrm{2}} {f}\left(\frac{\mathrm{1}}{{x}}\right)={x}^{\mathrm{2}} .\frac{\mathrm{1}}{{x}}\:={x}\:={f}\left({x}\right)\:{cond}.{b}\:{is}\:{verified}. \\ $$$$ \\ $$

Commented by rahul 19 last updated on 15/Apr/18

thank you sir !

$${thank}\:{you}\:{sir}\:! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com