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Let-f-R-R-be-polynomial-function-satisfying-f-x-f-1-x-f-x-f-1-x-and-f-3-28-then-f-x-is-

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Question Number 171435 by cortano1 last updated on 15/Jun/22
Let f:R→R be polynomial function satisfying f(x) f((1/x))=f(x)+f((1/x)) and f(3)=28, then f(x) is
$$\:\:{Let}\:{f}:{R}\rightarrow{R}\:{be}\:{polynomial} \\ $$$$\:{function}\:{satisfying}\: \\ $$$$\:{f}\left({x}\right)\:{f}\left(\frac{\mathrm{1}}{{x}}\right)={f}\left({x}\right)+{f}\left(\frac{\mathrm{1}}{{x}}\right)\:{and} \\ $$$$\:{f}\left(\mathrm{3}\right)=\mathrm{28},\:{then}\:{f}\left({x}\right)\:{is} \\ $$
Commented by infinityaction last updated on 15/Jun/22
now function is not define R→R
$${now}\:{function}\:{is}\:{not}\:{define}\:{R}\rightarrow{R} \\ $$
Commented by infinityaction last updated on 15/Jun/22
sir i have changed the question
$${sir}\:{i}\:{have}\:{changed}\:{the}\:{question} \\ $$
Commented by mr W last updated on 15/Jun/22
f is R→R means: for x∈R ⇒y=f(x) ∈ R. it doesn′t mean: for x∈R ⇒y=f((1/x)) ∈ R.
$${f}\:{is}\:{R}\rightarrow{R}\:{means}: \\ $$$${for}\:{x}\in{R}\:\Rightarrow{y}={f}\left({x}\right)\:\in\:{R}. \\ $$$${it}\:{doesn}'{t}\:{mean}: \\ $$$${for}\:{x}\in{R}\:\Rightarrow{y}={f}\left(\frac{\mathrm{1}}{{x}}\right)\:\in\:{R}. \\ $$
Commented by infinityaction last updated on 15/Jun/22
your question should be A polynomial function f(x) satisfying f(x) f((1/x))=f(x)+f((1/x)) and f(3)=28, then f(x) is then we know that f(x) = 1±x^n so f(3) = 1±3^n = 28 1+3^n = 28 then n= 3 f(x) = 1+x^3
$${your}\:{question}\:{should}\:{be} \\ $$$$\:\:{A}\:{polynomial}\:{function}\:{f}\left({x}\right)\:{satisfying}\: \\ $$$$\:{f}\left({x}\right)\:{f}\left(\frac{\mathrm{1}}{{x}}\right)={f}\left({x}\right)+{f}\left(\frac{\mathrm{1}}{{x}}\right)\:{and} \\ $$$$\:{f}\left(\mathrm{3}\right)=\mathrm{28},\:{then}\:{f}\left({x}\right)\:{is} \\ $$$${then}\:{we}\:{know}\:{that} \\ $$$$\:{f}\left({x}\right)\:=\:\:\mathrm{1}\pm{x}^{{n}} \\ $$$${so}\:\:{f}\left(\mathrm{3}\right)\:=\:\mathrm{1}\pm\mathrm{3}^{{n}} \:=\:\mathrm{28} \\ $$$$\:\:\:\:\:\:\mathrm{1}+\mathrm{3}^{{n}} \:=\:\mathrm{28} \\ $$$$\:\:{then}\:{n}=\:\mathrm{3} \\ $$$$\:\:\:{f}\left({x}\right)\:=\:\mathrm{1}+{x}^{\mathrm{3}} \\ $$
Commented by floor(10²Eta[1]) last updated on 15/Jun/22
f is not R→R because we have (1/x) there
$$\mathrm{f}\:\mathrm{is}\:\mathrm{not}\:\mathbb{R}\rightarrow\mathbb{R}\:\mathrm{because}\:\mathrm{we}\:\mathrm{have}\:\frac{\mathrm{1}}{\mathrm{x}}\:\mathrm{there} \\ $$
Commented by floor(10²Eta[1]) last updated on 15/Jun/22
x cant be zero⇒x∉R
$$\mathrm{x}\:\mathrm{cant}\:\mathrm{be}\:\mathrm{zero}\Rightarrow\mathrm{x}\notin\mathbb{R} \\ $$
Commented by mr W last updated on 15/Jun/22
this is not true sir. f((1/x)) doesn′t mean f is not R→R! example: f(x)=x f is R→R. f((1/x))=(1/x)
$${this}\:{is}\:{not}\:{true}\:{sir}. \\ $$$${f}\left(\frac{\mathrm{1}}{{x}}\right)\:{doesn}'{t}\:{mean}\:{f}\:{is}\:{not}\:{R}\rightarrow{R}! \\ $$$${example}: \\ $$$${f}\left({x}\right)={x} \\ $$$${f}\:{is}\:{R}\rightarrow{R}. \\ $$$${f}\left(\frac{\mathrm{1}}{{x}}\right)=\frac{\mathrm{1}}{{x}} \\ $$
Answered by aleks041103 last updated on 15/Jun/22
general solution f(x)= { ((f_1 (x),x>0)),((f_2 (x),x<0)) :} (1/(f(x))) +(1/(f(1/x)))=1 x=e^t (1/(f(e^t )))=g(t) ⇒g(t)+g(−t)=1 ⇒g(t)=(1/2)+o(t) ⇒f(x)=(1/((1/2)+o(ln(x)))) ⇒f(x)= { ((((1/2)+o_1 (ln(x)))^(−1) , x>0)),((((1/2)+o_2 (ln(−x)))^(−1) , x<0)),((f_0 , x=0)) :} where o_(1,2) (x) are odd functions, i.e. o_(1,2) (−x)=−o_(1,2) (x)
$${general}\:{solution} \\ $$$${f}\left({x}\right)=\begin{cases}{{f}_{\mathrm{1}} \left({x}\right),{x}>\mathrm{0}}\\{{f}_{\mathrm{2}} \left({x}\right),{x}<\mathrm{0}}\end{cases} \\ $$$$\frac{\mathrm{1}}{{f}\left({x}\right)}\:+\frac{\mathrm{1}}{{f}\left(\mathrm{1}/{x}\right)}=\mathrm{1} \\ $$$${x}={e}^{{t}} \\ $$$$\frac{\mathrm{1}}{{f}\left({e}^{{t}} \right)}={g}\left({t}\right) \\ $$$$\Rightarrow{g}\left({t}\right)+{g}\left(−{t}\right)=\mathrm{1} \\ $$$$\Rightarrow{g}\left({t}\right)=\frac{\mathrm{1}}{\mathrm{2}}+{o}\left({t}\right) \\ $$$$\Rightarrow{f}\left({x}\right)=\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{2}}+{o}\left({ln}\left({x}\right)\right)} \\ $$$$\Rightarrow{f}\left({x}\right)=\begin{cases}{\left(\frac{\mathrm{1}}{\mathrm{2}}+{o}_{\mathrm{1}} \left({ln}\left({x}\right)\right)\right)^{−\mathrm{1}} ,\:{x}>\mathrm{0}}\\{\left(\frac{\mathrm{1}}{\mathrm{2}}+{o}_{\mathrm{2}} \left({ln}\left(−{x}\right)\right)\right)^{−\mathrm{1}} ,\:{x}<\mathrm{0}}\\{{f}_{\mathrm{0}} \:,\:\:{x}=\mathrm{0}}\end{cases} \\ $$$${where}\:{o}_{\mathrm{1},\mathrm{2}} \left({x}\right)\:{are}\:{odd}\:{functions},\:{i}.{e}. \\ $$$${o}_{\mathrm{1},\mathrm{2}} \left(−{x}\right)=−{o}_{\mathrm{1},\mathrm{2}} \left({x}\right) \\ $$$$ \\ $$

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