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Question Number 17093 by 1234Hello last updated on 02/Jul/17

$$\mathrm{If}f\left(x\right)\mathrm{is}\mathrm{a}\mathrm{polynomial}\mathrm{function}$$$$\mathrm{satisfying}f\left(x\right).f\left(\frac{1}{x}\right)=f\left(x\right)+f\left(\frac{1}{x}\right);$$$$x\in R-\left\{0\right\}\mathrm{and}f\left(3\right)=28,\mathrm{then}f\left(4\right)\mathrm{is}$$$$\mathrm{equal}\mathrm{to}$$

Commented by prakash jain last updated on 30/Jun/17

$$f\left(x\right)=\frac{f\left(\frac{1}{x}\right)}{f\left(\frac{1}{x}\right)-1}$$$$f\left(x\right)=\frac{x-1}{x}$$$$f\left(\frac{1}{x}\right)=1-x$$$$f\left(x\right)+f\left(\frac{1}{x}\right)=\frac{x-1}{x}+1-x=\frac{-1+2x-x^2}{2}$$$$=\frac{-{(1-x)}^2}{x}$$$$f\left(x\right)\left(\frac{1}{x}\right)=\frac{-{(1-x)}^2}{x}$$$$f\left(x\right)=\frac{x-1}{x}satisfies$$$$f\left(x\right)f\left(\frac{1}{x}\right)=f\left(x\right)+f\left(\frac{1}{x}\right)$$$$f\left(3\right)=\frac{2}{3}$$$$f\left(4\right)=\frac{3}{4}$$$$f\left(3\right)=28?$$

Commented by mrW1 last updated on 01/Jul/17

$$\mathrm{I}\mathrm{think}\mathrm{functions}\mathrm{like}\mathrm{this}$$$$\mathrm{f}\left(\mathrm{x}\right)=1+{\mathrm{x}}^{\mathrm{k}}\mathrm{or}1-{\mathrm{x}}^{\mathrm{k}},\mathrm{k}\in \mathrm{R}$$$$\mathrm{fulfill}\mathrm{the}\mathrm{condiction}:$$$$\mathrm{f}\left(\frac{1}{\mathrm{x}}\right)=1\pm {\mathrm{x}}^{-\mathrm{k}}$$$$\mathrm{f}\left(\mathrm{x}\right)\mathrm{f}\left(\frac{1}{\mathrm{x}}\right)=(1\pm {\mathrm{x}}^{\mathrm{k}})(1\pm {\mathrm{x}}^{-\mathrm{k}})=1\pm {\mathrm{x}}^{\mathrm{k}}+1\pm {\mathrm{x}}^{-\mathrm{k}}$$$$=\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\frac{1}{\mathrm{x}}\right)$$$$$$$$\mathrm{for}\mathrm{f}\left(3\right)=28$$$$1\pm 3^{\mathrm{k}}=28?$$$$\Rightarrow 1+3^3=28$$$$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=1+{\mathrm{x}}^3$$$$\Rightarrow \mathrm{f}\left(4\right)=1+4^3=65$$

Commented by mrW1 last updated on 01/Jul/17

$$\mathrm{I}\mathrm{started}\mathrm{with}\mathrm{f}\left(\mathrm{x}\right)=\mathrm{a}+\mathrm{bx}\mathrm{and}\mathrm{got}$$$$\mathrm{a}=1\mathrm{and}\mathrm{b}=\pm 1\mathrm{and}\mathrm{then}\mathrm{I}\mathrm{replaced}\mathrm{x}$$$$\mathrm{with}{\mathrm{x}}^{\mathrm{k}}\mathrm{without}\mathrm{change}\mathrm{the}\mathrm{property}$$$$\mathrm{of}\mathrm{the}\mathrm{function}.\mathrm{I}\mathrm{am}\mathrm{not}\mathrm{sure}\mathrm{if}\mathrm{there}$$$$\mathrm{are}\mathrm{other}\mathrm{functions}\mathrm{besides}1\pm {\mathrm{x}}^{\mathrm{k}}.$$$$\mathrm{if}\mathrm{no},\mathrm{then}\mathrm{the}\mathrm{answer}\mathrm{for}\mathrm{the}\mathrm{question}\mathrm{is}$$$$\mathrm{unique}.$$$$\mathrm{e}.\mathrm{g}.\mathrm{if}\mathrm{f}\left(3\right)=-20\mathrm{then}$$$$1-3^{\mathrm{k}}=-20\Rightarrow \mathrm{k}=\frac{\ln 21}{\ln 3}\approx 2.77$$$$\Rightarrow \mathrm{f}\left(4\right)=1-4^{\mathrm{k}}=-45.607.$$$$\mathrm{if}\mathrm{f}\left(3\right)=6\Rightarrow 1+3^{\mathrm{k}}=6\Rightarrow \mathrm{k}=\frac{\ln 5}{\ln 3}$$$$\Rightarrow \mathrm{f}\left(4\right)=1+4^{\mathrm{k}}=8.62$$

Commented by Tinkutara last updated on 01/Jul/17

$$\mathrm{Thanks}\mathrm{Sirs}!$$

Commented by prakash jain last updated on 01/Jul/17

$$itsatisfiesg\left(x\right)g\left(\frac{1}{x}\right)=1$$$$\mathrm{and}\mathrm{as}\mathrm{you}\mathrm{described}$$$$f\left(x\right)=1\pm g\left(x\right)\mathrm{is}\mathrm{valid}$$$$\mathrm{solution}.$$$$\mathrm{coming}\mathrm{back}\mathrm{to}\mathrm{what}\mathrm{is}\mathrm{asked}\mathrm{in}$$$$\mathrm{question}.$$$$\mathrm{Try}\mathrm{an}\mathrm{alternative}f\left(x\right)$$$$g\left(x\right)=\frac{10x-3}{10-3x}$$$$f\left(x\right)=\frac{7x+7}{10-3x}\Rightarrow f\left(3\right)=28$$$$f\left(\frac{1}{x}\right)=\frac{7+7x}{10x-3}$$$$f\left(x\right)+f\left(\frac{1}{x}\right)=\frac{7x+7}{10-3x}+\frac{7+7x}{10x-3}$$$$=\frac{(7x+7)(10x-3+10-3x)}{(10-3x)(10x-3)}$$$$=\frac{(7x+7)(7x+7)}{(10-3x)(10x-3)}=f\left(x\right)f\left(\frac{1}{x}\right)$$$$f\left(4\right)=\frac{7\times 4+7}{10-3\times 4}=-\frac{35}{2}$$

Commented by prakash jain last updated on 01/Jul/17

$$\mathrm{Thanks},\mathrm{mrW}1!$$

Commented by mrW1 last updated on 01/Jul/17

$$\mathrm{Do}\mathrm{you}\mathrm{have}\mathrm{any}\mathrm{ideas}\mathrm{if}\mathrm{there}\mathrm{are}\mathrm{other}$$$$\mathrm{types}\mathrm{of}\mathrm{functions}\mathrm{with}\mathrm{this}\mathrm{property}?$$$$\mathrm{I}\mathrm{think}\mathrm{the}\mathrm{function}\mathrm{should}\mathrm{be}\mathrm{of}\mathrm{the}$$$$\mathrm{form}\mathrm{f}\left(\mathrm{x}\right)=1\pm \mathrm{g}\left(\mathrm{x}\right)\mathrm{then}$$$$\mathrm{f}\left(\frac{1}{\mathrm{x}}\right)=1\pm \mathrm{g}\left(\frac{1}{\mathrm{x}}\right)$$$$\mathrm{f}\left(\mathrm{x}\right)\mathrm{f}\left(\frac{1}{\mathrm{x}}\right)=[1\pm \mathrm{g}\left(\mathrm{x}\right)][1\pm \mathrm{g}\left(\frac{1}{\mathrm{x}}\right)]$$$$=1\pm \mathrm{g}\left(\mathrm{x}\right)+\mathrm{g}\left(\mathrm{x}\right)\mathrm{g}\left(\frac{1}{\mathrm{x}}\right)\pm \mathrm{g}\left(\frac{1}{\mathrm{x}}\right)$$$$=\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\frac{1}{\mathrm{x}}\right)\mathrm{if}\mathrm{g}\left(\mathrm{x}\right)\mathrm{g}\left(\frac{1}{\mathrm{x}}\right)=1$$$$$$$$\mathrm{g}\left(\mathrm{x}\right)={\mathrm{x}}^{\mathrm{k}}\mathrm{fulfills}\mathrm{this}\mathrm{condition}$$$$\mathrm{but}\mathrm{are}\mathrm{there}\mathrm{any}\mathrm{other}\mathrm{function}\mathrm{types}$$$$\mathrm{for}\mathrm{g}\left(\mathrm{x}\right)?$$

Commented by prakash jain last updated on 01/Jul/17

$$g\left(x\right)=\frac{{ax}^2+bx+c}{{cx}^2+bx+a}$$$$g\left(\frac{1}{x}\right)=\frac{\frac{a}{x^2}+\frac{b}{x}+c}{\frac{c}{x^2}+\frac{b}{x}+a}=\frac{a+bx+{cx}^2}{c+bx+{ax}^2}$$$$g\left(x\right)g\left(\frac{1}{x}\right)=1$$

Commented by mrW1 last updated on 01/Jul/17

$${\mathrm{that}}^{\prime }\mathrm{s}\mathrm{great}!$$$$\mathrm{can}\mathrm{we}\mathrm{generase}\mathrm{that}$$$$\mathrm{g}\left(\mathrm{x}\right)=\frac{\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}}}{\mathrm{a}}_{\mathrm{k}}{\mathrm{x}}^{\mathrm{k}}}{\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}}}{\mathrm{a}}_{\mathrm{n}-\mathrm{k}}{\mathrm{x}}^{\mathrm{k}}}$$$$\mathrm{fulfills}\mathrm{the}\mathrm{condition}?$$

Commented by mrW1 last updated on 01/Jul/17

$$\mathrm{that}\mathrm{means}\mathrm{for}\mathrm{f}\left(3\right)=28\mathrm{there}\mathrm{is}\mathrm{no}$$$$\mathrm{unique}\mathrm{solution}\mathrm{for}\mathrm{f}\left(4\right).$$$$\mathrm{thank}\mathrm{you}\mathrm{for}\mathrm{showing}\mathrm{me}\mathrm{the}$$$$\mathrm{right}\mathrm{thinking}\mathrm{direction}!$$

Commented by 1234Hello last updated on 02/Jul/17

$$\mathrm{Sir},\mathrm{I}\mathrm{changed}\mathrm{the}\mathrm{question}.\mathrm{Here}$$$$x\in R-\left\{0\right\}\mathrm{but}\mathrm{in}\mathrm{your}\mathrm{chosen}f\left(x\right)$$$$=\frac{7x+7}{10-3x},x\in R-\left\{\frac{10}{3}\right\}.\mathrm{Therefore}$$$$f\left(x\right)\mathrm{cannot}\mathrm{be}\mathrm{this}.\mathrm{But}\mathrm{it}\mathrm{can}\mathrm{be}\mathrm{when}$$$$x\in R-\left\{0\right\}.$$

Commented by prakash jain last updated on 02/Jul/17

$$g\left(x\right)=\left(\frac{{ax}^2+bx+1}{x^2+bx+a}\right)$$$$9a+3b+1=9\times 27+81b+27a$$$$18a+78b=-242$$$$9a+39b=-121$$$$a=\frac{-(121+39b)}{9}$$$$g\left(x\right)=\frac{\frac{230}{9}{x}^2-9x+1}{x^2-9x+\frac{230}{9}}$$$$g\left(3\right)=\frac{230-27+1}{9-27+\frac{230}{9}}=\frac{9(230-27+1)}{(81-243+230)}$$$$=\frac{9\left(204\right)}{68}=27$$$$g\left(x\right)g\left(\frac{1}{x}\right)=1$$$$f\left(3\right)=28$$$$f\left(4\right)=1+g\left(4\right)=1+\frac{230\cdot 4^2-81\cdot 4+9}{9\cdot 4^2-81\cdot 4+230}$$$$=1+\frac{673}{10}=\frac{683}{10}$$$$------$$$$9x^2-81x+230$$$$b^2-4ac={81}^2-4\times 9\times 230<0$$$$\mathrm{so}f\left(x\right)\mathrm{is}\mathrm{defined}.$$$$\mathrm{if}\mathrm{you}\mathrm{want}\mathrm{to}\mathrm{create}\mathrm{a}\mathrm{function}$$$$\mathrm{which}\mathrm{is}\mathrm{undefined}\mathrm{at}x=0$$$$\mathrm{choose}g\left(x\right)\mathrm{as}$$$$\frac{{ax}^2+bx+1}{x^2+bx+a}\times \frac{1}{x}$$$$\mathrm{g}\left(3\right)=27$$$$\frac{9a+3b+1}{9+3b+a}=81$$$$yougetanequationbetweenaand$$$$b.$$$$chooseavalueforbwhichgives$$$$b^2-4ac<0.$$

Commented by mrW1 last updated on 02/Jul/17

$$\mathrm{since}\mathrm{there}\mathrm{are}\mathrm{infinite}\mathrm{possible}\mathrm{functions}:$$$$\mathrm{f}\left(\mathrm{x}\right)=1\pm \frac{\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}}}{\mathrm{a}}_{\mathrm{k}}{\mathrm{x}}^{\mathrm{k}}}{\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}}}{\mathrm{a}}_{\mathrm{n}-\mathrm{k}}{\mathrm{x}}^{\mathrm{k}}}$$$$\mathrm{with}\mathrm{n}\in \mathbb{N}\mathrm{and}{\mathrm{a}}_0,{\mathrm{a}}_1,...,{\mathrm{a}}_{\mathrm{n}}\in \mathbb{R}$$$$$$$$\mathrm{there}\mathrm{is}\mathrm{no}\mathrm{unique}\mathrm{solution}\mathrm{for}$$$$(\mathrm{n},{\mathrm{a}}_0,{\mathrm{a}}_1,...,{\mathrm{a}}_{\mathrm{n}})\mathrm{from}\mathrm{the}\mathrm{only}\mathrm{one}\mathrm{equation}$$$$\mathrm{f}\left(3\right)=1\pm \frac{\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}}}{\mathrm{a}}_{\mathrm{k}}{3}^{\mathrm{k}}}{\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}}}{\mathrm{a}}_{\mathrm{n}-\mathrm{k}}{3}^{\mathrm{k}}}=28$$$$$$$$\mathrm{therefore}\mathrm{there}\mathrm{is}\mathrm{also}\mathrm{no}\mathrm{unique}$$$$\mathrm{value}\mathrm{for}\mathrm{f}\left(4\right):$$$$\mathrm{f}\left(4\right)=1\pm \frac{\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}}}{\mathrm{a}}_{\mathrm{k}}{4}^{\mathrm{k}}}{\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}}}{\mathrm{a}}_{\mathrm{n}-\mathrm{k}}{4}^{\mathrm{k}}}$$

Commented by prakash jain last updated on 02/Jul/17

$$\mathrm{Ok}.\mathrm{I}\mathrm{missed}\mathrm{some}\mathrm{part}\mathrm{in}\mathrm{the}$$$$\mathrm{question}.\mathrm{The}\mathrm{question}\mathrm{is}$$$$\mathrm{specifically}\mathrm{asking}\mathrm{for}\mathrm{polynomial}$$$$\mathrm{function}.$$$$f\left(x\right)=1+x^3\mathrm{is}\mathrm{only}\mathrm{correct}\mathrm{answer}.$$

Commented by mrW1 last updated on 02/Jul/17

$$\mathrm{to}1234\mathrm{hello}:$$$$\mathrm{I}\mathrm{think}\mathrm{the}\mathrm{original}\mathrm{question}\mathrm{was}\mathrm{much}$$$$\mathrm{more}\mathrm{interesting}\mathrm{and}\mathrm{challenging}$$$$\mathrm{than}\mathrm{now}.\mathrm{Even}\mathrm{though}\mathrm{there}\mathrm{was}\mathrm{no}$$$$\mathrm{unique}\mathrm{answer}\mathrm{but}\mathrm{one}\mathrm{may}\mathrm{think}\mathrm{in}$$$$\mathrm{all}\mathrm{directions}.\mathrm{I}\mathrm{see}\mathrm{it}\mathrm{as}\mathrm{a}\mathrm{great}\mathrm{gain}$$$$\mathrm{from}\mathrm{your}\mathrm{original}\mathrm{question}\mathrm{that}\mathrm{we}$$$$\mathrm{know}\mathrm{now}\mathrm{that}\mathrm{the}\mathrm{requested}$$$$\mathrm{function}\mathrm{which}\mathrm{satisfies}\mathrm{the}\mathrm{functional}$$$$\mathrm{equation}\mathrm{could}\mathrm{and}\mathrm{must}\mathrm{be}\mathrm{of}\mathrm{the}\mathrm{form}$$$$\mathrm{f}\left(\mathrm{x}\right)=1\pm \frac{\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}}}{\mathrm{a}}_{\mathrm{k}}{\mathrm{x}}^{\mathrm{k}}}{\underset{\mathrm{k}=0}{\sum ^{\mathrm{n}}}{\mathrm{a}}_{\mathrm{n}-\mathrm{k}}{\mathrm{x}}^{\mathrm{k}}}.$$$$$$$$\mathrm{We}{\mathrm{hadn}}^{\prime }\mathrm{t}\mathrm{lernt}\mathrm{so}\mathrm{much}\mathrm{if}\mathrm{you}\mathrm{had}$$$$\mathrm{said}\mathrm{from}\mathrm{very}\mathrm{begining}\mathrm{that}\mathrm{f}\left(\mathrm{x}\right)$$$$\mathrm{is}\mathrm{a}\mathrm{polynominal}\mathrm{function},\mathrm{because}$$$$\mathrm{that}\mathrm{would}\mathrm{have}\mathrm{restricted}\mathrm{the}\mathrm{direction}$$$$\mathrm{for}\mathrm{our}\mathrm{thinking}.$$

Commented by 1234Hello last updated on 03/Jul/17

$$\mathrm{But}\mathrm{this}\mathrm{was}\mathrm{the}\mathrm{original}\mathrm{question}!\mathrm{I}$$$$\mathrm{was}\mathrm{in}\mathrm{a}\mathrm{hurry}\mathrm{so}\mathrm{I}\mathrm{first}\mathrm{posted}\mathrm{that}$$$$\mathrm{version}.\mathrm{Thanks}\mathrm{all}\mathrm{for}\mathrm{your}\mathrm{efforts}!$$

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