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Question Number 110306 by Lekhraj last updated on 28/Aug/20

Commented by mr W last updated on 28/Aug/20

f(x)=((1±(√5))/2)x

$${f}\left({x}\right)=\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}}{x} \\ $$

Commented by Lekhraj last updated on 29/Aug/20

Thanks . How you get it?

$$\mathrm{Thanks}\:.\:\mathrm{How}\:\mathrm{you}\:\mathrm{get}\:\mathrm{it}? \\ $$

Answered by mr W last updated on 29/Aug/20

t=f(x) ⇒x=f^(−1) (t) ⇒f^(−1) (t)+t=f(t) or f(x)=f^(−1) (x)+x ...(I) since f^(−1) (x) and f(x) are symmetric about the line y=x, we can image that f(x) could be a line to fulfill (I). let f(x)=ax+b ⇒f^(−1) (x)=((x−b)/a) ax+b=((x−b)/a)+x ⇒(a−1−(1/a))x+b(1−(1/a))=0 ⇒b=0 ⇒a−1−(1/a)=0 ⇒a^2 −a−1=0 ⇒a=((1±(√5))/2) ⇒f(x)=((1±(√5))/2)x

$${t}={f}\left({x}\right) \\ $$$$\Rightarrow{x}={f}^{−\mathrm{1}} \left({t}\right) \\ $$$$\Rightarrow{f}^{−\mathrm{1}} \left({t}\right)+{t}={f}\left({t}\right) \\ $$$${or} \\ $$$${f}\left({x}\right)={f}^{−\mathrm{1}} \left({x}\right)+{x}\:\:\:\:...\left({I}\right) \\ $$$${since}\:{f}^{−\mathrm{1}} \left({x}\right)\:{and}\:{f}\left({x}\right)\:{are}\:{symmetric} \\ $$$${about}\:{the}\:{line}\:{y}={x},\:{we}\:{can}\:{image}\:{that} \\ $$$${f}\left({x}\right)\:{could}\:{be}\:{a}\:{line}\:{to}\:{fulfill}\:\left({I}\right). \\ $$$${let}\:{f}\left({x}\right)={ax}+{b} \\ $$$$\Rightarrow{f}^{−\mathrm{1}} \left({x}\right)=\frac{{x}−{b}}{{a}} \\ $$$${ax}+{b}=\frac{{x}−{b}}{{a}}+{x} \\ $$$$\Rightarrow\left({a}−\mathrm{1}−\frac{\mathrm{1}}{{a}}\right){x}+{b}\left(\mathrm{1}−\frac{\mathrm{1}}{{a}}\right)=\mathrm{0} \\ $$$$\Rightarrow{b}=\mathrm{0} \\ $$$$\Rightarrow{a}−\mathrm{1}−\frac{\mathrm{1}}{{a}}=\mathrm{0} \\ $$$$\Rightarrow{a}^{\mathrm{2}} −{a}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{a}=\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\Rightarrow{f}\left({x}\right)=\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}}{x} \\ $$

Commented by mr W last updated on 29/Aug/20

Commented by Lekhraj last updated on 02/Sep/20

Thanks . Why b = 0 ?

$$\mathrm{Thanks}\:.\:\mathrm{Why}\:{b}\:=\:\mathrm{0}\:? \\ $$

Commented by Lekhraj last updated on 02/Sep/20

How we could image that f(x) is a line , it may be any curve ?

$$\mathrm{How}\:\mathrm{we}\:\mathrm{could}\:\mathrm{image}\:\mathrm{that}\:{f}\left({x}\right)\:\mathrm{is}\:\mathrm{a}\: \\ $$$$\mathrm{line}\:,\:\mathrm{it}\:\mathrm{may}\:\mathrm{be}\:\mathrm{any}\:\mathrm{curve}\:? \\ $$

Commented by mr W last updated on 02/Sep/20

if b≠0 then a=1, this is not ok.

$${if}\:{b}\neq\mathrm{0}\:{then}\:{a}=\mathrm{1},\:{this}\:{is}\:{not}\:{ok}. \\ $$

Commented by mr W last updated on 02/Sep/20

if you can find a curve, that will be great. i can say that such a curve doesn′t exist.

$${if}\:{you}\:{can}\:{find}\:{a}\:{curve},\:{that}\:{will}\:{be} \\ $$$${great}.\:{i}\:{can}\:{say}\:{that}\:{such}\:{a}\:{curve} \\ $$$${doesn}'{t}\:{exist}. \\ $$

Commented by Lekhraj last updated on 02/Sep/20

ok I got it .Thanks

$$\mathrm{ok}\:\mathrm{I}\:\mathrm{got}\:\mathrm{it}\:.\mathrm{Thanks} \\ $$

Commented by Lekhraj last updated on 02/Sep/20

Is the following justification correct if the degree of f(x) > 1 then degree of f(f(x)) > f(x) ⇒ x +f(x) ≠ f(fx)) .

$${I}\mathrm{s}\:\mathrm{the}\:\mathrm{following}\:\mathrm{justification}\: \\ $$$$\mathrm{correct} \\ $$$$\mathrm{if}\:\mathrm{the}\:\mathrm{degree}\:\mathrm{of}\:\:{f}\left({x}\right)\:>\:\mathrm{1}\:\mathrm{then} \\ $$$$\mathrm{degree}\:\mathrm{of}\:{f}\left({f}\left({x}\right)\right)\:>\:{f}\left({x}\right)\: \\ $$$$\left.\Rightarrow\:{x}\:+{f}\left({x}\right)\:\neq\:\:{f}\left({fx}\right)\right)\:. \\ $$

Commented by mr W last updated on 02/Sep/20

correct

$${correct} \\ $$

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