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Question Number 122074 by bemath last updated on 13/Nov/20
Given f :R→R such that x^2 f(x)+f(1−x) = 2x−x^4 find f(x).
$$\:{Given}\:{f}\::{R}\rightarrow{R}\:{such}\:{that}\: \\ $$$$\:{x}^{\mathrm{2}} \:{f}\left({x}\right)+{f}\left(\mathrm{1}−{x}\right)\:=\:\mathrm{2}{x}−{x}^{\mathrm{4}} \\ $$$$\:{find}\:{f}\left({x}\right). \\ $$
Answered by bobhans last updated on 14/Nov/20
(1) x^2 f(x)+f(1−x)=2x−x^4 replacing x by 1−x give →(1−x)^2 f(1−x)+f(x)=2(1−x)−(1−x)^4 multiply eq (1) with (1−x)^2 give (2)→(1−x)^2 f(1−x)+(1−x)^2 x^2 f(x)=(1−x)^2 (2x−x^4 ) substract (1) by (2) give {1−x^2 (1−x)^2 } f(x)=(2−2x)−(1−x^4 )−(1−x)^2 (2x−x^4 ) ∵ f(x) (((2−2x)−(1−x^4 )−(1−x)^2 (2x−2x^4 ))/(1−x^2 (1−x)^2 ))
$$\left(\mathrm{1}\right)\:{x}^{\mathrm{2}} {f}\left({x}\right)+{f}\left(\mathrm{1}−{x}\right)=\mathrm{2}{x}−{x}^{\mathrm{4}} \\ $$$${replacing}\:{x}\:{by}\:\mathrm{1}−{x}\:{give} \\ $$$$\rightarrow\left(\mathrm{1}−{x}\right)^{\mathrm{2}} {f}\left(\mathrm{1}−{x}\right)+{f}\left({x}\right)=\mathrm{2}\left(\mathrm{1}−{x}\right)−\left(\mathrm{1}−{x}\right)^{\mathrm{4}} \\ $$$${multiply}\:{eq}\:\left(\mathrm{1}\right)\:{with}\:\left(\mathrm{1}−{x}\right)^{\mathrm{2}} \:{give} \\ $$$$\left(\mathrm{2}\right)\rightarrow\left(\mathrm{1}−{x}\right)^{\mathrm{2}} {f}\left(\mathrm{1}−{x}\right)+\left(\mathrm{1}−{x}\right)^{\mathrm{2}} {x}^{\mathrm{2}} {f}\left({x}\right)=\left(\mathrm{1}−{x}\right)^{\mathrm{2}} \left(\mathrm{2}{x}−{x}^{\mathrm{4}} \right) \\ $$$${substract}\:\left(\mathrm{1}\right)\:{by}\:\left(\mathrm{2}\right)\:{give} \\ $$$$\left\{\mathrm{1}−{x}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)^{\mathrm{2}} \right\}\:{f}\left({x}\right)=\left(\mathrm{2}−\mathrm{2}{x}\right)−\left(\mathrm{1}−{x}^{\mathrm{4}} \right)−\left(\mathrm{1}−{x}\right)^{\mathrm{2}} \left(\mathrm{2}{x}−{x}^{\mathrm{4}} \right) \\ $$$$\because\:{f}\left({x}\right)\:\frac{\left(\mathrm{2}−\mathrm{2}{x}\right)−\left(\mathrm{1}−{x}^{\mathrm{4}} \right)−\left(\mathrm{1}−{x}\right)^{\mathrm{2}} \left(\mathrm{2}{x}−\mathrm{2}{x}^{\mathrm{4}} \right)}{\mathrm{1}−{x}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)^{\mathrm{2}} } \\ $$
Answered by mathmax by abdo last updated on 14/Nov/20
⇒x^2 f(x)+f(1−x)=−x^4 +2x let change x by 1−x ⇒(1−x)^2 f(1−x)+f(x)=2(1−x)−(1−x)^4 we get the system { ((x^2 f(x)+f(1−x)=2x−x^4 )),((f(x)+(1−x)^2 f(1−x)=2(1−x)−(1−x)^4 )) :} Δ_s =x^2 (1−x)^2 −1 ⇒(x−x^2 )^2 −1 =x^2 −2x^3 +x^4 −1 ⇒f(x)=( determinant (((2x−x^4 1)),((2(1−x)−(1−x)^4 (1−x)^2 )))/(x^4 −2x^3 +x^2 −1))⇒ f(x)=(((2x−x^4 )(1−x)^2 −2(1−x)+(1−x)^4 )/(x^4 −2x^3 +x^2 −1))
$$\Rightarrow\mathrm{x}^{\mathrm{2}} \mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{1}−\mathrm{x}\right)=−\mathrm{x}^{\mathrm{4}} \:+\mathrm{2x} \\ $$$$\mathrm{let}\:\mathrm{change}\:\mathrm{x}\:\mathrm{by}\:\mathrm{1}−\mathrm{x}\:\Rightarrow\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{2}} \mathrm{f}\left(\mathrm{1}−\mathrm{x}\right)+\mathrm{f}\left(\mathrm{x}\right)=\mathrm{2}\left(\mathrm{1}−\mathrm{x}\right)−\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{4}} \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{the}\:\mathrm{system}\:\begin{cases}{\mathrm{x}^{\mathrm{2}} \mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\mathrm{1}−\mathrm{x}\right)=\mathrm{2x}−\mathrm{x}^{\mathrm{4}} }\\{\mathrm{f}\left(\mathrm{x}\right)+\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{2}} \mathrm{f}\left(\mathrm{1}−\mathrm{x}\right)=\mathrm{2}\left(\mathrm{1}−\mathrm{x}\right)−\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{4}} }\end{cases} \\ $$$$\Delta_{\mathrm{s}} =\mathrm{x}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{2}} −\mathrm{1}\:\Rightarrow\left(\mathrm{x}−\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{1}\:=\mathrm{x}^{\mathrm{2}} −\mathrm{2x}^{\mathrm{3}} \:+\mathrm{x}^{\mathrm{4}} −\mathrm{1} \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\frac{\begin{vmatrix}{\mathrm{2x}−\mathrm{x}^{\mathrm{4}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{2}\left(\mathrm{1}−\mathrm{x}\right)−\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{4}} \:\:\:\:\:\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{2}} }\end{vmatrix}}{\mathrm{x}^{\mathrm{4}} −\mathrm{2x}^{\mathrm{3}} \:+\mathrm{x}^{\mathrm{2}} −\mathrm{1}}\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\left(\mathrm{2x}−\mathrm{x}^{\mathrm{4}} \right)\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{1}−\mathrm{x}\right)+\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{4}} }{\mathrm{x}^{\mathrm{4}} −\mathrm{2x}^{\mathrm{3}} \:+\mathrm{x}^{\mathrm{2}} −\mathrm{1}} \\ $$
Answered by ajfour last updated on 14/Nov/20
x^2 f(x)+f(1−x)=2x−x^4 (1−x)^2 f(1−x)+f(x)=2(1−x)−(1−x)^4 f(x)=(((1−x)^2 (2x−x^4 )−2(1−x)+(1−x)^4 )/(x^2 (1−x)^2 −1)) =(((1−x){(1−x)(2x−x^4 )−2+(1−x)^3 })/({x(1−x)−1}{x(1−x)+1})) = (((1−x)(x^5 −x^4 −x^3 +x^2 −x−1))/({x(1−x)−1}{x(1−x)+1})) = (((1−x)(x^3 +1)(x^2 −x−1))/((x^2 −x+1)(x^2 −x−1))) f(x)=(((1−x)(1+x)(x^2 −x+1))/(x^2 −x+1)) ⇒ f(x)=1−x^2 ★
$${x}^{\mathrm{2}} {f}\left({x}\right)+{f}\left(\mathrm{1}−{x}\right)=\mathrm{2}{x}−{x}^{\mathrm{4}} \\ $$$$\left(\mathrm{1}−{x}\right)^{\mathrm{2}} {f}\left(\mathrm{1}−{x}\right)+{f}\left({x}\right)=\mathrm{2}\left(\mathrm{1}−{x}\right)−\left(\mathrm{1}−{x}\right)^{\mathrm{4}} \\ $$$$\:{f}\left({x}\right)=\frac{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} \left(\mathrm{2}{x}−{x}^{\mathrm{4}} \right)−\mathrm{2}\left(\mathrm{1}−{x}\right)+\left(\mathrm{1}−{x}\right)^{\mathrm{4}} }{{x}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)^{\mathrm{2}} −\mathrm{1}} \\ $$$$\:=\frac{\left(\mathrm{1}−{x}\right)\left\{\left(\mathrm{1}−{x}\right)\left(\mathrm{2}{x}−{x}^{\mathrm{4}} \right)−\mathrm{2}+\left(\mathrm{1}−{x}\right)^{\mathrm{3}} \right\}}{\left\{{x}\left(\mathrm{1}−{x}\right)−\mathrm{1}\right\}\left\{{x}\left(\mathrm{1}−{x}\right)+\mathrm{1}\right\}} \\ $$$$\:=\:\frac{\left(\mathrm{1}−{x}\right)\left({x}^{\mathrm{5}} −{x}^{\mathrm{4}} −{x}^{\mathrm{3}} +{x}^{\mathrm{2}} −{x}−\mathrm{1}\right)}{\left\{{x}\left(\mathrm{1}−{x}\right)−\mathrm{1}\right\}\left\{{x}\left(\mathrm{1}−{x}\right)+\mathrm{1}\right\}} \\ $$$$\:=\:\frac{\left(\mathrm{1}−{x}\right)\left({x}^{\mathrm{3}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}−\mathrm{1}\right)}{\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}−\mathrm{1}\right)} \\ $$$$\:{f}\left({x}\right)=\frac{\left(\mathrm{1}−{x}\right)\left(\mathrm{1}+{x}\right)\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)}{{x}^{\mathrm{2}} −{x}+\mathrm{1}} \\ $$$$\Rightarrow\:\:{f}\left({x}\right)=\mathrm{1}−{x}^{\mathrm{2}} \:\:\:\bigstar \\ $$

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