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Question Number 79532 by naka3546 last updated on 26/Jan/20
Given function f : R ⇒ R x^2 f(x) + f(1 − x) = 2x − x^4 f(2019) = ?
$${Given}\:\:{function}\:\:{f}\::\:\mathbb{R}\:\:\Rightarrow\:\:\mathbb{R} \\ $$$$\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} \:{f}\left({x}\right)\:+\:{f}\left(\mathrm{1}\:−\:{x}\right)\:\:=\:\:\mathrm{2}{x}\:−\:{x}^{\mathrm{4}} \\ $$$${f}\left(\mathrm{2019}\right)\:\:=\:\:? \\ $$
Commented by john santu last updated on 26/Jan/20
(1) x = 2019 (2019)^2 f(2019)+f(−2018)=2×(2019)−(2019)^4 (2) x=−2018 (−2018)^2 f(−2018)+f(2019)=2×(−2018)−(−2018)^4 now you get the result
$$\left(\mathrm{1}\right)\:{x}\:=\:\mathrm{2019}\: \\ $$$$\left(\mathrm{2019}\right)^{\mathrm{2}} {f}\left(\mathrm{2019}\right)+{f}\left(−\mathrm{2018}\right)=\mathrm{2}×\left(\mathrm{2019}\right)−\left(\mathrm{2019}\right)^{\mathrm{4}} \\ $$$$\left(\mathrm{2}\right)\:{x}=−\mathrm{2018}\: \\ $$$$\left(−\mathrm{2018}\right)^{\mathrm{2}} {f}\left(−\mathrm{2018}\right)+{f}\left(\mathrm{2019}\right)=\mathrm{2}×\left(−\mathrm{2018}\right)−\left(−\mathrm{2018}\right)^{\mathrm{4}} \\ $$$${now}\:{you}\:{get}\:{the}\:{result} \\ $$
Commented by mathmax by abdo last updated on 26/Jan/20
x^2 f(x)+f(1−x)=2x−x^4 let change x by 1−x ⇒ (1−x)f(1−x)+f(x)=2(1−x)−(1−x)^4 so we get the systeme { ((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 = determinant (((x^2 1)),((1 (1−x)^2 )))=x^2 (1−x)^2 −1 ⇒ f(x)=( determinant (((2x−x^4 1)),((2(1−x)−(1−x)^4 ( 1−x)^2 )))/(x^2 −x^3 −1)) =(((2x−x^4 )(1−x)^2 −2(1−x)+(1−x)^4 )/(x^2 (1−x)^2 −1)) ⇒ f(2019) =(((2×2019−(2019)^4 )(2018)^2 +2×(2018)+(2018)^4 )/((2019)^2 ×2018^2 −1))
$${x}^{\mathrm{2}} {f}\left({x}\right)+{f}\left(\mathrm{1}−{x}\right)=\mathrm{2}{x}−{x}^{\mathrm{4}} \:{let}\:{change}\:{x}\:{by}\:\mathrm{1}−{x}\:\Rightarrow \\ $$$$\left(\mathrm{1}−{x}\right){f}\left(\mathrm{1}−{x}\right)+{f}\left({x}\right)=\mathrm{2}\left(\mathrm{1}−{x}\right)−\left(\mathrm{1}−{x}\right)^{\mathrm{4}} \:\:{so}\:{we}\:{get}\:{the}\:{systeme} \\ $$$$\begin{cases}{{x}^{\mathrm{2}} {f}\left({x}\right)+{f}\left(\mathrm{1}−{x}\right)=\mathrm{2}{x}−{x}^{\mathrm{4}} }\\{{f}\left({x}\right)\:+\left(\mathrm{1}−{x}\right)^{\mathrm{2}} {f}\left(\mathrm{1}−{x}\right)=\mathrm{2}\left(\mathrm{1}−{x}\right)−\left(\mathrm{1}−{x}\right)^{\mathrm{4}} }\end{cases} \\ $$$$\Delta_{{s}} =\begin{vmatrix}{{x}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\end{vmatrix}={x}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)^{\mathrm{2}} −\mathrm{1}\:\Rightarrow\:{f}\left({x}\right)=\frac{\begin{vmatrix}{\mathrm{2}{x}−{x}^{\mathrm{4}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{2}\left(\mathrm{1}−{x}\right)−\left(\mathrm{1}−{x}\right)^{\mathrm{4}} \:\:\:\:\:\:\:\:\left(\:\mathrm{1}−{x}\right)^{\mathrm{2}} }\end{vmatrix}}{{x}^{\mathrm{2}} −{x}^{\mathrm{3}} −\mathrm{1}} \\ $$$$=\frac{\left(\mathrm{2}{x}−{x}^{\mathrm{4}} \right)\left(\mathrm{1}−{x}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{1}−{x}\right)+\left(\mathrm{1}−{x}\right)^{\mathrm{4}} }{{x}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)^{\mathrm{2}} −\mathrm{1}}\:\Rightarrow \\ $$$${f}\left(\mathrm{2019}\right)\:=\frac{\left(\mathrm{2}×\mathrm{2019}−\left(\mathrm{2019}\right)^{\mathrm{4}} \right)\left(\mathrm{2018}\right)^{\mathrm{2}} +\mathrm{2}×\left(\mathrm{2018}\right)+\left(\mathrm{2018}\right)^{\mathrm{4}} }{\left(\mathrm{2019}\right)^{\mathrm{2}} ×\mathrm{2018}^{\mathrm{2}} −\mathrm{1}} \\ $$$$ \\ $$
Answered by mr W last updated on 26/Jan/20
x^2 f(x) + f(1 − x) = 2x − x^4 ...(i) let t=1−x ⇒x=1−t (1−t)^2 f(1−t) + f(t) = 2(1−t) −(1−t )^4 or (1−x)^2 f(1−x) + f(x) = 2(1−x) −(1−x )^4 ...(ii) (i)×(1−x)^2 −(ii): [x^2 (1−x)^2 −1]f(x) = 2x(1−x)^2 −2(1−x)+(1−x)^4 − x^4 (1−x)^2 [x^2 (1−x)^2 −1]f(x) =(x−1){2+2(x−1)x+(x−1)^3 −(x−1)x^4 } ⇒f(x) =(((x−1){2+2(x−1)x+(x−1)^3 −(x−1)x^4 })/(x^2 (1−x)^2 −1)) f(2019) =((2018{2+2×2018×2019+2018^3 −2018×2019^4 })/(2018^2 ×2019^2 −1)) f(2019)=−4076360
$${x}^{\mathrm{2}} \:{f}\left({x}\right)\:+\:{f}\left(\mathrm{1}\:−\:{x}\right)\:\:=\:\:\mathrm{2}{x}\:−\:{x}^{\mathrm{4}} \:\:\:…\left({i}\right) \\ $$$${let}\:{t}=\mathrm{1}−{x}\:\Rightarrow{x}=\mathrm{1}−{t} \\ $$$$\left(\mathrm{1}−{t}\right)^{\mathrm{2}} \:{f}\left(\mathrm{1}−{t}\right)\:+\:{f}\left({t}\right)\:\:=\:\:\mathrm{2}\left(\mathrm{1}−{t}\right)\:−\left(\mathrm{1}−{t}\:\right)^{\mathrm{4}} \\ $$$${or} \\ $$$$\left(\mathrm{1}−{x}\right)^{\mathrm{2}} \:{f}\left(\mathrm{1}−{x}\right)\:+\:{f}\left({x}\right)\:\:=\:\:\mathrm{2}\left(\mathrm{1}−{x}\right)\:−\left(\mathrm{1}−{x}\:\right)^{\mathrm{4}} \:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)×\left(\mathrm{1}−{x}\right)^{\mathrm{2}} −\left({ii}\right): \\ $$$$\left[{x}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)^{\mathrm{2}} −\mathrm{1}\right]{f}\left({x}\right)\:=\:\:\mathrm{2}{x}\left(\mathrm{1}−{x}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{1}−{x}\right)+\left(\mathrm{1}−{x}\right)^{\mathrm{4}} \:−\:{x}^{\mathrm{4}} \left(\mathrm{1}−{x}\right)^{\mathrm{2}} \\ $$$$\left[{x}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)^{\mathrm{2}} −\mathrm{1}\right]{f}\left({x}\right)\:=\left({x}−\mathrm{1}\right)\left\{\mathrm{2}+\mathrm{2}\left({x}−\mathrm{1}\right){x}+\left({x}−\mathrm{1}\right)^{\mathrm{3}} −\left({x}−\mathrm{1}\right){x}^{\mathrm{4}} \right\} \\ $$$$\Rightarrow{f}\left({x}\right)\:=\frac{\left({x}−\mathrm{1}\right)\left\{\mathrm{2}+\mathrm{2}\left({x}−\mathrm{1}\right){x}+\left({x}−\mathrm{1}\right)^{\mathrm{3}} −\left({x}−\mathrm{1}\right){x}^{\mathrm{4}} \right\}}{{x}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)^{\mathrm{2}} −\mathrm{1}} \\ $$$${f}\left(\mathrm{2019}\right)\:=\frac{\mathrm{2018}\left\{\mathrm{2}+\mathrm{2}×\mathrm{2018}×\mathrm{2019}+\mathrm{2018}^{\mathrm{3}} −\mathrm{2018}×\mathrm{2019}^{\mathrm{4}} \right\}}{\mathrm{2018}^{\mathrm{2}} ×\mathrm{2019}^{\mathrm{2}} −\mathrm{1}} \\ $$$${f}\left(\mathrm{2019}\right)=−\mathrm{4076360} \\ $$

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