Given function f : R ⇒ R x^2 f(x) + f(1 − x) = 2x − x^4 f(2019) = ?


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Question Number 79532 by naka3546 last updated on 26/Jan/20

$G i v e n f u n c t i o n f : R \Rightarrow R$ $x^{2} f (x) + f (1 - x) = 2 x - x^{4}$ $f (2019) = ?$
Commented by john santu last updated on 26/Jan/20

$(1) x = 2019$ ${(2019)}^{2} f (2019) + f (- 2018) = 2 \times (2019) - {(2019)}^{4}$ $(2) x = - 2018$ ${(- 2018)}^{2} f (- 2018) + f (2019) = 2 \times (- 2018) - {(- 2018)}^{4}$ $n o w y o u g e t t h e r e s u l t$
Commented by mathmax by abdo last updated on 26/Jan/20
$x^2 f(x)+f(1−x)=2x−x^4 let change x by 1−x ⇒ (1−x)f(1−x)+f(x)=2(1−x)−(1−x)^4 so we get the systeme { ((x^2 f(x)+f(1−x)=2x−x^4 )),((f(x) +(1−x)^2 f(1−x)=2(1−x)−(1−x)^4 )) :} Δ_s = determinant (((x^2 1)),((1 (1−x)^2 )))=x^2 (1−x)^2 −1 ⇒ f(x)=( determinant (((2x−x^4 1)),((2(1−x)−(1−x)^4 ( 1−x)^2 )))/(x^2 −x^3 −1)) =(((2x−x^4 )(1−x)^2 −2(1−x)+(1−x)^4 )/(x^2 (1−x)^2 −1)) ⇒ f(2019) =(((2×2019−(2019)^4 )(2018)^2 +2×(2018)+(2018)^4 )/((2019)^2 ×2018^2 −1))$
$x^{2} f (x) + f (1 - x) = 2 x - x^{4} l e t c h a n g e x b y 1 - x \Rightarrow$ $(1 - x) f (1 - x) + f (x) = 2 (1 - x) - {(1 - x)}^{4} s o w e g e t t h e s y s t e m e$ ${\begin{cases} x^{2} f (x) + f (1 - x) = 2 x - x^{4} \\ f (x) + {(1 - x)}^{2} f (1 - x) = 2 (1 - x) - {(1 - x)}^{4} \end{cases}$ $Δ_{s} = \| \begin{matrix} x^{2} 1 \\ 1 {(1 - x)}^{2} \end{matrix} \| = x^{2} {(1 - x)}^{2} - 1 \Rightarrow f (x) = \frac{\| \begin{matrix} 2 x - x^{4} 1 \\ 2 (1 - x) - {(1 - x)}^{4} {(1 - x)}^{2} \end{matrix} \|}{x^{2} - x^{3} - 1}$ $= \frac{(2 x - x^{4}) {(1 - x)}^{2} - 2 (1 - x) + {(1 - x)}^{4}}{x^{2} {(1 - x)}^{2} - 1} \Rightarrow$ $f (2019) = \frac{(2 \times 2019 - {(2019)}^{4}) {(2018)}^{2} + 2 \times (2018) + {(2018)}^{4}}{{(2019)}^{2} \times 2018^{2} - 1}$
	Answered by mr W last updated on 26/Jan/20
	$x^2 f(x) + f(1 − x) = 2x − x^4 ...(i) let t=1−x ⇒x=1−t (1−t)^2 f(1−t) + f(t) = 2(1−t) −(1−t )^4 or (1−x)^2 f(1−x) + f(x) = 2(1−x) −(1−x )^4 ...(ii) (i)×(1−x)^2 −(ii): [x^2 (1−x)^2 −1]f(x) = 2x(1−x)^2 −2(1−x)+(1−x)^4 − x^4 (1−x)^2 [x^2 (1−x)^2 −1]f(x) =(x−1){2+2(x−1)x+(x−1)^3 −(x−1)x^4 } ⇒f(x) =(((x−1){2+2(x−1)x+(x−1)^3 −(x−1)x^4 })/(x^2 (1−x)^2 −1)) f(2019) =((2018{2+2×2018×2019+2018^3 −2018×2019^4 })/(2018^2 ×2019^2 −1)) f(2019)=−4076360$
	$x^{2} f (x) + f (1 - x) = 2 x - x^{4} . . . (i)$ $l e t t = 1 - x \Rightarrow x = 1 - t$ ${(1 - t)}^{2} f (1 - t) + f (t) = 2 (1 - t) - {(1 - t)}^{4}$ $o r$ ${(1 - x)}^{2} f (1 - x) + f (x) = 2 (1 - x) - {(1 - x)}^{4} . . . (i i)$ $(i) \times {(1 - x)}^{2} - (i i) :$ $[x^{2} {(1 - x)}^{2} - 1] f (x) = 2 x {(1 - x)}^{2} - 2 (1 - x) + {(1 - x)}^{4} - x^{4} {(1 - x)}^{2}$ $[x^{2} {(1 - x)}^{2} - 1] f (x) = (x - 1) {2 + 2 (x - 1) x + {(x - 1)}^{3} - (x - 1) x^{4}}$ $\Rightarrow f (x) = \frac{(x - 1) {2 + 2 (x - 1) x + {(x - 1)}^{3} - (x - 1) x^{4}}}{x^{2} {(1 - x)}^{2} - 1}$ $f (2019) = \frac{2018 {2 + 2 \times 2018 \times 2019 + 2018^{3} - 2018 \times 2019^{4}}}{2018^{2} \times 2019^{2} - 1}$ $f (2019) = - 4076360$
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