Given-f-R-R-such-that-x-2-f-x-f-1-x-2x-x-4-find-f-x-

Question Number 122074 by bemath last updated on 13/Nov/20

G i v e n f : R \to R s u c h t h a t

x^{2} f (x) + f (1 - x) = 2 x - x^{4}

f i n d f (x) .

Answered by bobhans last updated on 14/Nov/20

(1) x^{2} f (x) + f (1 - x) = 2 x - x^{4}

r e p l a c i n g x b y 1 - x g i v e

\to {(1 - x)}^{2} f (1 - x) + f (x) = 2 (1 - x) - {(1 - x)}^{4}

m u l t i p l y e q (1) w i t h {(1 - x)}^{2} g i v e

(2) \to {(1 - x)}^{2} f (1 - x) + {(1 - x)}^{2} x^{2} f (x) = {(1 - x)}^{2} (2 x - x^{4})

s u b s t r a c t (1) b y (2) g i v e

{1 - x^{2} {(1 - x)}^{2}} f (x) = (2 - 2 x) - (1 - x^{4}) - {(1 - x)}^{2} (2 x - x^{4})

∵ f (x) \frac{(2 - 2 x) - (1 - x^{4}) - {(1 - x)}^{2} (2 x - 2 x^{4})}{1 - x^{2} {(1 - x)}^{2}}

Answered by mathmax by abdo last updated on 14/Nov/20

\Rightarrow x^{2} f (x) + f (1 - x) = - x^{4} + 2 x

let change x by 1 - x \Rightarrow {(1 - x)}^{2} f (1 - x) + f (x) = 2 (1 - x) - {(1 - x)}^{4}

we get the system {\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} = x^{2} {(1 - x)}^{2} - 1 \Rightarrow {(x - x^{2})}^{2} - 1 = x^{2} - {2 x}^{3} + x^{4} - 1

\Rightarrow f (x) = \frac{| \begin{matrix} 2 x - x^{4} 1 \\ 2 (1 - x) - {(1 - x)}^{4} {(1 - x)}^{2} \end{matrix} |}{x^{4} - {2 x}^{3} + x^{2} - 1} \Rightarrow

f (x) = \frac{(2 x - x^{4}) {(1 - x)}^{2} - 2 (1 - x) + {(1 - x)}^{4}}{x^{4} - {2 x}^{3} + x^{2} - 1}

Answered by ajfour last updated on 14/Nov/20

x^{2} f (x) + f (1 - x) = 2 x - x^{4}

{(1 - x)}^{2} f (1 - x) + f (x) = 2 (1 - x) - {(1 - x)}^{4}

f (x) = \frac{{(1 - x)}^{2} (2 x - x^{4}) - 2 (1 - x) + {(1 - x)}^{4}}{x^{2} {(1 - x)}^{2} - 1}

= \frac{(1 - x) {(1 - x) (2 x - x^{4}) - 2 + {(1 - x)}^{3}}}{{x (1 - x) - 1} {x (1 - x) + 1}}

= \frac{(1 - x) (x^{5} - x^{4} - x^{3} + x^{2} - x - 1)}{{x (1 - x) - 1} {x (1 - x) + 1}}

= \frac{(1 - x) (x^{3} + 1) (x^{2} - x - 1)}{(x^{2} - x + 1) (x^{2} - x - 1)}

f (x) = \frac{(1 - x) (1 + x) (x^{2} - x + 1)}{x^{2} - x + 1}

\Rightarrow f (x) = 1 - x^{2} ★

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