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Question Number 62340 by Tawa1 last updated on 19/Jun/19

Commented by maxmathsup by imad last updated on 20/Jun/19
$3) f(x)+xf(−x) =x ⇒f(−x)−xf(x) =−x we get the systeme { ((f(x)+xf(−x) =x)),((xf(x)−f(−x)=x (with unknown f(x) and f(−x)))) :} Δ = determinant (((1 x)),((x −1)))=−1−x^2 f(x) =(Δ_(f(x)) /Δ) we have Δ_(f(x)) = determinant (((x x)),((x −1)))=−x−x^2 ⇒f(x)=((−x−x^2 )/(−1−x^2 )) =((x^2 +x)/(x^2 +1)) ⇒f(x) =((x^2 +x)/(x^2 +1))$
$3) f (x) + x f (- x) = x \Rightarrow f (- x) - x f (x) = - x w e g e t t h e s y s t e m e$ ${\begin{cases} f (x) + x f (- x) = x \\ x f (x) - f (- x) = x (w i t h u n k n o w n f (x) a n d f (- x)) \end{cases}$ $Δ = \| \begin{matrix} 1 x \\ x - 1 \end{matrix} \| = - 1 - x^{2}$ $f (x) = \frac{Δ_{f (x)}}{Δ} w e h a v e Δ_{f (x)} = \| \begin{matrix} x x \\ x - 1 \end{matrix} \| = - x - x^{2} \Rightarrow f (x) = \frac{- x - x^{2}}{- 1 - x^{2}} = \frac{x^{2} + x}{x^{2} + 1}$ $\Rightarrow f (x) = \frac{x^{2} + x}{x^{2} + 1}$
Commented by Tawa1 last updated on 20/Jun/19

$God bless you sir$
Commented by prof Abdo imad last updated on 20/Jun/19

$y o u a r e w e l c o m e .$
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