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Determinați legea de corespondență a funcției f:R->R, știind că 2f(x)-f(1-x)=3-x, oricare x aparține R.

Răspuns :

Functia liniara are ecuatia f(x)=ax+b
f(1-x)=a(1-x)+b
Inlocuim in prima relatie:
2(ax+b)-[tex] \left[\begin{array}{ccc}a(1-x)+b\end{array}\right] =3-x[/tex]
2ax+2b-(a-ax+b)=3-x
2ax+2b-a+ax-b=3-x⇒3ax+2b-a-b=-x+3⇒3ax+b-a=-x+3
Egalam coeficientii: 3a=-1⇒a=-1/3 si din b-a=3⇒b+1/3=3⇒b=8/3
Legea de corespondenta a functiei liniare f este:f(x)=-1/3x+8/3
Rayzen
[tex]2f(x) - f(1-x) = 3-x \\ f(1-x) = 2f(x) +x-3 \quad (1)\\ \\ \\ 2f(x)-f(1-x)=3-x \\ 2f(x) = 3-x+f(1-x) \\ f(x) = \dfrac{3-x+f(1-x)}{2} \quad (\text{Facem: } x \rightarrow 1-x) \\ \\ \begin{array}{rcl} f(1-x) &=& \dfrac{3-(1-x) +f\Big(1-(1-x)\Big)}{2} \\ \\&=& \dfrac{2+x+f(1-1+x)}{2}\\ \\&=& \dfrac{2+x+f(x)}{2}\quad (2)\end{array} \\ \\ \\\text{Din (1) si (2) } \Rightarrow \\ \\ 2f(x) + x-3 = \dfrac{2+x+f(x)}{2} \\ 4f(x)+2x-6 = 2+x+f(x) \\ 4f(x) - f(x) = 2+x-2x+6 \\ 3f(x) = -x+8[/tex]
[tex] f(x) = \dfrac{-x+8}{3}
\Rightarrow \boxed{f(x) = -\dfrac{1}{3}x +\dfrac{8}{3}}[/tex]

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