Question

Se considera functiile f,g :R→R, f(x)=ax+b, g(x)= x²-x. Determinati numerele reale a si b, astfel incat f compus g= g compus f.

Answer 1

[tex]f(x)=ax+b \\ \\ g(x)=x^2-x \\ \\ f(g(x))=f(x^2-x)=a(x^2-x)+b = ax^2-ax+b \\ \\ g(f(x))=g(ax+b)=(ax+b)^2-(ax+b) =a^2x^2+2abx+b^2-ax-b \\ \\ \text {Stim ca f(g(x)) = g(f(x)), deci: } \\ \\ ax^2-ax+b = a^2x^2+2abx+b^2-ax-b \\ \\ (a^2-a)x^2+2abx+b^2-2b=0 \ \textless \ =\ \textgreater \ \text{ a=0 si b=2 sau a=1 si b=0.} [/tex]

Answer 2

$f,g:\mathbb_{R} \rightarrow \mathbb_{R}, \quad f(x) = ax+b,\quad g(x) = x^2-x \\ \\ f \circ g = g \circ f \\ \\ f\Big(g(x)\Big) = g\Big(f(x)\Big) \\ \\ a(x^2-x)+b = (ax+b)^2-(ax+b)\\ \\ ax^2-ax+b = a^2x^2+2abx+b^2-ax-b \\ \\ ax^2-ax+b = a^2x^2+(2ab-a)x+b^2-b \\ \\ Egalam coeficientii lui x^2, lui x si termenii liberi:$

$\left\{ \begin{array}{ll} a = a^2 \\ -a = 2ab-a \\ b = b^2-b \end{array} \right \Rightarrow \left\{ \begin{array}{ll} a^2-a = 0 \\ 2ab-a+a = 0 \\ b^2-2b = 0 \end{array} \right \Rightarrow \left\{ \begin{array}{ll} a(a-1) = 0 \\ 2ab = 0 \\ b(b-2) = 0 \end{array} \right \Rightarrow \\ \\ \\ \boxed{I} a = 0: \left\{ \begin{array}{ll} 0(a-1) = 0 \\ 2\cdot 0\cdot b = 0 \\ b(b-2) = 0 \end{array} \right \Rightarrow \left\{ \begin{array}{ll} 0= 0 \\ 0 = 0 \\ b(b-2) = 0 \end{array} \right$
$~~~~~\boxed{I_1}\quad b = 0 \Rightarrow (a,b) = (0,0)\\ ~~~~~~\boxed{I_2} \quad b = 2 \Rightarrow (a,b) = (0,2) \\ \\ \\ \boxed{II} a= 1: \left\{ \begin{array}{ll} 1(1-1) = 0 \\ 2\cdot 1\cdot b = 0 \\ b(b-2) = 0 \end{array} \right \Rightarrow \left\{ \begin{array}{ll} 1\cdot 0 = 0 \\ 2b = 0 \\ b(b-2) = 0 \end{array} \right \Rightarrow$

$\Rightarrow \left\{ \begin{array}{ll} 0= 0 \\ b = 0 \\ 0(b-2) = 0 \end{array} \right \Rightarrow \left\{ \begin{array}{ll} 0= 0 \\ b = 0 \\ 0= 0 \end{array} \right \Rightarrow (a,b) = (1,0) \\ \\ \\ \boxed{III} b=2: \left\{ \begin{array}{ll} a(a-1) = 0 \\ 2a\cdot 2 = 0 \\ 2(2-2) = 0 \end{array} \right \Rightarrow \left\{ \begin{array}{ll} a(a-1) = 0 \\ a = 0 \\ 0 = 0 \end{array} \right \Rightarrow$

$\Rightarrow \left\{ \begin{array}{ll} 0 = 0 \\ a = 0 \\ 0 = 0 \end{array} \right \Rightarrow (a,b) = (0,2) \\ \\ \\ \boxed{IV} b = 0: \left\{ \begin{array}{ll} a(a-1) = 0 \\ 2a\cdot 0 = 0 \\ 0\cdot (0-2) = 0 \end{array} \right \Rightarrow \left\{ \begin{array}{ll} a(a-1) = 0 \\ 0= 0 \\ 0 = 0 \end{array} \right \Rightarrow \\ \\ \Rightarrow (a,b) = \Big\{(0,0);(1,0)\Big\} \\ \\ \\ Din I \cup II \cup III \cup IV \Rightarrow (a,b) = \Big\{(0,0); (0,2); (1,0)\Big\}$