Question

Nu știu cum sa îl aflu pe g(x) la subpunctul d de la 14... dau 30p pentru rezolvare

Answer 1

$\displaystyle\it\\\boxed{\it o~functie~este~injectiva~daca~f(x)= f(y) \implies x=y}~.\\\boxed{\it (f\circ g)=f(g(x))}~.\\------------------------\\f(x)=3x-1,~x\in\mathbb{R}.\\fie~functia~g(x)=ax+b,~a,b\in\mathbb{R}.\\asadar,~(f\circ g)(x)=2x-1 \Leftrightarrow f(g(x))=2x-1 \Leftrightarrow\\f(ax+b)=2x-1 \Leftrightarrow 3(ax+b)-1=2x-1 \Leftrightarrow 3ax+b=2x \Leftrightarrow\\\underline{3a}x+\boxed{b}=\underline{2}x+\boxed{0} \implies 3a=2 \implies \boxed{\it a=\frac{2}{3}},~iar~\boxed{\it b=0}~.$

$\displaystyle\it\\deci,~\boxed{\it g(x)=\frac{2x}{3}}~,~este~aceasta~functie~injectiva~?\\fie~x_1,~x_2 \in\mathbb{R},~a.i.~g(x_1)=g(x_2),~este~x_1=x_2~?\\g(x_1)=g(x_2) \Leftrightarrow \frac{2x_1}{3}=\frac{2x_3}{3} \Leftrightarrow 2x_1=2x_2 \implies x_1=x_2,~asadar,~functia\\g(x)~este~injectiva.$