Question

Exercitiul 2.........​

Answer 1

$\it f(x)=ax+b\\ \\ (f\circ f)(x)=f(f(x))=f(ax+b)=a(ax+b)+b=a^2x+ab+b\ \ \ \ (1)\\ \\ (f\circ f)(x)==3f(x)+5=3(ax+b)+5=3ax+3b+5\ \ \ \ \ \ (2)\\ \\ (1),\ (2) \Rightarrow ax^2+ab+b=3ax+3b+5\ \Rightarrow \\ \\ \\ \Rightarrow \begin{cases} \it a^2=3a \Rightarrow a=3\\ \\ \it ab+b=3b+5 \Rightarrow 3b+b=3b+5 \Rightarrow b=5\end{cases}\\ \\ \\ f(x)=3x+5$