Question

ajutor,va rog............

Answer 1

$f(x) = \begin{cases}1-x^2,\,\,\,\, x\geq 0\\ 5x+1,\,\,\,\, x<0\end{cases},\quad g(x) =\begin{cases}x^2,\,\,\,\,x\leq -2\\ 2x-1,\,\,\,\, x>-2\end{cases}\\ \\ \\ f \circ g = f\Big(g(x)\Big) =\begin{cases} 1-g^2(x),\,\,\,\, g(x)\geq 0 \\ 5g(x)+1,\,\,\,\,g(x)<0\end{cases} =$

$= \begin{cases} 1-(x^2)^2,\,\,\,\, x^2\geq 0\quad \text{si}\quad x\leq -2 \\ 5(x^2)+1,\,\,\,\,x^2<0\quad \text{si}\quad x\leq -2\\ \\1-(2x-1)^2,\,\,\,\,2x-1\geq 0\quad \text{si}\quad x > -2 \\ 5(2x-1)+1,\,\,\,\, 2x-1 < 0\quad \text{si}\quad x>-2\end{cases} = \\ \\ \\= \begin{cases} 1-x^4,\,\,\,\, x\leq -2 \\ \text{nu are sens}\\\\1-(4x^2-4x+1),\,\,\,\,x\geq \dfrac{1}{2}\\ 10x-4,\,\,\,\,-2<x<\dfrac{1}{2}\end{array} =$

$=\begin{cases}1-x^4,\,\,\,\, x\leq -2 \\ 4x(1-x),\,\,\,\, x\geq \dfrac{1}{2} \\ 2(5x-2),\,\,\,\, -2<x<\dfrac{1}{2}\end{cases} \Rightarrow \boxed{B}\to\text{corect}$