Question

am nevoie urgenta la tema la matematica

Answer 1

$\displaystyle\bf\\f(x)=\begin{cases} 2x-1,~x\in(-\infty,-1] \\ 3x,~x\in(-1,2) \\ -x+5,~x\in[2,+\infty) \end{cases}\\\\f(-3),~cum~-3\in[-\infty,-1) \implies \boxed{\bf f(-3)=-7}~.\\f(1),~cum~1\in(-1,2) \implies \boxed{\bf f(1)=3\cdot1=3}~.\\f(0),~cum~0\in(-1,2) \implies \boxed{\bf f(0)=0}~.\\f(2),~cum~2\in[2,+\infty) \implies \boxed{\bf f(2)=-2+5=3}~.\\f(3),~cum~3\in[2,+\infty) \implies \boxed{\bf f(3)= -3+5=2}~.\\f(-2),~cum~-2\in(-\infty,-1) \implies \boxed{\bf f(-2)=2(-2)-1=-5}~.$

$\displaystyle\bf\\S=f(-3)f(1)+f(0)f(2)-3\bigg[f(3)-f(2)\bigg]=\\-21+6=~\boxed{\bf-15}~.$