[tex]1)\\f(x)=3x-4\\|\ \ \ x\ \ \ |\ \ -3\ \ \ |\ \ -1\ \ \ |\ \ \ 0\ \ \ |\ \ \ 2\ \ \ |\ \ \ 5\ \ \ |\\=========================\\|\ f(x)\ |\ \ -13\ \ |\ \ -7\ \ |\ -4\ \ |\ \ \ 2\ \ \ |\ \ \ 11\ \ |\\\\f(-3)=3\cdot(-3)-4=-9-4=-13\\f(-1)=3\cdot(-1)-4=-3-4=-7\\f(0)=3\cdot0-4=0-4=-4\\f(2)=3\cdot2-4=6-4=2\\f(5)=3\cdot5-4=15-4=11[/tex]
[tex]g(x)=x^2-2\\|\ \ \ x\ \ \ |\ \ -3\ \ \ |\ \ -1\ \ \ |\ \ \ 0\ \ \ |\ \ \ 2\ \ \ |\ \ \ 5\ \ \ |\\=========================\\|\ g(x)\ |\ \ \ \ \ 7\ \ \ \ |\ \ -1\ \ \ |\ -2\ \ |\ \ \ 2\ \ |\ \ \ 23\ \ |\\\\g(-3)=(-3)^2-2=9-2=7\\g(-1)=(-1)^2-2=1-2=-1\\g(0)=0^2-2=0-2=-2\\g(2)=2^2-2=4-2=2\\g(5)=5^2-2=25-2=23\\\\f(x)=g(x)\ \ \ \Leftrightarrow\ \ \ x=2,\ \ \ \ because\ \ \ \ f(2)=2\ \ \ and\ \ \ g(2)=2[/tex]
[tex]2)\\the\ relation:\ \{(-4, 3), (-1, 0), (0, -2), (2,1), (4, 3)\}.\\\\the\ domain:\ D=\{-4,-1,0,2,4\}\\the\ range:\ R=\{3,0,-2,1\}\\\\This\ relation\ is\ the\ function,\ because\ \ each\ number\\ of\ the\ domain\ D\ has\ exactly\ one\ value\ in\ the\ range\ R. [/tex]
[tex]3)\\f(x)=|x+2|\\\\|x+2|= \left \{ {\big{x+2\ \ \ \ \ if\ \ \ x \geq -2} \atop \big{-x-2\ \ \ if\ \ \ x<-2}} \right. [/tex]
