Question

max/min problem
pic attached

Answer 1

\[ 1.f \left( x \right)=-\left( x-2 \right)-\left( 3x-2 \right)+\left( x+2 \right)=-3x+6 ,-9\le x <-2\]
\[2.f \left( x \right)=-\left( x-2 \right)-\left( 3x-2 \right)-\left( x+2 \right)=-5x+2, -2\le x < \frac{ 2 }{ 3 }\]
\[3.f \left( x \right)=-\left( x-2 \right)+\left( 3x-2 \right)-\left( x+2 \right)=x-2,for \frac{ 2 }{3 }\le x < 2\]
\[4.f \left( x \right)=\left( x-2 \right)+\left( 3x-2 \right)-\left( x+2 \right),for 2\le x \le 10\]

Answer 2

addition in 4th.
\[4.f \left( x \right)=3x-6\]

Answer 3

can you explain how why you chose those domains for each of the 4?

Answer 4

\[\left| x-2 \right|=-\left( x-2 \right) if x-2<0, or x<2\]
\[\left| x-2 \right|=\left( x-2 \right),if x-2>0, or x \ge 2\]
\[similarly \left| 3x-2 \right|=-\left( 3x-2 \right),if 3x-2<0, or x <\frac{ 2 }{ 3 }\]
\[\left| 3x-2 \right|=\left( 3x-2 \right),if 3x \ge 2, or x \ge \frac{ 2 }{3 }\]
\[\left| x+2 \right|=-\left( x+2 \right),if x+2<0,x<-2\]
\[\left| x+2 \right|=\left( x+2 \right),if x \ge -2\]