Redefine the function f(x) = |x – 2| + |2 + x|,

Solution:

According to the question,

function f(x) = |x–2| + |2 + x|, –3≤ x≤ 3

We know that,

when x>0,

|x – 2| is (x–2), x≥2

|2 + x| is (2 + x), x≥–2

when x>0

|x – 2| is –(x–2), x<2

|2 + x| is –(2 + x), x<–2

Given that, f(x) = |x–2| + |2 + x|, –3≤ x≤ 3

It can be rewritten as,

$f(x)=\left\{\begin{array}{c}-(x-2)-(2+x),-3 \leq x<-2 \\ -(x-2)+(2+x),-2 \leq x<2 \\ (x-2)+(2+x), 2 \leq x \leq 3\end{array}\right.$

Or

$f(x)=\left\{\begin{array}{c}-x+2-2-x,-3 \leq x<-2 \\ -x+2+2+x,-2 \leq x<2 \\ x-2+2+x, 2 \leq x \leq 3\end{array}\right.$

Or,

$f(x)=\left\{\begin{aligned}-2 x,-3 & \leq x<-2 \\ 4,-2 & \leq x<2 \\ 2 x, 2 & \leq x \leq 3 \end{aligned}\right.$