Question

Split the function into piece wise defined function if f(x) = 2|x| + |x+2| - ||x+2|-2|x||

Answer 1

You're missing a \(|\)

Answer 2

@FoolAroundMath @joemath314159

Answer 3

\(||x+2|-2|x|| = \left\{\begin{matrix}
|x+2|-2|x| \;,\;\;\;\; |x+2|-2|x| \ge 0 \text{ or }-2/3 \le x \le 2\\
2|x|-|x+2| \;,\;\;\;\; |x+2|-2|x| < 0 \text{ or } x<-2/3 \;\bigcup\;x>2
\end{matrix}\right.\)


Substituting in the original function,

\(f(x)=\left\{\begin{matrix} 4|x|, \;\;\;\;\; -2/3 \le x \le 2
\\2|x+2|,\;\; \;\;x<-2/3 \;\bigcup\;x>2\end{matrix}\right.\)

Now \(|x| = \left\{\begin{matrix}x \;\;\; x \ge 0 \\ -x\;\;\; \;\;x < 0\end{matrix}\right. \text { and } |x+2|=\left\{\begin{matrix} x+2\;\;\; x\ge -2\\-(x+2) \;\;\;\;\; x<-2 \end{matrix}\right.\)

So,
\(f(x)=\left\{\begin{matrix} -2(x+2) \;\;\;\;\;\;\;\;\;\;\;\;\;\; x<-2\\2(x+2) \;\;\;\;-2\le x<-2/3 \\ -4x\;\;\;\;\; -2/3 \le x <0\\4x\;\;\;\;\;\;\;\;\;\; 0 \le x < 2\\ 2(x+2)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; x\ge 2 \end{matrix}\right.\)