Let `f(x)` be defined on `[-2,2]` and be given by `f(x)`= -1 , -2`<= ` x `<=` 0 1-x , 0`<=`x`<=`2 . and `g(x)=f(|x|)+|f(x)|dot` Then find `g(x)dot`

Explanation

We have
`f(x)={(-1",",-2 le x le 0),(x-1",",0 lt x le 2):}`
or `f(|x|)={(-1",",-2 le |x| le 0),(|x|-1",",0 le |x| le 2):}`
`=|x|-1,0 le |x| le 2`
` " " `(As `-2 le |x| lt 0` is not possible)
`={(-x-1",",-2 le x le 0),(x-1",",0 lt x le 2):} " (1)" `
Again, `f(x)={(-1",",-2 le x le 0),(x-1",",0 lt x le 2):}`
or `|f(x)|={(|-1|",",-2 le x le 0),(|x-1|",",0 lt x le 2):}`
or `|f(x)|={(1",",-2 le x le 0),(-(x-1)",",0 lt x le 1),(+(x-1)",",1 lt x le 2):} " (2)" `
Therefore, `g(x)=f(|x|)+|f(x)|` can be expressed as
`g(x)={((-x-1)+1",",-2 le x le 0),((x-1)+(1-x)",",0 lt x le 1),((x-1)+(x+1)",",1 lt x le 2):}` [Using (1) and (2) ]
`={(-x",",-2 le x le 0),(0",",0 lt x le 1),(2(x-1)",",1 lt x le 2):}`