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If f(x)=|x-1|+|x-2|+|x-3| where 2 < x <3...

If `f(x)=|x-1|+|x-2|+|x-3|` where `2 < x <3` is

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Let f(x)=f_1(x)-2f_2 (x) , where ,where f_1(x)={((min{x^2,|x|},|x|le 1),(max{x^2,|x|},|x| le 1)) and f_2(x)={((min{x^2,|x|},|x| lt 1),({x^2,|x|},|x| le 1)) and let g(x)={ ((min{f(t):-3letlex,-3 le x le 0}),(max{f(t):0 le t le x,0 le x le 3})) for -3 le x le -1 the range of g(x) is

Let f(x)=f_1(x)-2f_2 (x), where ,where f_1(x)={((min{x^2,|x|},|x|le 1),(max{x^2,|x|},|x| le 1)) and f_2(x)={((min{x^2,|x|},|x| lt 1),({x^2,|x|},|x| le 1)) and let g(x)={ ((min{f(t):-3letlex,-3 le x le 0}),(max{f(t):0 le t le x,0 le x le 3})) for -3 le x le -1 the range of g(x) is

Let f(x)=f_1(x)-2f_2 (x) , where ,where f_1(x)={((min{x^2,|x|},|x|le 1),(max{x^2,|x|},|x| le 1)) and f_2(x)={((min{x^2,|x|},|x| lt 1),({x^2,|x|},|x| le 1)) and let g(x)={ ((min{f(t):-3letlex,-3 le x le 0}),(max{f(t):0 le t le x,0 le x le 3})) for -3 le x le -1 the range of g(x) is

Let f(x)=f_1(x)-2f_2 (x) , where ,where f_1(x)={((min{x^2,|x|},|x|le 1),(max{x^2,|x|},|x| le 1)) and f_2(x)={((min{x^2,|x|},|x| lt 1),({x^2,|x|},|x| le 1)) and let g(x)={ ((min{f(t):-3letlex,-3 le x le 0}),(max{f(t):0 le t le x,0 le x le 3})) for -3 le x le -1 the range of g(x) is