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 f_(1)(x)={{:(min{x^(2),|x|}",",|x|le1),(max{x^(2),|x|}",",|x|gt1):} "and "f_(2)(x)={{:(min {x^(2),|x|}",",|x|gt1),(max{x^(2),|x|}",",|x|le1):} "and let "g(x)={{:(min{f(t),-3letlex,-3lexlt0}),(max{f(t),0letltx,0lexle3}):} For x in(-1,00),f(x)+g(x) is

Let f(x)=f_(1)(x)-2f_(2)(x), where f_(1)(x)={{:(min{x^(2),|x|}",",|x|le1),(max{x^(2),|x|}",",|x|gt1):} "and "f_(2)(x)={{:(min {x^(2),|x|}",",|x|gt1),(max{x^(2),|x|}",",|x|le1):} "and let "g(x)={{:(min{f(t),-3letlex,-3lexlt0}),(max{f(t),0letltx,0lexle3}):} For x in(-1,00),f(x)+g(x) is

Let f(x)=f_(1)(x)-2f_(2)(x), where f_(1)(x)={{:(min{x^(2),|x|}",",|x|le1),(max{x^(2),|x|}",",|x|gt1):} "and "f_(2)(x)={{:(min {x^(2),|x|}",",|x|gt1),(max{x^(2),|x|}",",|x|le1):} "and let "g(x)={{:(min{f(t),-3letlex,-3lexlt0}),(max{f(t),0letltx,0lexle3}):} The graph of y=g(x) in its domain is broken at

Let f(x)=f_(1)(x)-2f_(2)(x), where f_(1)(x)={{:(min{x^(2),|x|}",",|x|le1),(max{x^(2),|x|}",",|x|gt1):} "and "f_(2)(x)={{:(min {x^(2),|x|}",",|x|gt1),(max{x^(2),|x|}",",|x|le1):} "and let "g(x)={{:(min{f(t),-3letlex,-3lexlt0}),(max{f(t),0letltx,0lexle3}):} The graph of y=g(x) in its domain is broken at

Let f(x)=x^(3)-x^(2)+x+1 and g(x) be a function defined by g(x)={{:(,"Max"{f(t):0le t le x},0 le x le 1),(,3-x,1 le x le 2):} Then, g(x) is

Let f(x) ={:{(x, "for", 0 le x lt1),( 3-x,"for", 1 le x le2):} Then f(x) is