`f(x)=x^2-4|x| and g(x)={min {f(t):-6letlex},x in [-6,0] and max{f(t):0 < tlex},x in (0,6],` then `g(x)` has

f(x)=x^(2)-4|x| and g(x)={min{f(t):-6<=t<=x},x in[-6,0] and max{f(t):0

f(x)=x^(2)-4|x|andg(x)={{:("min"[f(t):-6letlex}","x in[-6,0],),("max"{f(t):0letlex}","x in (0,6],):} Minimum value of g(x) is

f(x)=x^(2)-4|x|andg(x)={{:("min"[f(t):-6letlex}","x in[-6,0],),("max"{f(t):0letlex}","x in (0,6],):} g(x) is constant in

f(x)=x^(2)-4|x|andg(x)={{:("min"[f(t):-6letlex}","x in[-6,0],),("max"{f(t):0letlex}","x in (0,6],):} g(x) is strictly increasing in

f(x)=x^(2)-2|x|,g(x)=min f(t):0<=t<=x,-2<=x<=0 and max f(t):0<=t<=x,0<=x<=3 Sketch the graph of g(x) and discuss its differentiability

Let f(x)=|x-1|+|x-2| and g(x)={min{f(t):0<=t<=x,0<=x<=3 and x-2,x then g(x) is not differentiable at

If f(x) = x^(2) - 2|x| and g(x) = {{:("min{f(t)":-2 le t le x",",-2 le x le "}"),("max {f(t):"0 le t le x",",0 le x le 3"}"):} (i) Draw the graph of f(x) and discuss its continuity and differentiablity. (ii) Find and draw the graph of g(x Also, discuss the continuity.

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