Let `f(x)=x^(3)-x^(2)+x+1` and `g(x)={("max "f(t) 0letlex 0lexle1),(3-x 1ltxle2):}` then

Let f(x) = x^(3) - x^(2) + x + 1 and g(x) = {{:(max f(t)",", 0 le t le x,"for",0 le x le 1),(3-x",",1 lt x le 2,,):} Then, g(x) in [0, 2] is

Let f(x) = x^(3) - x^(2) + x + 1 and g(x) = {{:(max f(t)",", 0 le t le x,"for",0 le x le 1),(3-x",",1 lt x le 2,,):} Then, g(x) in [0, 2] is

Let f(x) = x^(3) - x^(2) + x + 1 and g(x) = {{:(max f(t)",", 0 le t le x,"for",0 le x le 1),(3-x",",1 lt x le 2,,):} Then, g(x) in [0, 2] is

Let f(x) = x^(3) - x^(2) + x + 1 and g(x) = {{:(max f(t)",", 0 le t le x,"for",0 le x le 1),(3-x",",1 lt x le 2,,):} Then, g(x) in [0, 2] is a. continuous for x in [0, 2] - {1} b. continuous for x in [0, 2] c. differentiable for all x in [0, 2] d. differentiable for all x in [0, 2] - {1}

Let f(x)=x^(3)-x^(2)-x+1 and g(x)={max{f(t);0<=t<=x},0<=x<=1,3-x,1<=x<=2 Discuss the continuity and differentiability of the function g(x) in the interval (0,2).

If f(x)=4x^3-x^2-2x+1 and g(x)={min f(t): 0<=t<=x; 0<=x<=1, 3-x : 1ltxle2} then g(1/4)+g(3/4)+g(5/4) is equal to

Let f(x)={(x+1, -1lexle0),(-x,0ltxle1):} then