For `0 le x le 1`, the function
`f(x)=|x|+|x-1|` is

The function f(x)=cos x, 0 le x le pi is

Draw the graph of the function: f(x) = |1-x| + |1+x|, -2 le x le 2

The range of the function f(x)=|x-1|+|x-2|, -1 le x le 3, is

Draw the graph of the function f(x)= x- |x-x^(2)|, -1 le x le 1 and find the points of non-differentiability.

If f(x)=2x+3 , where 1 le x le 2 , then the range of the function f is given by

If f(x)=4x-3 , where -3 le x le 1 then the range of the function f is given by

Draw the graph of the function f(x) = x - |x - x^(2)|, -1 le x le 1 and discuss the continuity or discontinuity of f in the interval -1 le x le 1

Let f (x) be defined as f (x) ={{:(|x|, 0 le x lt1),(|x-1|+|x-2|, 1 le x lt2),(|x-3|, 2 le x lt 3):} The range of function g (x)= sin (7 (f (x)) is :

Let f(x) and g(x) be two functions given by f(x)=-1|x-1|,-1 le x le 3 and g(x)=2-|x+1|,-2 le x le 2 Then,

Consider the functions f(x)={(x+1",",x le 1),(2x+1",",1lt x le 2):} and g(x)={(x^(2)",", -1 le x lt2),(x+2",",2le x le 3):} The domain of the function f(g(x)) is