If `f(x)=|x-1| ([x]-[-x])`, then find `f'(1^+) & f'(1^-)` where [x] denotes greatest integer function

Let f(x)=(-1)^([x]) where [.] denotes the greatest integer function),then

The function,f(x)=[|x|]-|[x]| where [] denotes greatest integer function:

The function, f(x)=[|x|]-|[x]| where [] denotes greatest integer function:

Let f(x)=[|x|] where [.] denotes the greatest integer function, then f'(-1) is

Let f(x)=[|x|] where [.] denotes the greatest integer function, then f'(-1) is

f(x)=1+[cos x]x, in 0<=x<=(x)/(2) (where [.] denotes greatest integer function)

f(x)=1+[cos x]x in 0<=x<=(pi)/(2) (where [.] denotes greatest integer function)

If domain of f(x) is [-1,2] then domain of f(x]-x^(2)+4) where [.] denotes the greatest integer function is

f(x) = [x-1] + {x)^{x}, x in (1,3), then f^-1(x) is- (where [.] denotes greatest integer function and (J denotes fractional part function)

Domain of f(x)=log(x^2+5x+6)/([x]-1) where [.] denotes greatest integer function: