1(1) for any integer n, n <= x < n+1
y = [x] + √{x} = n + √(x-n)
so the curve will be a set of part of parabolas (y-n)2=(x-n)
it will be continuous but non-differentiable at integral points (a pointed curve at integral points)
(2) If we replace x by -x there is no change in the function, so the curve is symmetrical about y-axis
so draw the graph as in (1) in 1st quadrant and plot its reflection in 2nd quadrant.
(3) and (4) are same as fractional part is non-negative. Try it
