Based on comments from the other answer, the intended function was #"floor"(x/2)#, that is, the function on #RR# whose output is the greatest integer less than or equal to the input. Let #f(x) = "floor"(x)#. Then a graph of #f(x)# looks like this:

Notice that points of discontinuity occur every time we approach an integral value. When we reach that value, we *jump* from the lesser integer less than that value to that value (notice these are jump discontinuities) If we instead apply #f# to #x/2#, then those integral values will occur only half as often, at #x in {...-2, 0, 2, 4, ...}#. This will cause the line segments to extend:

So the points of discontinuity occur for #f(x/2)# at each even integer.

Now, #f(x/2)# is clearly continuous at every #x!in{...-2,0,2,4,...}#, so we will now examine whether it is left or right continuous at those points.

Notice that if we approach an even value of #x# from the right, we are traveling along a constant line whose value matches #f(x/2)#. For example, #lim_(x->2^+)f(x/2) = lim_(x->2^+)1 = 1#, as #f(x/2)# matches the constant function #1# on the right side of #x=2#. In fact, this is the case at every even integer, meaning we have #lim_(x->x_0^+)f(x/2) = f(x_0/2)# for all #x in RR#. Thus, #f(x/2)# is right-continuous.

If we approach from the left, however, we get a different result. Let's look at #x=2# again.

#lim_(x->2^-)f(x/2) = lim_(x->2^-)0 = 0#, as #f(x/2)# matches the constant function #0# on the left side of #x=2#. As #lim_(x->2^-)!=f(2/2)#, this means #f(x/2)# is not left-continuous at #x=2#. Indeed, this is the case at every even integer. So, taken together, we get, using the notation #2ZZ# to represent the even integers:

#f(x)# has jump discontinuities at #x in 2ZZ#

#f(x)# is right-continuous on #RR#

#f(x)# is left-continuous on #RR-2ZZ#