We can calculate the first derivative separately for #x<0# and #x>0#:
1) For #x<0# we have:
#y=x-ln(-x)#
#(dy)/(dx) = 1-1/x#
2) For #x>0# we have:
#y=x-ln(x)#
#(dy)/(dx) = 1-1/x#
Thus the derivative is the same in the two intervals #(-oo,0)# and #(0,+oo)#. The second derivative is:
#(d^2y)/(dy^2) = d/(dx) (1-1/x)= 1/x^2#
We can therefore see that the function has only a critical point for #x=1# and that
#y'(x) < 0# for #x in (-oo,0) uu (0,1)#
#y'(x) >0# for #x in (1,+oo)#
so this critical point is a local minimum. It has no inflection points and is concave up in its domain.
We can also note that:
#lim_(x->-oo) y(x) = -oo#
#lim_(x->+oo) y(x) = +oo#
#lim_(x->0) y(x) = +oo#
So #y(x)# starts from #-oo# strictly increases approaching #+oo# for #x->0^-#, then decraeses starting from #+oo# as #x->0^+#, reaches a minimum for #x=1# and then starts increasing again approaching #+oo# for #x->+oo#.
