When you take the first derivative (#d/(dx)#) and set it equal to 0, you can find where the slope is 0, and that lets you determine at what points on the graph that there are concavities. Taking the second derivative (#d^2/(dx^2)#) tells you whether it is concave up (positive) or down (negative).
#f(x) = 1 + x^-1#
#d/(dx)[f(x)] = 1 + (-1x^-2) = 1 - 1/(x^2) = 0#
#1 = 1/(x^2) => x^2 = 1 => x = 1, x = -1#
Plug it back into #f(x)# to get:
#f(1) = 2#
#f(-1) = -2#
Thus, your two coordinates for the extrema are #(-1,-2)# and #(1,2)#. Now, let's take the second derivative to find their concavity directions.
#d^2/(dx^2)[f(x)] = -(-2x^(-3)) = 2/(x^3)#
We don't have to set it to 0 here. If we plug in #1# and #-1# from the earlier first derivative, we get:
#(d^2)/(dx^2)[f(1)] = 2#
#(d^2)/(dx^2)[f(-1)] = -2#
Since the first one is positive, the coordinate #(1, 2)# is where the concave-up part of the curve changes direction, and since the second one is negative, the coordinate #(-1,-2)# is where the concave-down part of the curve changes direction.
Naturally, an asymptote is #x = 0#. Another asymptote is #x = y#. Let's try it:
#y = y + 1/y => 0 = 1/y => y ne 1/0 = "undefined"#
Anyways, since there's a vertical asymptote at #x = 0#, the domain is (interval notation):
#(-oo, 0) uu (0, oo)#
So, the graph is therefore concave down at #(-oo, 0)#, and concave up at #(0, oo)#.
Overall, you know what #y = x# and #y = 1/x# look like; adding them together has #y = x# transform #y = 1/x# by giving it a #y = x# asymptote, like so:
graph{x+1/x [-10, 10, -5, 5]}
