What does it mean?
To determine concavity and points of inflection we need to find second derivative.
If #y# is continuous at some point and:
#y''>0<=>#function is convex (or concave up)
#y''<0<=>#function is concave (or concave down)
#y''=0# and it changes sign#<=>#inflection point.
We will discover various features of this function and sketch the curve as we go.
If you want to see how I got any of the following limit let me know in the comments.
Before deriving
#y=xe^(1/x)#
Domain: #x in RR"\"{0}# or #x!=0# - no y intercept
#y!=0# for #x!=0# - no x intercept
Finding limits:
#lim_(x->0^-)xe^(1/x)=0#
#lim_(x->0^+)xe^(1/x)=+oo#
#lim_(x->0^-)y!=lim_(x->0^+)y# so the limit at 0 doesn't exist and we have discontinuity. Assuming #x!=0# from now on.
#lim_(x->+-oo)xe^(1/x)=+-oo#
Because these limits are infinite, there could be up to 2 diagonal asymptotes #y=ax+b#. We find them by formulas
#a=lim_(x->+-oo)y/x=lim_(x->+-oo)e^(1/x)=1#
and
#b=lim_(x->+-oo)y-ax=lim_(x->+-oo)xe^(1/x)-x=1#.
It appears that both asymptotes exist and are the same.
First derivative
#y=xe^(1/x)#
#y'=e^(1/x)+xe^(1/x)*(-1)/x^2# (product rule and chain rule)
#y'=(1-1/x)e^(1/x)#
Setting it to 0 to find extrema
#(1-1/x)e^(1/x)=0#
#(1-1/x)=0#
#1=1/x#
#x=1# - candidate for extremum.
It must be a minimum, because here #y'# changes sign from negative to positive.
Most limits of first derivative are already given from general function behavior.
#lim_(x->0^+)dy/dx=-oo#
#lim_(x->+-oo)dy/dx=1#
One left to calculate
#lim_(x->0^-)(x-1)/xe^(1/x)=0#
Second derivative
#y'=(1-1/x)e^(1/x)#
#y''=1/x^2e^(1/x)+(1-1/x)e^(1/x)*(-1)/x^2# (product rule and chain rule)
#y''=1/x^2e^(1/x)(1-(1-1/x))#
#y''=e^(1/x)/x^3!=0# (thus no inflection points)
Looking at sign we see that #y''# has the same sign as #x#, so your function #y=xe^(1/x)# is convex for #x>0# and concave for #x<0#.
Graph of #xe^(1/x)# together with its diagonal asymptote:
graph{(y-xe^(1/x))(y-x-1)sqrt(|x|-0.03)=0 [-10, 10, -5, 5]}
