**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]}