Question

How do you find points of inflection and determine the intervals of concavity given `y=xe^(1/x)`?

Answer 1

The function `y=xe^(1/x)` is convex for `x>0` and concave for `x<0`. It has no inflection points. It has discontinuity at `x=0`

Explanation:

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]}