#f(x) = xe^-x#

#f'(x) = (1)e^-x + x[e^-x(-1)]#

# = e^-x-xe^-x#

# = -e^-x(x-1)#

So,

#f''(x) = [-e^-x(-1)] (x-1)+ (-e^-x)(1)#

# = e^-x (x-1)-e^-x#

# = e^-x(x-2)#

Now, #f''(x) = e^-x(x-2)# is continuous on its domain, #(-oo, oo)#, so the only way it can change sign is by passing through zero. (The only partition numbers are the zeros of #f''(x)#)

#f''(x) = 0# if and only if either #e^-x=0# or #x-2 = 0#

#e# to any (real) power is positive, so the only way for #f''# to be #0# is for #x# to be #2#.

We partition the number line:

#(-oo, 2)# and #(2,oo)#

On the interval #(-oo,2)#, we have #f''(x) < 0# so #f# is concave down.

On #(2,oo)#, we get #f''(x) >0#, so #f# is concave up.

**Inflection point**

The point #(2, f(2)) = (2,2/e^2)# is the only inflection point for the graph of this function.