To determine concavity, analyze the sign of ##f”(x)##.
##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.
