How do you find the maximum, minimum and inflection points and concavity for the function #y = xe^(-x)#?

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Answer 1 · 2017-07-11T01:08:41

There is a local maximum at #(1,e^(-1))#

There is a non-stationary point of inflection at #(2, 2e^(-2))#.

We have:

# y = xe^(-x) #
graph{xe^(-x) [-5, 10, -5, 5]}

Firstly, let us look for critical points, that is coordinates where #dy/dx=0#:

# dy/dx = (x)(-e^(-x)) + (1)(e^(-x)) #
# " " = e^(-x) - xe^(-x) #
# " " = (1-x)e^(-x) #

For a critical point:

# dy/dx = 0 => (1-x)e^(-x) = 0 #

Which has a single solution #x=1#, as #e^a gt 0 AA a in RR#, and

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

Thus there is a single critical point at #(1,e^(-1))#. To determine the nature of the critical point we must examine the second derivative:

# (d^2y)/(dx^2) = -e^(-x) - {(1-x)e^(-x)} #
# " " = -(2-x)e^(-x) #
# " " = (x-2)e^(-x) #

When #x=1 => (d^2y)/(dx^2) = (1-2)e^(-x) lt 0 #

As the second derivative is negative at the critical point then we can conclude the critical point #(1,e^(-1))# is a maximum.

Secondly, we look for inflection points, which are coordinates where the second derivative vanishes:

# (d^2y)/(dx^2) = 0 => (x-2)e^(-x) = 0 #

Which has a single solution #x=2#, for which we have:

# y = 2e^(-2) #

We already know that #x=2# does not correspond to a critical point and thus we have a non-stationary point of inflection at #(2, 2e^(-2))#.