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

Question

Evelyn Agard · Accepted Answer

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) # Regarding a crucial 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. Next, we search for coordinates known as inflection points, which are the locations where the second derivative disappears: # (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))#.

Aurora Alvarado · Answer

undefined To find the maximum and minimum points:
1. Find the critical points by setting the derivative equal to zero and solving for x.
2. Use the second derivative test to determine whether each critical point corresponds to a maximum, minimum, or neither.

To find the inflection points and concavity:
1. Find the second derivative of the function.
2. Set the second derivative equal to zero and solve for x to find any inflection points.
3. Determine the intervals where the second derivative is positive (concave up) or negative (concave down) to identify concavity.