Question

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

Answer 1

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

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

Explanation:

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))`.