Question

How do you determine whether the function `f(x) = xe^-x` is concave up or concave down and its intervals?

Answer 1

To determine concavity, analyze the sign of `f''(x)`.

Explanation:

`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.