Question

What are the local extrema of `f(x)= xe^-x`?

Answer 1

`(1,e^-1)`

Explanation:

We need to use the product rule: `d/dx(uv) = u(dv)/dx+v(du)/dx`

`:. f'(x) = xd/dx(e^-x) + e^-x d/dx(x)`
`:. f'(x) = x(-e^-x) + e^-x (1)`
`:. f'(x) = e^-x-xe^-x`

At a min/max `f'(x)=0`
`f'(x)=0 => e^-x(1-x) = 0`
Now, `e^x > 0 AA x in RR`
`:. f'(x)=0 => (1-x) = 0 => x=1`

`x=1 => f(1)=1e^-1 = e^-1`

Hence, there a single turning point at `(1,e^-1)`

graph{xe^-x [-10, 10, -5, 5]}