Question

How do you find the critical numbers for `f(x)= (x-1) e^x` to determine the maximum and minimum?

Answer 1

The only critical point is for `x=0` and the function has a local minimum in that point.

Explanation:

Critical points are found by equating the first derivative to zero:

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

Solving:

`xe^x = 0`

the only critical point is `x=0`

Now consider that:

`f'(x) = xe^x < 0` for `x<0`

`f'(x) = xe^x > 0` for `x>0`

We have then that `f(x)` is decreasing in `(-oo,0)` and increasing in `(0,+oo)`, therefore `x=0` is a local minimum.

graph{(x-1)e^x [-10, 10, -5, 5]}