Please help: calculus prob

Question

Where is the second derivative of y = 2xe^-x equal to 0? (5 points)

Options:

1. 0
2. 1
3. 2
4. 4

Answer 1

The first derivative would be 2x(-e^-x) + 2(e^-x) = -2xe^-x + 2e^-x

The second derivative would be: -2x(-e^-x) - 2(e^-x) + 2(-e^-x)

which equals 2xe^-x - 4e^-x

2xe^-x - 4e^-x = 0

Factor our e^-x:

e^-x(2x - 4) = 0

e^-x can never equal 0

2x - 4 = 0 when x = 2