Question

Determine the coordinates of all points at which the tangent to the curve defined by y = x^2(e^-x) is horizontal.

Answer 1

tangents r horizontal
slope=0
dy/dx=0

Answer 2

Ok so this is the derivative that I got:
y' = -1x^2 * e^-x

Answer 3

Making it equal 0:
0 = -1x^2 * e^-x <-- Now Im getting confused.

Answer 4

y'=2x(e^-x)-x^2(e^-x)

Answer 5

How did you get that? Oh wait You used product rule?

Answer 6

yes product rule

Answer 7

\[e ^{-x}x[2-x]=0\]

Answer 8

x=0 x=2

Answer 9

Thank you