Find the equation of the line tangent to

Question

f(x) = xe^-x

at the point where x = 0. What does this tell you about the behavior of the graph when x = 0?

Answer 1

first of all , when x = 0 , f(0) = 0
so the point is (0,0)

f'(x) = x(-e^-x) + e^-x
when x = 0
f'(0) = 0 + 1 = 1
equation of tangent : y = x

at (0,0) the function is increasing