posted by Tima on .
Find the equation of the line tangent to
f(x) = xe^-x
at the point where x = 0. What does this tell you about the behavior of the graph when x = 0?
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