f = xe^-x^2
f' = (1-2x^2) e^-x^2
so, where is f' = 0?
Find the absolute extrema of the function. (Round your answer to three decimal places.)
f(x) = xe-x2 on [0,2]
Absolute maximum value:
at x =
Absolute minimum value:
at x =
1 answer
1 answer