Find the absolute maximum and absolute minimum values of f on the given interval.

Question

f(x) = xe^{-x^2 / 128} , [-4,16]

Answer 1

At x=-4, y = -3.53

At x = 8, y = 4.852

At x=16, y = 2.165

the derivative is x * exp(-x^2/128) (-x/64) + exp(-x^2/128)

= exp("-x^2/128) [ -x^2/64 + 1]

exponential is never zero, so -x^2/64 + 1 = 0 ---> x^2 = 64 ---> x = +or- 8 are the critical values