AB/BC 5.
f is decreasing where f ′(x) < 0, which occurs for −2 < x < 1.
(b) f is decreasing on the interval −2 < x < 1, so there is a minimum at (1, −e^2 ). Note that, as
x approaches ±∞, f (x) = e^2 x(x^2 − 2) is always positive. Hence (1,−e^2 ) is the global
minimum.
(c) As x approaches +∞, f (x) = e^2 x(x^2 − 2) also approaches +∞. There is no global maximum.