AB 5.
(b) (i) Use washers; then(ii) See the figure above.(a) f(x) = e^2 x(x^2 − 2),
f ′(x) = e^2 x(2x) + 2e2x(x^2 − 2)
= 2e^2 x(x + 2)(x − 1)
= 0 at x = −2, 1.
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.