e , when – ∞ < x < – 2 In this case, we have x – 1 < 0, x + 2 < 0 and x – 3 < 0 (In particular, observe that for x = –3, f ′(x) = (x – 1) (x + 2) (x – 3) = (– 4) (– 1)
(– 6) < 0)
Therefore,
f ′(x) < 0 when – ∞ < x < – 2