, when  – 2 < x < 1 In this case, we have x – 1 < 0, x + 2 > 0 and x – 3 < 0
(In particular, observe that for x = 0, f ′(x) = (x – 1) (x + 2) (x – 3) = (–1) (2) (–3)
= 6 > 0)
So
f ′(x) > 0 when – 2 < x < 1 Thus,
f is increasing in (– 2, 1) Now consider the interval (1, 3), i