Solution We have
f (x)
= 4x3 – 6x2 – 72x + 30
or
f ′(x) = 12x2 – 12x – 72
= 12(x2 – x – 6)
= 12(x – 3) (x + 2)
Therefore, f ′(x) = 0 gives x = – 2, 3 The
points x = – 2 and  x = 3 divides the real line into
three disjoint intervals, namely, (– ∞, – 2), (– 2, 3)
and (3, ∞) Fig 6 3
Rationalised 2023-24
 MATHEMATICS
156
In the intervals (– ∞, – 2) and (3, ∞), f ′(x) is positive while in the interval (– 2, 3),
f ′(x) is negative