Question

Find the intervals of monotonicities and hence find the local extremum for the following functions:

f(x) = 2x³ + 3x² – 12x

Answer 1

f'(x) = 6x² + 6x – 12

f'(x) = 0

⇒ 6(x² + x – 2) = 0

(x + 2)(x – 1) = 0

Stationary points x = – 2, 1

Now, the intervals of monotonicity are
`(- oo, -2), (-2, 1) and (1, oo)`

In `(- oo, -2)`, f'(x) > 0 ⇒ f(x) is strictly increasing.

In (– 2, 1), f'(x) < 0 ⇒ f(x) is strictly decreasing.

In `(1, oo)`, f'(x) > 0 ⇒ f(x) is strictly increasing.

f(x) attains local maximum as f'(x) changes its sign from positive to negative when passing through x = – 2.

Local maximum

f(– 2) = 2(– 8) + 3(4) – 12(– 2)

= – 16 + 12 + 24

= 20

f(x) attains local minimum as f'(x) changes its sign from negative to positive when passing through x = 1.

∴ Local minimum f(1) = 2 + 3 – 12 = – 7