Question
Find the intervals in which the following functions are strictly increasing or strictly decreasing:
f (x) = 2x3 – 3x2 – 36x + 7
Solution
Here f (x) = 2x3 – 3x2 – 36x + 7
∴ f ' (x) = 6x2 – 6x – 36 = 6 (x2 – x – 6) = 6 (x + 2) (x – 3)
f ' (x) = 0 gives us 6 (x + 2) (x – 3)
∴ x = – 2, 3
The points x = – 2, 3 divide the real line into three intervals (– ∞, – 2), (– 2, 3), (3, ∞).
In the interval (– ∞, – 2), f ' (x) > 0
strictly ∴ f (x) is strictly increasing in (– ∞, – 2)
In the interval (– 2, 3), f ' (x) < 0 ∞ f (x) is strictly decreasing in ( – 2, 3)
In the interval (3, ∞), f ' (x) > 0
∴ f (x) is strictly increasing in (3, ∞)
∴ we see that f (x) is strictly increasing in (– 8, – 2) ∪ (3, ∞) and strictly decreasing in ( – 2, 3)
