Question
Find the intervals in which the function
f(x) = x3 – 12x2 + 36x + 17 is
(a) strictly increasing (b) strictly decreasing
Solution
(i) Here f (x) = x3 – 12x2 + 36x + 17
∴ f '(x) = 3x2 – 24x + 36 = 3 (x2 – 8x + 12) = 3 (x – 2) (x – 6)
f ' (x) = 0 gives us 3 (a – 2) (a – 6)
∴ x = 2, 6
The points x = 2, 6 divide the real line into three intervals (– ∞ , 2), (2, 6), (6, ∞)
(1) In the interval (– ∞, 2), f ' (x) > 0
∴ f (x) is increasing in (– ∞, 2)
(2) In the interval (2, 6), f ' (x) < 0
∴ f ' (x) is decreasing in (2, 6).
(3) In the interval (6, ∞), f ' > 0
∴ f (x) is increasing in (6, ∞).
