Find the intervals of monotonicity and hence find the local extrema for the function f(x) = x² - 4x +4.
Answers
Step-by-step explanation:
(i) f(x) = 2x3 + 3x2 – 12x f'(x) = 6x2 + 6x – 12 f'(x) = 0 ⇒ 6(x2 + x – 2) = 0 (i.e.,) 6(x + 2)(x – 1) = 0 ⇒ x = -2 or 1 Taking the points in the number line The intervals are (-∞, -2), (-2, 1), (1, ∞) when x ∈ (-∞, -2), f'(x) = 6 (-1) (-4) = +ve say x = – 3 ⇒ f(x) is strictly increasing in the interval (-∞, -2) when x ∈ (-2, 1), f’ (x) = 6 (2) (-1) = -ve say x = 0 ⇒ f(x) is strictly decreasing in the interval (-2, 1) when x ∈ (1, ∞), f'(x) = 6 (4) (+1) = + ve say x = 2 ⇒ f(x) is strictly increasing in (1, ∞) Since f(x) changes from +ve to – ve when passing through -2, the first derivative, test tells us there is a local maximum at x = -2 and the local maximum value is f(-2) = 20. Again f'(x) changes from – ve to +ve when passing through 1 ⇒ there is a local minimum at x = 1 and the local minimum value is f( 1) = -7. So (1 )f(x) is strictly increasing on (-∞, -2) and (1, ∞). And (2)f(x) is strictly decreasing on (-2, 1) The local maximum = 20 and the local minimum = -7 (ii) f(x) = x/x - 5 f(x) is strictly decreasing on (-∞, 5) and (5, ∞) And there is no local extremum (iii) f(x) = ex/(1 - ex) For all x values, so f(x) is strictly increasing in (-∞, ∞) and there is no local extremum. (iv) f(x) = (x3/3) - log x ⇒ f(x) is strictly decreasing in (0, 1) say x = 2 ⇒ f(x) is strictly increasing in (1, ∞) Since f'(x) changes from -ve to +ve at x = 1, there is a local minimum at x = 1 and the local minimum values is f(1) = (1/3) − 0 = 1/3 So the function is strictly decreasing on (0, 1) and strictly increasing on (1, ∞) and the local minimum value is (1/3) (v) f(x) = sin x cos x + 5 So the interval are Read more on Sarthaks.com - https://www.sarthaks.com/879952/find-the-intervals-monotonicities-hence-find-local-extremum-for-the-following-functions