Question

Find the intervals in which the function f given by f(x) = 2x^3 − 3x^2 − 36x + 7 is (a) strictly increasing (b) strictly decreasing

Answer 1

$f(x) = 2 {x}^{3} - 3 {x}^{2} - 36x + 7 \\ f'(x) = 6 {x}^{2} - 6x - 36 \\ taking \: f'(x) = 0 \\ 6 {x}^{2} - 6x - 36 = 0 \\ {x}^{2} - x - 6 = 0 \\ (x - 3)(x + 2) = 0 \\ \\ critical \: points \: for \: f'(x) \: are \: - 2 \: and \: 3 \\ \\ f'(x) \geqslant 0 \: xϵ[3, \infty ) \\ or \\ f'(x) \geqslant 0 \: xϵ[ - 2, - \infty ) \\ \\ and \: f'(x) \leqslant 0 \: xϵ[ - 2, 3] \\ so \: \\ function \: is \: increasing \: in \\ \\ = [3, \infty )∪[ - 2, - \infty ) \: \\ decreasing \: in \\ [ - 2, 3]$
hope it will help you!!

Answer 2

given, f(x) = 2x³ - 3x² - 36x + 7
differentiate f(x) with respect to x,
f'(x) = 2.3x² - 3.2x - 36 + 0
f'(x) = 6x² - 6x -36 = 6(x² - x - 6)
f'(x) = 6(x² - x - 6) -------(1)

(a) when f is strictly increasing function :
f'(x) > 0
from equation (1),
6(x² - x - 6) > 0
x² - 3x + 2x - 6 > 0
x(x - 3) + 2(x - 3) > 0
(x + 2)(x - 3) > 0
x > 3 or, x < -2
e.g., x∈(-∞ , -2) U (3, ∞)


(b) when f is strictly decreasing function :
f'(x) < 0
from equation(1),
6(x² - x - 6) < 0
(x + 2)(x - 3) < 0
-2 < x < 3
e.g., x∈ (-2, 3)