Question

Find intervals in which f(x)=3/10x⁴-4/5x³-3x²+36/5x+11 is increasing or intervals in which it is which it is increasing.

Answer 1

function, f(x) = (3/10)x⁴ - (4/5)x³ - 3x² + (36/5)x + 11

differentiating f(x) with respect to x,

f'(x) = (3/10) 4x³ - (4/5) 3x² - 6x + (36/5)

= (12/10)x³ - (12/5)x² - 6x + (36/5)

= (12x³ - 24x² - 60x + 72)/10

= 12(x³ - 2x² - 5x + 6)/10

f(x) is increasing only when f'(x) > 0

so, 12(x³ - 2x² - 5x + 6)/10 > 0

⇒x³ - x² - x² + x - 6x + 6 > 0

⇒x²(x - 1) - x(x - 1) - 6(x - 1) > 0

⇒(x² - x - 6)(x - 1) > 0

⇒(x² - 3x + 2x - 6)(x - 1) > 0

⇒(x - 3)(x + 2)(x - 1) > 0

x ∈ (-2, 3) U (3, ∞)

and f(x) is decreasing only when f'(x) < 0

i.e., (x - 3)(x + 2)(x - 1) < 0

x ∈ (-∞, -2) U (1, 3)

hence, f(x) is increasing in intervals (-2, 3) or (3, ∞) while f(x) is decreasing in intervals (-∞, -2) or (1, 3)