(i) Given
f(x) = 2x3 – 15x2 – 84x + 7
f'(x) = 6x2 – 30x – 84
= 6(x2 – 5x – 14) = 6(x – 7) (x + 2)
f(x) is increasing if f'(x) > 0
(x – 7) (x + 2) > 0
Case – 1
x – 7 > 0 & x + 2 > 0
x > 7 & x > -2
x > 7 ⇒ (7, ∞)
Case – 2:
x – 7 < 0 & x + 2 < 0
x < 7 & x < -2
x < -2 ⇒ (-∞ , -2) U (7,∞)
∴ Interval is (-∞ , -2) U (7,∞)
f(x) is decreasing if f'(x) < 0
(x – 7) (x + 2)
x – 7 < 0 & x + 2 > 0
Case -1:
x < 7 & x > -2
∴ -2 < x < 7
Case – 2
x – 7 > 0 & x + 2 < 0
x > 7 & x < -2 (not possible)
(ii) Given f(x) = x4 – 2x3 + 1
f'(x) = 4x3 – 6x2
= 2x2 (2x – 3) Here 2x2 > 0
f(x) is increasing if f'(x) > 0
⇒ 2x – 3 > 0
x > \(\frac{3}{2}\)⇒ (\(\frac{3}{2}\), ∞) is the interval
f(x) is decreasing if f‘(x) < 0
2x – 3 < 0 x < \(\frac{3}{2}\)(-∞, \(\frac{3}{2}\)) is the interval`
(iii) Given f(x) = x3 – 3x2 + 3x – 100
f'(x) = 3x2 – 6x + 3
= 3(x2 -2x + 1)
= 3(x – 1)2
f(x) is increasing
f‘(x) > 0 ⇒ (x – 1)2 > 0
increasing for all
∵ f'(x) > 0 it will not be decreasing
(iv) Given f(x) = 2x2 – 96x + 5
f ‘(x) = 4x – 96 = 4(x – 24)
f(x) is increasing if f’(x) > 0
x – 24 > 0
x > 24
f(x) is decreasing if
f'(x) < 0 x – 24 < 0 x < 24
(v) f(x) = 10 – 6x – 2x2
f'(x) = -6 -4x = -(6 + 4x) = -(6 + 4x)
f(x) is increasing if -(6 + 4x) > 0
⇒ 6 + 4x < 0
⇒ 4x < -6
⇒ x > – –\(\frac{3}{2}\)
f(x) is decreasing : if -(6 + 4x) < 0
⇒ 6 + 4x > 0 ⇒ 4x > -6
⇒ x > –\(\frac{3}{2}\)
(vi) Given
f(x) = 2x3 + 9x2 + 12x + 20
f'(x) = 6x2 + 18x + 12
= 6(x2 + 3x + 2)
= 6(x + 2) (x + 1)
f(x) is increasing if f ‘(x) > 0
(x + 1) (x + 2) > 0
Case 1:
x + 2 > 0 and x + 1 > 0
x > -2 and x > -1
⇒ x > -1 ⇒ (-1, ∞)
Case 2:
x + 2 < 0 & x + 1 < 0
x < -2 & x < -1
x < -2 ⇒ (-∞, -2)
f(x) is decreasing if f ‘(x) < 0
(x + 2) (x + 1) < 0
Case 1:
x + 2 < 0 and x + 1 > 0 x < -2 and x > -1 Not possible
Case – 2:
x + 2 > 0 and x + 1 < 0
x > -2 and x < -1
x > -2 and x < -1 ⇒ -2 < x < -1.
