Given f (x) = 5 + 36x + 3x2 –2x3
⇒ f’(x)= 36 + 6x – 6x2
For f(x) now we have to find critical point, we must have
⇒ f’(x)= 0
⇒ 36+ 6x – 6x2 = 0
⇒ 6(–x2 +x + 6) = 0
⇒ 6(–x2 +3x – 2x + 6) = 0
⇒ –x2 +3x – 2x + 6 = 0
⇒ x2 –3x + 2x – 6 = 0
⇒ (x– 3) (x + 2) = 0
⇒ x= 3, – 2
Clearly, f’(x) > 0 if –2< x < 3 and f’(x) < 0 if x< –2 and x > 3
Thus, f(x) increases on x ∈ (–2, 3) and f(x) is decreasing on interval (–∞, –2) ∪ (3, ∞)
(vi)
Given f (x) = 8 + 36x + 3x2 – 2x3
Now differentiating with respect to x
⇒ 
⇒ f’(x)= 36 + 6x – 6x2
For f(x) we have to find critical point, we must have
⇒ f’(x)= 0
⇒ 36+ 6x – 6x2 = 0
⇒ 6(–x2 +x + 6) = 0
⇒ 6(–x2 +3x – 2x + 6) = 0
⇒ –x2 +3x – 2x + 6 = 0
⇒ x2 –3x + 2x – 6 = 0
⇒ (x– 3) (x + 2) = 0
⇒ x= 3, – 2
Clearly, f’(x) > 0 if –2 < x < 3 and f’(x) < 0 if x< –2 and x > 3
Thus, f(x) increases on x ∈ (–2, 3) and f(x) is decreasing on interval (–∞, 2) ∪ (3, ∞)
(vii)
Given f(x) = 5x3 – 15x2 – 120x +3
Now by differentiating above equation with respect x, we get
⇒ 
⇒ f’(x)= 15x2 – 30x – 120
For f(x) we have to find critical point, we must have
⇒ f’(x)= 0
⇒ 15x2 –30x – 120 = 0
⇒ 15(x2 –2x – 8) = 0
⇒ 15(x2 –4x + 2x – 8) = 0
⇒ x2 –4x + 2x – 8 = 0
⇒ (x– 4) (x + 2) = 0
⇒ x= 4, – 2
Clearly, f’(x) > 0 if x < –2 and x > 4 and f’(x) < 0if –2 < x < 4
Thus, f(x) increases on (–∞,–2) ∪ (4, ∞) and f(x) is decreasing oninterval x ∈ (–2,4)
(viii)
Given f (x) = x3 – 6x2 – 36x + 2
⇒ 
⇒ f’(x)= 3x2 – 12x – 36
For f(x) we have to find critical point, we must have
⇒ f’(x)= 0
⇒ 3x2 –12x – 36 = 0
⇒ 3(x2 –4x – 12) = 0
⇒ 3(x2 –6x + 2x – 12) = 0
⇒ x2 –6x + 2x – 12 = 0
⇒ (x– 6) (x + 2) = 0
⇒ x= 6, – 2
Clearly, f’(x) > 0 if x < –2 and x > 6 and f’(x) < 0if –2< x < 6
Thus, f(x) increases on (–∞,–2) ∪ (6, ∞) and f(x) is decreasing oninterval x ∈ (–2,6)
(ix)
Given f (x) = 2x3 – 15x2 + 36x +1
Now by differentiating above equation with respect x, we get
⇒ 
⇒ f’(x)= 6x2 – 30x + 36
For f(x) we have to find critical point, we must have
⇒ f’(x)= 0
⇒ 6x2 –30x + 36 = 0
⇒ 6(x2 – 5x + 6) = 0
⇒ 6(x2 –3x – 2x + 6) = 0
⇒ x2 –3x – 2x + 6 = 0
⇒ (x– 3) (x – 2) = 0
⇒ x= 3, 2
Clearly, f’(x) > 0 if x < 2 and x > 3 and f’(x) < 0if 2 < x < 3
Thus, f(x) increases on (–∞, 2) ∪ (3, ∞) and f(x) is decreasing oninterval x ∈ (2,3)
(x)
Given f (x) = 2x3 + 9x2 + 12x +20
Differentiating above equation we get
⇒ 
⇒ f’(x)= 6x2 + 18x + 12
For f(x) we have to find critical point, we must have
⇒ f’(x)= 0
⇒ 6x2 +18x + 12 = 0
⇒ 6(x2 +3x + 2) = 0
⇒ 6(x2 +2x + x + 2) = 0
⇒ x2 +2x + x + 2 = 0
⇒ (x+ 2) (x + 1) = 0
⇒ x= –1, –2
Clearly, f’(x) > 0 if –2 < x < –1 and f’(x) < 0 if x< –1 and x > –2
Thus, f(x) increases on x ∈ (–2,–1) and f(x) is decreasing on interval (–∞, –2) ∪ (–2, ∞)