(i) Given f(x) = 9x² + 12x + 2 ….(1) 

f'(x) = 18x + 12 …(2) 

f”(x) = 18 > 0 ……(3) 

⇒ f(x) attains minimum 

f'(x) = 0 ⇒ 18x+ 12 = 0 ⇒ x = -\(\frac{2}{3}\)

& f”(-\(\frac{2}{3}\) )18 > 0 ⇒ f(x) is minimum & the minimum value is 

f(-\(\frac{2}{3}\)) = 9\(\frac{4}{9}\) + 12(-\(\frac{2}{3}\)) + 2 

= 4 – 8 + 2 = -2 

(ii) f(x) = 2x³ – 15x² + 36x + 10 …….(1) 

f‘(x) = 6x² – 30x + 36 = 6 (x² – 5x + 6) ……..(2) 

f”(x) = 12x – 30 …..(3) 

f'(x) = 0 ⇒ x 2 -5x + 6 = 0 

⇒ (x – 3)(x – 2) = 0 ⇒ x = 3 or 2 

when x = 3 f “(x) = 12x – 30 

f“(3) = 36 – 30 = 6 > 0 

⇒ f(x) has minimum 

Minimum value is f(3) = 2(3)³ – 15(3)² + 36(3) + 10

f(3) = 54 – 135 + 108 + 10 = 37 

when x = 2, f”(2) = 24 – 30 = -6 < 0 

⇒ f(x) has maximum 

maximum value is f(2) = 2(2)³ – 15(2) + 36(2) + 10 f(2) = 16 – 60 + 72 + 10 = 38. 

(iii) Given f(x) = 2x³ – 21x² + 36x – 20 ….. (1) 

f'(x) = 6x² – 42x + 36 …… (2) 

= 6(x² – 7x + 6) 

f'(x) = 6(x – 1) (x – 6) = 0 ⇒ x = 1 or 6 

f”(x) = 12x – 42 … (3) 

when x = 1, f “(1) = 12 – 42 = -30 < 0 ⇒ f(x) is maximum 

maximum value is f(1) = 2 – 21 + 36 – 20 = -3 when x = 6, f“(6) = 72 – 42 = 30 > 0 ⇒ f(x) is minimum 

minimum value is f(6) = 2(6)3³ – 21 (6)² + 36(6) – 20 

f(6) = 432 – 756 + 216 – 20 = -128 

(iv) Given f(x) = 12x⁵ – 45x⁴ + 40 x³ + 6 ….(1) 

f'(x) = 60x⁴ – 180x³ + 120x² ….(2) 

= 60x² (x² – 3x + 2) = 60x² (x – 1) (x – 2) 

f‘(x) = 0 ⇒ 60x²(x – 1) (x – 2) = 0 ⇒ x = 0, 1, 2 f “(x) = 60 (4x³ – 6x + 4x) ……. (3) 

when x = 0, f”(x) = 0 ⇒ f(x) has neither maximum nor minimum 

when x = 1, f”(x) = -1 < 0 ⇒ f(x) has a maximum & maximum value is 

f(1) = 12 – 45 + 40 + 6 = 13 

When x =2 f”(x) = 4 > 0 

∴ f(x) has a minimum 

minimum value is f(2) = 12(32) – 45(16) + 40(8) + 6 

f(2) = 384 – 720 + 320 + 6 = -10.