(a) Let y = 2x3 – 24x + 107, x ∈ [1,3]
dy/dx = 6x2 – 24
For maxima or minima
dy/dx = 0 ⇒ 6x2 – 24 = 0
⇒ x = ± 2.
∵ x ∈ [1, 3]
∴ x = 2
Now, y1 = 2(1)3 – 24(1) + 107
= 2 – 24 + 107 = 85
y2 = 2(2)3 – 24(2) +107
= 16 – 48 + 107 = 75
y3 = 2(3)3 – 24(3) + 107
= 54 – 72 + 107 = 89
Hence, maximum value is 89 at x = 3 where as minimum value is 75 at x = 2.
(b) Let y = 3x4 – 2x3 – 6x2 + 6x + 1,
x ∈ (0, 2)
dy/dx = 12x3 – 6x2 – 12x + 6
= 6(2x3 – x2 – 2x + 1)
For maxima or minima
Now y1 = 3(1)4 – 2(1)3 – 6(1)2 + 6(1) + 1
⇒ y1 = 3 – 2 – 6 + 6 + 1 = 2
⇒ y-1 = 3 (- 1)4 – 2 (-1)3 – 6(- 1)2 + 6(- 1) + 1
⇒ y-1 = 3 + 2 – 6 – 6 + 1 = -6
⇒ y0 = 3(0)4 – 2(0)3 – 6(0)2 + 6(0) + 1
⇒ y0 = 1
⇒ y2 = 3(2)4 – 2(2)3 – 6(2)2 + 6(2) + 1
= 48 – 16 – 24 + 12 + 1 = 21
∴ Hence, maximum value of function is 2π at x = 2π and minimum value of function is 0 at x = 0.
(d) x3 – 18x2 + 96x, x ∈ [0, 9]
y4 = (0)3 – 18(0)2 + 96(0)
= 64 – 288 + 384 = 160
y8 = (8)3 – 18(8)2 + 96 × 8
= 512 – 1152 + 768
= 128
y9 = (9)3 – 18(9)2 + 96(9)
= 729 – 1458 + 864= 135
So, at x = 0 minimum value = 0
and at x = 4 maximum value = 160