The smallest value of the polynomial x3 – 18x2 + 96x in [0, 9] is ______.

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The smallest value of the polynomial x3 – 18x2 + 96x in [0, 9] is 0. Explanation: Let f(x) = x3 – 18x2 + 96x So, f'(x) = 3x2 – 36x + 96 For local maxima and local minima f'(x) = 0 ∴ 3x2 – 36x + 96 = 0 ⇒ x2 – 12x + 32 = 0 ⇒ x2 – 8x – 4x + 32 = 0 ⇒ x(x – 8) – 4(x – 8) = 0 ⇒ (x – 8)(x – 4) = 0 ∴ x = 8, 4 ∈ [0, 9] So, x = 4, 8 are the points of local maxima and local minima. Now we will calculate the absolute maxima or absolute minima at x = 0, 4, 8, 9 ∴ f(x)= x3 – 18x2 + 96x "f"(x)_(x = 0) = 0 – 0 + 0 = 0 "f"(x)_(x = 4) = (4)3 – 18(4)2 + 96(4) = 64 – 288 + 384 = 448 – 288 = 160 "f"(x)_(x = 8) = (8)3 – 18(8)2 + 96(8) = 512 – 1152 + 768 = 1280 – 1152 = 128 "f"(x)_(x = 9) = (9)3 – 18(9)2 + 96(9) = 729 – 1458 + 864 = 1593 – 1458 = 135 So, the absolute minimum value of f is 0 at x = 0

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