Question

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

विकल्प

• 126

• 0

• 135

• 160

Answer 1

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