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

Let f(x)= x³ – 18x² + 96x

Applying the first derivative we get

Applying the sum rule of differentiation, so we get

Applying the derivative,

⇒ f' (x)=3x²-36x+96

Putting f’(x)=0,we get critical points

3x²-36x+96=0

⇒ 3(x²-12x+32)=0

⇒ x²-12x+32=0

Splitting the middle term, we get

⇒ x²-8x-4x+32=0

⇒ x(x-8)-4(x-8)=0

⇒ (x-8)(x-4)=0

⇒ x-8=0 or x-4=0

⇒ x=8 or x=4

⇒ x∈[0,9]

Now we will find the value of f(x) at x=0, 4, 8, 9

f(x)= x³ – 18x² + 96x

f(0)= 0³ – 18(0)² + 96(0)=0

f(4)= 4³ – 18(4)² + 96(4)=64-288+384=160

f(8)= 8³ – 18(8)² + 96(8)=512-1152+768=128

f(9)= 9³ – 18(9)² + 96(9)=729-1458+864=135

Hence we find that the absolute minimum value of f(x) in [0,9] is 0 at x=0.

So the correct option is option B.