# The smallest value of the polynomial x3 – 18x2 + 96x in [0, 9] isA. 126B. 0C. 135D. 160

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.

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