Answer :

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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