#### The smallest value of the polynomial $x^3 - 18x^2 + 96x$ in [0, 9] is A. 126 B. 0 C. 135 D. 160

Let  $f(x)=x^3 - 18x^2 + 96x$

Apply first derivative and get

\begin{aligned} &f^{\prime}(x)=\frac{d\left(x^{3}-18 x^{2}+96 x\right)}{d x}\\ &\text { Apply sum rule and get }\\ &\Rightarrow f^{\prime}(x)=\frac{d\left(x^{3}\right)}{d x}-18 \frac{d\left(x^{2}\right)}{d x}+96 \frac{d(x)}{d x} \end{aligned}

Apply derivative,
$\Rightarrow f'(x) = 3x^2 - 36x + 96 \\ \text{Put } f'(x) = 0 \text{ and get critical points} \\ 3x^2 - 36x + 96 = 0 \\ \Rightarrow 3(x^2 - 12x + 32) = 0 \\ \Rightarrow x^2 - 12x + 32 = 0$

Split middle term and get

$\\ { \Rightarrow x\textsuperscript{2}-8x-4x+32=0}\\ { \Rightarrow x(x-8)-4(x-8)=0}\\ { \Rightarrow (x-8)(x-4)=0}\\ { \Rightarrow x-8=0 or x-4=0}\\ { \Rightarrow x=8 or x=4}\\ { \Rightarrow x \in [0,9]}\\$

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

$\\ {f(x)= x\textsuperscript{3} - 18x\textsuperscript{2} + 96x}\\ {f(0)= 0\textsuperscript{3} - 18(0)\textsuperscript{2} + 96(0)=0}\\ {f(4)= 4\textsuperscript{3} - 18(4)\textsuperscript{2} + 96(4)=64-288+384=160}\\ {f(8)= 8\textsuperscript{3} - 18(8)\textsuperscript{2} + 96(8)=512-1152+768=128}\\ {f(9)= 9\textsuperscript{3} - 18(9)\textsuperscript{2} + 96(9)=729-1458+864=135}\\$

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

So the correct answer is option B.