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

The correct option is 0.

To find the smallest value of the polynomial $f\left(x\right)=x^3-18x^2+96x$ in the closed interval $[0,9]$, we need to evaluate the function at its critical points within the interval and at the boundary points of the interval.

First, let's find the derivative of $f\left(x\right)$ with respect to $x$ to locate the critical points:
$f^{\text{'}}\left(x\right)=\frac{d}{dx}(x^3-18x^2+96x)=3x^2-36x+96$

Set the derivative to zero to find the critical points:
$3x^2-36x+96=0$

Dividing the entire equation by 3 gives:
$x^2-12x+32=0$

Factoring the quadratic equation:
$(x-4)(x-8)=0$

This yields the critical points:
$x=4$ and $x=8$

Both critical points $x=4$ and $x=8$ lie within the interval $[0,9]$. Now, we evaluate $f\left(x\right)$ at the critical points and the boundary points $x=0$ and $x=9$.

1. At the boundary point $x=0$:
$f\left(0\right)=0^3-18\left(0^2\right)+96\left(0\right)=0$

2. At the critical point $x=4$:

3. At the critical point $x=8$:
$f\left(8\right)=8^3-18\left(8^2\right)+96\left(8\right)=512-18\left(64\right)+768=512-1152+768=128$

4. At the boundary point $x=9$:
$f\left(9\right)=9^3-18\left(9^2\right)+96\left(9\right)=729-18\left(81\right)+864=729-1458+864=135$

Comparing the values:
- $f\left(0\right)=0$
- $f\left(4\right)=160$
- $f\left(8\right)=128$
- $f\left(9\right)=135$

The smallest value among these is 0, which occurs at the boundary point $x=0$.