The smallest value of polynomial $x^3 - 18x^2 + 96x$ in $[0, 9]$ is

$0$

The correct answer is Option (2) → $0$ ##

We have, $f(x) = x^3 - 18x^2 + 96x$

$∴f'(x) = 3x^2 - 36x + 96$

So, $f'(x) = 0$ Gives, $3x^2 - 36x + 96 = 0$

$\Rightarrow 3(x^2 - 12x + 32) = 0 \Rightarrow (x - 8)(x - 4) = 0$

$x = 8, 4 \in [0, 9]$

We shall now evaluate the value of $f$ at these points and at the end points of the interval $[0, 9]$ i.e., at $x = 4$ and $x = 8$ and at $x = 0$ and at $x = 9$.

$∴f(4) = 4^3 - 18 \cdot 4^2 + 96 \cdot 4 = 64 - 288 + 384 = 160$

$f(8) = 8^3 - 18 \cdot 8^2 + 96 \cdot 8 = 512 - 1152 + 768 = 128$

$f(9) = 9^3 - 18 \cdot 9^2 + 96 \cdot 9 = 729 - 1458 + 864 = 135$

and $f(0) = 0^3 - 18 \cdot 0^2 + 96 \cdot 0 = 0$

Thus, we conclude that absolute minimum value of $f$ on $[0, 9]$ is $0$ occurring at $x = 0$.