Question

Find the smallest value of the polynomial $x^{3}-18x^{2}+96x$ in the interval $\left[0 , 9\right]$ .

Answer 1

Let, $f\left(x\right)=x^{3}-18x^{2}+96x$
Differentiating it w.r.t. $x$
$f^{'}\left(x\right)=3x^{2}-36x+96$
$=3\left(x^{2} - 12 x + 32\right)$
$=3\left(x - 4\right)\left(x - 8\right)$

Clearly at $x=8f\left(x\right)$ is minimum.
$\min f\left(x\right)=8\left(64 - 144 + 96\right)=128$
At $x=0,f\left(x\right)=0$