The least value of the function

Solution:

(d) 0

Given: $f(x)=x^{3}-18 x^{2}+96 x$

$\Rightarrow f^{\prime}(x)=3 x^{2}-36 x+96$

For a local maxima or a local minima, we must have

$f^{\prime}(x)=0$

$\Rightarrow 3 x^{2}-36 x+96=0$

$\Rightarrow x^{2}-12 x+32=0$

$\Rightarrow(x-4)(x-8)=0$

$\Rightarrow x=4,8$

So,

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

$f(4)=(4)^{3}-18(4)^{2}+96(4)=64-288+384=160$

$f(0)=(0)^{3}-18(0)^{2}+96(0)=0$

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

Hence, 0 is the minimum value in the range $[0,9]$.