Show that the least value of the function `f(x)=x^3-18 x^2+96 x` in the interval `[0,9]` is 135.

Explanation

We have, `f(x)=x^(3)-18x^(2)+96x`
`therefore f^(')(x)=3x^(2)-36x+96`
So, `f^(')(x)=0`
Gives, `x^(2)-36x + 96=0`
`rArr 3(x^(2)-12x+32)=0`
`rArr (x-8)(x-4)=-0`
`rArr 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)-18.4^(2)+96.4`
`=64-288+384=160`
`f(8)=8^(3)-18.8^(2)+96.8=128`
`f(9)= 9^(3)-18.9^(2)+96.9`
`=729-1458+864=135`
and `f(0)=0^(3)-18.0^(2)+96.0=0`
Thus, we conclude that absolute minimum value of f on [0,9] is 0 occurring at x=0.