Question

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

• A. $126$
• B. $0$
• C. $135$
• D. $160$

Answer 1

Answer: (B)

$0$

Given $f\left(x\right)=x^3-18x^2+96x$

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

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

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

$=3(x^2-12x+32)=0$

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

$=3\left(x\right(x-4)-8(x-4\left)\right)=0$

$=3(x-4)(x-8)=0$

$x=4,8$

For least value of $f\left(x\right)$ in $[0,9]$ we should use the value of $f\left(x\right)$ at $x=0,4,8,9$

$f\left(0\right)=0$

$f\left(4\right)=4^3-18\times 4^2+96\times 4=64-288+384=160$

$f\left(3\right)=8^3-18\times 8^2+96\times 8=128$

$f\left(9\right)=9^3-18\times 9^2+96\times 9=279-1458+864=156$

The minimum value of $f\left(x\right)$ is at $0$

$f\left(0\right)=0$

Ans $=0$