Question

The greatest value of $x^3-18x^2+96x$ in the interval $(0,9)$ is-

• A. $128$
• B. $60$
• C. $160$
• D. $120$

Answer 1

The correct option is C $160$

$f\left(x\right)=x^3-18x^2+96x$
$f^{\prime }\left(x\right)=3x^2-36x+96$
$f^{\prime }\left(x\right)=x^2-12x+32=0$
$=x^2-8x-4x+32=0$
$=x(x-8)-4(x-8)$
$f^{\prime }\left(x\right)=(x-4)(x-8)=0$
$x=4,8$
$f^{\prime }\left(x\right)=2x-12$
$f^{\prime }\left(4\right)=-4<0$
$x=4$ point of maxima
$f\left(4\right)=64-16\times 18+96\times 4$
$=448-288=160$