Question

Let $f\left(x\right)=6-12x+9x^2-2x^3,1\le x\le 4.$ Then the absolute maximum value of the function in the given interval is

• A. $2$
• B. $1$
• C. $4$
• D. none of these

Answer 1

Answer: (A)

$2$

$f\left(x\right)=6-12x+9x^2-2x^3$
$f^{\prime }\left(x\right)=-12+18x-6x^2$
For maxima or minima of $f\left(x\right)$
$f^{\prime }\left(x\right)=0=-6(x^2-3x+2)\Rightarrow x=1,2$
Now $f^{\prime \prime }\left(x\right)=18-12x$
Clearly $f^{\prime \prime }\left(1\right)>0$ and $f^{\prime \prime }\left(2\right)<0$
hence $f\left(x\right)$ will attain it's maxima at $x=2$ which is 2.