Question

The largest value of $2x^3-3x^2-12x+5$ for $-2\le x\le 4$ occurs at $x$ is equal to :

• A. -4
• B. 4
• C. 1
• D. 0

Answer 1

Answer: (B)

$f(x)=2x^3-3x^2-12x+5$ and $-2\le x\le 4$
To find maxima, differentiate $f(x)$ & put it equal to $o$
$f(x)=6x^2-6x-12=0$
$\Rightarrow x^2-x-2=0$
$\Rightarrow x^2-2x+x-2=0$
$\Rightarrow x(x-2)+1(x-2)=0$
$(x-2)(x+1)=0$
$x=2,-1$
$f^{\prime \prime }(x)=12x-6$
$f(2)=18>0$
$\therefore$ At $x=2,$ value of $f(x)$ is minimum
$f(-1)=-18<0$
$\therefore x=-1$ can be point of maxima
$\therefore$ We check value of $f(x)$ at $x=-2,2,-1,4$
$f(-2)=-16-12+24+5=1$
$f(2)=16-12-24+5=-15$
$f(-1)=-2-3+12+5=12$
$f(4)=128-48-48+5=37$
$\therefore$ At $x=4,f(x)$ is maximum.