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. $0$
• C. $1$
• D. $4$

Answer 1

The correct option is D $4$

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