Question

The function $y=2x^3-3x^2-12x+8$ has a local minimum at $x=?$

• A. $-1$
• B. $2$
• C. $-\frac{1}{2}$
• D. $\frac{3}{2}$

Answer 1

The correct option is B $2$

$y={2x}^3-{3x}^2-12x+8$

Extreme values of $y$ occur at $y^1=0$

$y^{{}^{\prime }}={6x}^2-6x-12=0$

$x^2-x-2=0$

$x^2-2x+x-2=0$

$x(x-1)+1(x-2)=0$

$x=-1,2$

$y^{{}^{\prime \prime }}=12x-6$

$y^{{}^{\prime \prime }}$ at $x=-1$ is $y^{{}^{\prime \prime }}=12(-1)-6=-18<0$

$y^{{}^{\prime \prime }}$ at $x=2$ is $y^{{}^{\prime \prime }}=12\left(2\right)-6=18>0$

$\because$ $y^1$ at $x=2$ is $0$ and $y^{11}>0$.