Question

The image of the interval [-1, 3] under the mapping $f\left(x\right)=4x^3-12x$ is

• A. [-2, 0]
• B. [-8, 72]
• C. [-8, 0]
• D. [-8, -2]

Answer 1

The correct option is B [-8, 72]

$f^{\prime }\left(x\right)=12(x^2-1)$

$f^{\prime }\left(x\right)=0$ at $x=\pm 1$

$f^{\prime }\left(x\right)<0$ If $\mid x\mid <1$

$f^{\prime }\left(x\right)>0$ If $\mid x\mid >1$

$f\left(x\right)$ is min when $x=1$

$f\left(1\right)=-8$

$f\left(x\right)$ is max either at $x=-1$ or $x=3$

$f(-1)=8$ , $f\left(3\right)=72$

So image [-8, 72] in [-1, 3]