Question

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

• A. $\left[f\right(+1),f(-1\left)\right]$
• B. $\left[f\right(-1),f(3\left)\right]$
• C. $[-8,16]$
• D. $[-8,72]$

Answer 1

The correct option is A $\left[f\right(+1),f(-1\left)\right]$

$f\left(x\right)=4x^3-12x$
$f^{{}^{\prime }}\left(x\right)=12x^2-12$
$=12\{x^2-1\}$
$f^{{}^{\prime }}\left(x\right)=0$
$12(x^2-1)=0$
$x^2-1=0$
$x=\pm 1$
$f^{{}^{\prime \prime }}\left(x\right)=24x$
$f^{{}^{\prime \prime }}\left(1\right)=24>0\Rightarrow x=1$ is point of local minima
$f^{{}^{\prime \prime }}(-1)=-24<0\Rightarrow x=-1$ is point of local maxima
$\therefore$ Image of interval is $\left[f\right(+1),f(-1\left)\right]$