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. none of these

Answer 1

The correct option is B $[-8,72]$

To find the image of the given interval, we
must find the set of values of $f\left(x\right)$ for $x\in [-1,3]$.
By virtue of the continuity of $f\left(x\right)$, the image is the interval $\left[{min}_{x\in [-1,3]}f\right(x),{max}_{x\in [-1,3]}f(x\left)\right]$
The critical points of $f\left(x\right)$ are given by $f^{\prime }\left(x\right)=12x^2-12=12(x^2-1)=0$.
That is, $x=\pm 1$, so that $f\left(1\right)=4\cdot 1-12=-8$, $f(-1)=-4+12=8$ and $f\left(3\right)=4\cdot 27-12\cdot 3=108-36=72$.
$\therefore {max}_{x\in [-1,3]}f\left(x\right)=f\left(3\right)=72$
and ${min}_{x\in [-1,3]}f\left(x\right)=f\left(1\right)=-8$
Hence the image of $[-1,3]$ under the mapping $f\left(x\right)$ is $[-8,72]$.
Ans: B