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

To find the image of the given interval, we must find the set 
of values of $f\left(x\right)forx\in [-1,3]$ .  By virtue of the continuity 
of f(x) , the image is the interval $\left[\underset{x\in [-1,3]}{\min }f\left(x\right),\underset{x\in [-1,3]}{\max }f\left(x\right)\right]$

$\begin{array}{l}\text{ The critical points of }\mathrm{f}\left(\mathrm{x}\right)\text{ are given by }\\ f^{\mathrm{\prime }}\left(x\right)=12x^2-12=12\left(x^2-1\right)=0\text{. That is, }\\ x=\pm 1,\text{ sothat }f\left(1\right)=4\cdot 1-12=-8,f(-1)=-4+12=8\\ \text{ and }f\left(3\right)=4\cdot 27-12\cdot 3=108-36=72\\ \therefore \underset{x\in [-1,3]}{max} f\left(x\right)=f\left(3\right)=72\end{array}$

$\begin{array}{l}\text{ And }\underset{x\in [-1,3]}{min} f\left(x\right)=f\left(1\right)=-8\\ \text{ Hence the image of }[-1,3]\text{ under the mapping }\\ f\left(x\right)\text{ is }[-8,72]\end{array}$