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. $[8,72]$
• D. $[1,3]$

Answer 1

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

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