Question

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

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

Answer 1

Answer: (C)

$[-8,72]$

Given curve is
$f\left(x\right)=4x^3-12x,f:R\to R$
and interval $[-1,3]$.
$\therefore f(-1)=4(-1{)}^3-12(-1)=-4+12=8$
$f\left(0\right)=4(0{)}^3-12(0)=0$
$f\left(1\right)=4(1{)}^3-12(1)=4-12=-8$
$f\left(2\right)=4(2{)}^3-12(2)=32-24=8$
and $f\left(3\right)=4(3{)}^3-12(3)=108-36=72$
So, the required image is $[-8,72]$.