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)

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