The image of the interval [-1, 3] under the mapping f:R to R given by f(x)=4x³-12x is ? (a)[8, 72] (b)[-8, 72] (c)[0,8] (d)none
{ANSWER IS [-8, 72]}

Question

The image of the interval [-1, 3] under the mapping f:R to R given
by f(x)=4x​3-12x is ? (a)[8, 72] (b)[-8, 72]
(c)[0,8] (d)none
{ANSWER IS [-8, 72]}

Vishvesh Kumar · Accepted Answer

f(x) =4x3-12x I=[-1,3] I is an interval and f is continuous (because it is polynomial function)So the image of the interval is also a interval 
and also note that f is a increasing function in [1,3](beacuse f'(x)=12x2-12 ≥0 for all x∈[1,3]) So f([-1,3])=[f(1) , f(3)]  { since f(-1) =8 and f(1)=-8 which is less than f(-1) 
 that is Why we f(1) as starting point of interval}But f(1)=4×(1)3-12(1) =-8and  f(3)=4×33-12×3=72So Image of I=[-1,3] under f is [-8,72]

The image of the interval [-1, 3] under the mapping f:R to R given by f(x)=4x​3-12x is ? (a)[8, 72] (b)[-8, 72] (c)[0,8] (d)none {ANSWER IS [-8, 72]}

The image of the interval [-1, 3] under the mapping f:R to R given by f(x)=4x³-12x is ? (a)[8, 72] (b)[-8, 72] (c)[0,8] (d)none
{ANSWER IS [-8, 72]}