The image of the interval [−1,3] under the maping f(x)=4x3−12x is

To find the image of the given interval, we must find the set of values of f(x) for  x ∈  [−1,  3] .  By virtue of the continuity of f(x) , the image is the interval minx∈[−1, 3]  f(x) ,   maxx∈[−1, 3]  f(x) The critical points of f(x) are given by f′(x)=12x2−12=12x2−1=0. That is, x=±1, sothat f(1)=4⋅1−12=−8,f(−1)=−4+12=8 and f(3)=4⋅27−12⋅3=108−36=72∴ maxx∈[−1,3] f(x)=f(3)=72 And minx∈[−1,3] f(x)=f(1)=−8 Hence the image of [−1,3] under the mapping f(x) is [−8,72]