The image of interval [–1, 3] under the mapping f : R → R given by f…

$f (x) =4x^3- 12x$

$f'(x)=12x^2-12≥0$ for $x∈[-1,3]$

so $f_{min}=f(-1)=8,\,f_{max}=f(3)=72$

so for $x∈[-1,3]⇒f(x)∈[-8,72]$