The Mean Value Theorem assures us that if
#f# is continuous on closed interval #[a, b]#,
and if #f# is also differentiable on the open interval #(a,b)#,
then there is a #c# in #(a,b)# for which:
#f'(c) = (f(b)-f(a))/(b-a)#.
Applied to this function:
We are told that #f# in this question is continuous on #[2,6]#.
The condition #-4<=f'(x)<=4# for all #x# in #(2,6)# tells us that #f'# exists all #x# in #(2,6)# and so #f# is differentiable on the interval #(2,6)# .
Therefore, the Mean Value Theorem assures us that there is a #c# in #(2,6)# for which
#f'(c) = (f(6)-f(2))/(6-2) = (f(6)-f(2))/4)# .
So we can be certain that
#f(6)-f(2) = 4f'(c)# for some #c# in #(2,6)# .
Given the restrictions of #f'(c)# on the interval, we reason:
#-4<=f'(x)<=4# implies that #-16<= 4f'(c) <=16# .
So we conclude: #-16<= f(6)-f(2) <=16# .