# How do you use the Mean Value Theorem to estimate f(6)−f(2) if suppose f(x) is continuous on [2,6] and −4≤f′(x)≤4 for all x in (2,6)?

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## .