Question

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)?

Answer 1

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` .