Suppose that 2 ≤ f ' (x) ≤ 3 for all values of x. what are the minimum and maximum possible values of f (6) - f (4) ?

If we let m represent the average value of f' (x) over the interval x ∈ [4, 6], then the value of f (6) will be

f (6) = f (4) + m (6 - 4)

f (6) = f (4) + 2m

And the difference f (6) - f (4) is

f (6) - f (4) = (f (4) + 2m) - f (4) = 2m

The problem statement tells us that m must be in the range 2 ≤ m ≤ 3, so 2m is in the range 4 ≤ 2m ≤ 6.

The minimum possible value of f (6) - f (4) is 4.

The maximum possible value of f (6) - f (4) is 6.