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.