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.