Suppose f (x) is continuous on [2,8] and - 4≤f′ (x) ≤5 for all x in (2,8). Use the Mean Value Theorem to estimate f (8) - f (2).

2≤x≤8 - 4≤ f′ (x) ≤5 f is continuous in differentable on [2,8] so we know there is c btw 2 and 8 such that f′ (c) = f (8) - f (2) / 8-2 And we know - 4≤ f′ (c) ≤5 = > we have - 4≤ f (8) - f (2) / 8-2 ≤ 5