Use the Mean Value Theorem to estimate f(6)-f(3)

Question

suppose f(x) is continuous on [3,6] and -5≤f'(x)≤4 for all x in (3,6)

_____≤f(6)-f(3)≤_____

Answer 1

A sketch might look like this:

The Mean Value Theorem says that there is at least one place along f(x) between x = 3 and x = 6 where the slope of the tangent line equals the slope of the secant line (given that f(x) is continuous and differentiable).

The slope of the secant line is [f(6) - f(3)]/(6 - 3) = [f(6) - f(3)]/3

So, by the MVT, somewhere on (3, 6), f '(x) must equal [f(6) - f(3)]/3 but f '(x) is between -5 and 4 so:

-5 ≤ [f(6) - f(3)]/3 and [f(6) - f(3)]/3 ≤ 4 which means -15 ≤ f(6) - f(3) ≤ 12