Mean Value Theorem to find a and A?

Question

Suppose that f is a differentiable function, that f(2) = 3 and that 4 < f ' (x) < 6 for all x. Use the Mean Value Theorem to find numbers a and A such that a < f(6) < A.

Could really use some help with understanding how to go about this problem. Thanks.

Answer 1

Well let's suppose the function is as shallow as it could possibly be and has a slope of 4 everywhere. Then we would have that f(6) = f(2) + 4*(6-2) = 3 + 16 = 19. Conversely, if it is everywhere as steep as it could possibly be, and has a slope of 6 everywhere, then we would have f(6) = f(2) + 6*(6-2) = 3 + 24 = 27. Now, since the slope is actually everywhere strictly less than 6 and strictly greater than 4, we would expect that 19 < f(6) < 27. And indeed, using the mean value theorem, we can prove that this is true.

We proceed by contradiction: suppose that ¬(19 < f(6) < 27). Then either f(6) ≤ 19 or f(6) ≥ 27. If f(6) ≤ 19, then by the mean value theorem, ∃c∈(2, 6): f'(c) = (f(6) - f(2))/(6-2) = (f(6) - 3)/4 ≤ (19 - 3)/4 = 4. But this contradicts the fact that ∀x f'(x)>4. Conversely, if f(6)≥27, then by the mean value theorem we have ∃c∈(2, 6): f'(c) = (f(6) - f(2))/(6-2) ≥ (27 - 3)/4 = 6, contradicting the fact that ∀x f'(x)<6. So neither f(6) ≤ 19 nor f(6) ≥ 27 can hold, which means that 19 < f(6) < 27. Q.E.D.