mean value theorem

Question

Suppose that we know that f(x) is continuous and differentiable on [7, 19]. Suppose that we also know that f(19) = 31 and that f 0 (x) ≥ 9. What is the largest possible value for f(7)?

Answer 1

If you mean f'(x)>=9 (when you write f 0 (x), I guess its a typo) then by the mean value theorem we have that there exists a c in the interval (7,19) such that f'(c)=(f(19)-f(7))/(19-7)=(31-f(7))/12>=9. Equivalently we have that 31-f(7)>=108 or f(7)<= -77.