Calculus Question

Question

Suppose that 2 ≤ f '(x) ≤ 3 for all values of x. What are the minimum and maximum possible values of f(8) − f(3)?

 ≤ f(8) − f(3) ≤

Answer 1

Apply the Mean Value Theorem. f(8)-f(3)= f'(c)(8-3)=5f'(c) for some number c.

We are given that 2<=f'(x)<=3. So multiplying both sizes of this inequality by 5, we have 10<= f(8)-f(3)<= 15.