## anonymous one year ago If f(1) = 15 and f '(x) ≥ 3 for 1 ≤ x ≤ 4, how small can f(4) possibly be? Please help! I don't understand how to solve it.

1. Zarkon

mean value theorem

2. jim_thompson5910

As Zarkon pointed out, you use the mean value theorem http://www.sosmath.com/calculus/diff/der11/der11.html $\Large f'(c) = \frac{f(b)-f(a)}{b-a}$ $\Large f'(c) = \frac{f(4)-f(1)}{4-1}$ $\Large 3 = \frac{x-15}{3}$ I replaced f'(c) with 3 since I want f'(x) to be as small as possible. Solve the equation for x to get your answer.

3. anonymous

I got that x=24?

4. anonymous

Thanks!!

5. Zarkon

I would write it like this $\Large \frac{f(4)-f(1)}{4-1}=f'(c)\ge 3$ $\Large \frac{f(4)-15}{3}\ge 3$ $\Large f(4)-15\ge 9$ $\Large f(4)\ge 24$