Question

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.

Answer 1

mean value theorem

Answer 2

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.

Answer 3

I got that x=24?

Answer 4

Thanks!!

Answer 5

I would write it like this

\[\Large \frac{f(4)-f(1)}{4-1}=f'(c)\ge 3\]

\[\Large \frac{f(4)-15}{3}\ge 3\]

\[\Large f(4)-15\ge 9\]

\[\Large f(4)\ge 24\]