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.

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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.

Mathematics
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mean value theorem
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.
I got that x=24?

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

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