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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.
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.

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jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1As Zarkon pointed out, you use the mean value theorem http://www.sosmath.com/calculus/diff/der11/der11.html \[\Large f'(c) = \frac{f(b)f(a)}{ba}\] \[\Large f'(c) = \frac{f(4)f(1)}{41}\] \[\Large 3 = \frac{x15}{3}\] 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.

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.1I would write it like this \[\Large \frac{f(4)f(1)}{41}=f'(c)\ge 3\] \[\Large \frac{f(4)15}{3}\ge 3\] \[\Large f(4)15\ge 9\] \[\Large f(4)\ge 24\]
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