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If $ f(1) = 10 $ and $ f'(x) \geqslant 2 $ for $ 1 \leqslant x \leqslant 4 $, how small can $ f(4) $ possibly be?

By the Mean Value Theorem, $f(4)-f(1)=f^{\prime}(c)(4-1)$ for some $c \in(1,4) .$ But for every $c \in(1,4)$ we have

$f^{\prime}(c) \geq 2 .$ Putting $f^{\prime}(c) \geq 2$ into the above equation and substituting $f(1)=10,$ we get

$f(4)=f(1)+f^{\prime}(c)(4-1)=10+3 f^{\prime}(c) \geq 10+3 \cdot 2=16 .$ So the smallest possible value of $f(4)$ is 16

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Harvey Mudd College

Baylor University

University of Nottingham

Boston College

if half of one is equal to ten and a half, Prime of axe is greater than equal to for one last, any quarterbacks and last thing he could afford. How small can effort for possibly do so We're essentially looking for the small value of core And were we know the interval Because this tells us that this is in the interval one and the closing number from one to four because it includes the equal sign. So we can actually use the means at its serums to solve this. So we know that f crime. See, it is equal to effort for minus half of one all over four minus one. We're looking the smallest possible by your effort for and we know the value of life of one. So we got your song for after four. And so this will be a time see equals after four institution outbreak manipulation. Since this part a mystery, we could just bring the three on this side. And then what we will do is for is equal to three times a prime C worth one. And since since F primary backs is greater than equal to, we know that if when after prime is included, too. The smallest possible value of effort for can be soft for, because this is the maximum value this will tell us to smallish. Well, it's possible value for effort for so all we have to really do is just step itude and two for a crime of C. It was just a very times two and then we know what f of one is, which is ten three, ten to six foot ten is sixteen, so the smallest value possible for effort for is sixteen.