A differentiable function f on the interval -2

Question

A differentiable function f on the interval -2<x<7 (equal signs included so it is a closed interval) is such that f ' (c)=0 where -2 <c<7.  If f(c)<f(7) then which of the statements must be true?

precal · Answer

A.  f(7) is the absolute maximum value of f on the interval [-2,7]
B. f(c) is the absolute minimum value of f on the interval [-2,7]
C. Either f(-2) or f(2) is the absolute minimum on the interval [-2,7]
D. Either f(c) or f(7) is the absolute minimum on the interval [-2,7]
E. The value of f ' (x)<0 on the interval [c,7]

precal · Answer

I know the solution is C but I want to know why

anonymous · Answer

So $f$ is differentiable on $[-2,7]$ (this should be an open interval; a function can't be differentiable at an endpoint - in any case, we know the function is continuous) and $f'(c)=0$ for some $-2<c<7$.

Do you recall the mean value theorem?

precal · Answer

yes where f ' (c) = f(7)-f(-2) over 9

precal · Answer

in this case

anonymous · Answer

Ooh, actually Rolle's theorem is the one I meant to say

anonymous · Answer

Rolle's is just a specific case of the MVT, at any rate.

We know that for this $c$, we have
$$\frac{f(7)-f(-2)}{9}=0~~\iff~~f(7)=f(-2)$$

precal · Answer

thanks I can't seem to get those thm straight

anonymous · Answer

No problem. I don't see how C could be right though. How could we say anything definitive about $f(2)$?

precal · Answer

who knows?  Could be a typo on the part of the author

anonymous · Answer

Hmm, I would have gone with B.
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