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So my teacher gave me an excercise like this:

If f '(5) = 8 and f '(10) = -3 What can we conclude about minimus and maximums?

My answer was:

We cannot conclude any maximum or minimum because in order to be a minimum or maximum slope must be 0 at that current point. if f'(C) = 0 then there is a minimum or maximum at (C, f(C)) point.

And he put it wrong. So how to know the maximums/minimums with that little information?

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    $\begingroup$ You know that there is a (local) maximum in between 5 and 10. $\endgroup$
    – Anton V.
    May 16 at 5:15
  • $\begingroup$ @AntonV. How? explain please? $\endgroup$
    – bau8312
    May 16 at 5:17
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    $\begingroup$ Like Bertrand's answer, if you first go up and later go down, there is a point where f' switches from positive to negative through 0. $\endgroup$
    – Anton V.
    May 16 at 5:20
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    $\begingroup$ Without further assumptions about $f$ you can't really say anything, so the question was not very well formulated in my opinion. $\endgroup$
    – Hans Lundmark
    May 16 at 5:43
  • $\begingroup$ I'm assuming that the assumption is that the f is differential everywhere, and thus continuous everywhere so we can make the assumption that there is a local maximum? $\endgroup$
    – bob
    May 16 at 13:42

2 Answers 2

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Note that, if $f$ is differentiable on an interval, $f'$ has the intermediate value property on that interval (cf. Darboux's theorem). I imagine this is being considered in the context of, say, a Calculus $1$ course, though, so you may find it simpler to think that $f'$ is continuous and can invoke the intermediate value theorem (but that is a little stronger than necessary).

I especially imagine this is the assumption since, strictly speaking, you won't be able to get all of that information without Darboux's theorem, or without an assumption of continuity, as far as I know. The derivative could exist at those points, but not necessarily elsewhere.

Somewhat of a pedagogical issue with Calculus $1$ in trying to boil things down and simplify them, but what can you do.

Anyhow.

Notice that $f'(5) > 0$ and $f'(10) < 0$. By the intermediate value property, then, we know that $f'(\xi) = 0$ for some $\xi \in (5,10)$.

However, since this $\xi$ may not be unique, we cannot conclude much more than that (in particular, we cannot conclude what $\xi$ may manifest as).

Even so, though, we may conclude that - since $f$ is ultimately increasing and changes to decreasing - at some point it certainly hits a local maximum. (Compare this idea with the first derivative test.) Not necessarily just one, and minima may be involved, but a local maximum is ensured.

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    $\begingroup$ Just to nitpick, the assumption of $\,f'\,$ continuity is not required. It's enough to assume that $\,f'\,$ exists on $\,[5,10]\,$ then the existence of a zero follows from the intermediate value property which all derivatives have by Darboux' theorem. $\endgroup$
    – dxiv
    May 16 at 5:35
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    $\begingroup$ @dxiv: Actually it's enough to assume just that $f$ is continuous on $[5,10]$ to get at least one maximum in $(5,10)$ (from the extreme value theorem, together with the given derivatives which show that the endpoints can't be maxima). $\endgroup$
    – Hans Lundmark
    May 16 at 5:47
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The teacher is technically wrong

Assuming OP didn't leave out any info, we really can't conclude anything about the existence of a local maximum between 5 and 10 by just knowing that f is differentiable at 5 and 10. We have to assume differentiability or continuity on the interval [5,10] to make that conclusion, which isn't how the question was posed according to OP. So the teacher is wrong unless OP left information out (e.g. the teacher verbally told the students "assume all functions are differentiable"). It's obvious what the teacher meant, but that doesn't appear to be what the teacher said.

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  • $\begingroup$ I didn't leave anything out. The teacher didn't say anything verbally since it was an online test. $\endgroup$
    – bau8312
    May 16 at 18:14
  • $\begingroup$ Makes sense. I just wanted to make sure I allowed for that possibility in my answer. $\endgroup$
    – bob
    May 16 at 18:20

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