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Given a function $f$ continuous on $[-4, 1]$ with its maximum at $(-3, 5)$ and its minimum at $(1/2, -6)$, is it not correct to say that both statements (B) and (D) must be false?

(A) The graph of $f$ crosses both axes.

(B) $f$ is always decreasing on $[-4, 1]$.

(C) $f(-2)=0$,

(D) $f(-1)=6$,

(E) $f(0)=2$.

If the maximum (on this interval at least) is $(-3, 5)$, then $(-1, 6)$ cannot be a point on the graph (D). That the function is "always decreasing" (B) on the interval seems to contradict both the stated maximum and stated minimum. The other options all seem either certainly true or possible. Am I missing something?

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    $\begingroup$ You are right about (D) if "maximum" means "global maximum" but wrong if "maximum" means "local maximum". $\endgroup$
    – WillO
    Jul 16, 2015 at 2:38
  • $\begingroup$ The original phrasing of the problem is: "A function f is continuous on [-4, 1] and has its maximum at (-3, 5) and its minimum at (1/2, -6). . ." I'll edit the question to reflect that -- seems to me the intention is global maximum, at least on the interval (though I also thought there was supposed to be only one answer) $\endgroup$
    – Rasputin
    Jul 16, 2015 at 2:41
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    $\begingroup$ @WillO, for the following reason, I think the only reasonable convention is that maximum (unqualified) means global maximum. Here's why. If I have a function $f : X \rightarrow P$ where $X$ is a (mere) set and $P$ is a poset, then it may or may not be the case that $f$ has a maximum. But the notion of a "local maximum" doesn't even make sense in this context. Hence, "global maxima" are the more fundamental notion, and I therefore think that maximum (unqualified) should mean "global maximum." $\endgroup$ Jul 16, 2015 at 3:23
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    $\begingroup$ @goblin: Your conclusion might or might not be correct, but your argument is not at all convincing. In this case we are given that $f$ is a continuous function on a closed interval. That's a specific situation. It's not uncommon for well-established conventions to apply to some specific situations and not others. $\endgroup$
    – WillO
    Jul 16, 2015 at 4:48
  • $\begingroup$ @WillO, at the risk of comparing something very small to the something very large: in much the same way as I hold that the human rights abuses under Saddam Hussein were unacceptable irrespective of whether or not they were uncommon, and irrespective of whether the regime held that that they were acceptable or not, similarly I hold that its unacceptable to change the meaning of terms in this way, irrespective of whether or not it is uncommon and irrespective of whether others who fancy themselves to be authorities on this issue deem it unacceptable or not… $\endgroup$ Jul 16, 2015 at 5:18

3 Answers 3

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If you assume that decreasing means $x<y \Rightarrow f(x)> f(y)$ then $(B)$ "must" be false

If you assume that decreasing means $x\leq y \Rightarrow f(x)\geq f(y)$ then $(B)$ might be true

If you assume that $f$ has a global maximum in the point $(-3,5)$ then $(D)$ "must" be false

If you assume that $f$ has a local maximum in the point $(-3,5)$ then $(D)$ might be true

Note that only (A) "must" be true.

EDIT: Goblin is right to say that the question leaves margin to doubt. Since there is ambiguity, it's up to you to decide which ways to go.

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  • $\begingroup$ Yeah the question seems to definitely not be "well-posed" in the words of goblin, because when I read it I got the sense that your first and third conditions (going down) are the ones they intended. Unless of course they meant for there to be two answers (which seems strange for a multiple choice question) . . . $\endgroup$
    – Rasputin
    Jul 16, 2015 at 15:05
  • $\begingroup$ One should be aware that any human language is bound to have ambiguity, so in math, one often finds tacit consensus and imprecisions. I would also think that the first and third conditions (going down) are the ones they intended but one cannot be sure so it is well done to note the imprecisions. $\endgroup$ Jul 16, 2015 at 15:10
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In my opinion, the conditions on $f$ aren't well-posed. I'll put the offending words in bold:

Given a function $f$ continuous on $[-4, 1]$ with its maximum at (-3, 5) and its minimum at (1/2, -6)...

I feel that the phrasing mistakenly assumes that $f$ has a unique maximum and a unique minimum, which (obviously) doesn't follow from the assumption that $f$ is continuous on $[-4,1]$. The question should read:

Given a function $f$ continuous on $[-4, 1]$ with a maximum at (-3, 5) and a minimum at (1/2, -6)...

When phrased in this way, it becomes more clear that $f(x)$ may equal $5$ for all $x \in [-4,-3]$ and may equal $-6$ for all $x \in [1/2,1]$.

There's a further ambiguity: does decreasing mean $x < x' \rightarrow f(x) < f(x'),$ or does it mean the weaker $x \leq x' \rightarrow f(x) \leq f(x')$? If it means the weaker of the two conditions, then (B) needn't be false. So (D) is definitely "more false" than (B). But still, the question simply isn't clear.

I have another issue with this question (not your question, but rather, the question you've been asked.) Namely that it indulges in a common yet unfortunate convention. In particular, notice that the question never tells the reader that the codomain of $f$ is meant to be $\mathbb{R}$. And sure, you can infer it from the context; nonetheless I think its sloppy not to tell the reader these things explicitly.

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  • $\begingroup$ your answer would be much better if you wrote on a different tone, I found it aggressive, maybe I am wrong, but that is the way I felt it. I will be glad to up vote it if you would rewrite it in a kinder style. $\endgroup$ Jul 16, 2015 at 3:56
  • $\begingroup$ @ConradoCosta, sorry, but if I don't criticize this kind of thing rather aggressively, then quite frankly I'm not doing my job. The OP needs to know that its not their fault that the question seems so difficult, and the writer of the question needs to know that the community does not approve of their sloppiness. $\endgroup$ Jul 16, 2015 at 4:02
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    $\begingroup$ The use of the word "its" rather than "a" implies the maximum is unique. $\endgroup$ Jul 16, 2015 at 5:41
  • $\begingroup$ @columbus8myhwm, I'm going to have to disagree with you on this one. In my opinion, it is simply wrong. The mathematical basis for my opinion on this issue is that letting $\imath$ denote the definite-descriptor quantifier, if $\varphi(x)$ is true for many different $x : X$, then I prefer to regard $(\imath x:X) \varphi(x)$ as being $\bot_X$, the "improper point" of $X$. Of course, these "improper points" or "bottom elements" aren't mentioned very often in present-day mathematics, but I think this is to its detriment. $\endgroup$ Jul 16, 2015 at 5:56
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There is no reason the function needs to be monotonic between (-3,5) and (1/2,-6). Many authors,e.g. Bourbaki,use decreasing to mean non-increasing, rather than strictly decreasing. It cannot be strictly decreasing if its max and min occur strictly between the end points of its domain. The values in (C) and (E) are consistent but cannot be inferred, which means they are possible.

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