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Today, I took my Calculus I final. One of the questions on the final was thus (paraphrased):

Which of the following must be true in order for a function to have global extrema along the interval $[a, b]$?

  1. The function must be continuous along the interval
  2. The function must be positive along the interval
  3. The function must be increasing along the interval
  4. There is no guarantee of global extrema along the interval

(These were with letters, but I can't find a way to duplicate a list with letters easily in markdown.)

For me, numbers two and three were toss-outs. The first seemed correct until I thought of a possible counterexample: $f(x) = k$, where $k$ is some constant.

In this case, it's just a straight line across the interval. There are no global extrema, and the endpoints of the interval aren't extrema either. Which is the correct answer? The one that was on the key was the first, which I did mark by a stroke of luck (she said that 21 of the answers were A and I had 20).

Is this simply a matter of semantics? Or did the creators of this exam simply not consider the possibility of a horizontal line? Or something else?

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A horizontal line does have global extrema: the unique value is both a global maximum and a global minimum. Look at the definition of "global maximum" and verify that if $f(x) = k$, then $k$ satisfies the definition. – Arturo Magidin May 11 '11 at 21:09
Your example $f(x)=k$ is in fact continuous, and every point in the interval $(a,b)$ is a global maximum. It just isn't a strict global maximum. – Leonardo Fontoura May 11 '11 at 21:12
Your title and the question are asking two very different questions: the title asks for a "guarantee" for global extremes; both 1 and 3 guarantee the existence of global extremes. But the question says "must be true in order for..." that is the converse of a guarantee, it's a requirement. None of 1, 2, nor 3 are requirements, and number 4 does not talk about requirements, but about guarantees, so it's not applicable. So, which one is it? Were you asked which conditions guarantee global extremes (1 and 3), or were you asked which conditions are required (in which case, a rephrase of 4)? – Arturo Magidin May 11 '11 at 21:16
@Arturo: I see now -- I simply assumed (oops) that a global maximum would not be present if every y-value was the same. As for the wording of the question, it was almost identical to the above quote. But I specifically remember the words "no guarantee" in the last sentence, so I believe it's your latter interpretation. Apologies for the useless question. If you post your comment as an answer, I would be glad to accept it. – Reid May 11 '11 at 21:23
up vote 5 down vote accepted

The title of your question and the question itself are asking two very different questions.

When we ask "What condition guarantee that a function has global extremes in $[a,b]$?" we are asking for conditions that are sufficient: if the condition is met, then the function will definitely have global extremes, but it is possible for a function to have global extremes without meeting the conditions. This can also be phrased as saying "which conditions ensure that a function has global extremes?".

However, when we ask "Which condition must be true in order for a function to have global extremes?" we are asking for conditions that are necessary; these are conditions that every function that has global extremes must satisfy, though it is possible for a function to satisfy the conditions and yet not have global extremes.

So the title of your question asks one thing. The quoted text asks something different.

To further complicate things, the final option in your quoted text does not match the question quoted, because the final option again talks about "guarantees" rather than "requirements".

If the question was meant to ask about guarantees (that is, conditions that, if met, will ensure that the function has global extremes), then the correct answers are (1) and (3); (1) guarantees the existence of global extremes by the Extreme Value Theorem. Condition (3) guarantees the existence of global extremes because for all $x$ in $[a,b]$, we have $a\leq x\leq b$, so if $f$ is increasing on the interval this implies $f(a)\leq f(x)\leq f(b)$; so $f(a)$ is a global minimum and $f(b)$ is a global maximum. Condition (2) does not ensure the existence of global extremes. For example, the following function defined on $[0,1]$ is positive at all points, but has neither a maximum nor a minimum: $$f(x) = \left\{\begin{array}{ll} 10 &\mbox{if $x=0$;}\\ \frac{1}{x} &\mbox{if $0\lt x\lt\frac{1}{2}$;}\\ 10 &\mbox{if $x=\frac{1}{2}$;}\\ \frac{1}{1-x} &\mbox{if $\frac{1}{2}\lt x\lt 1$;}\\ 10&\mbox{if $x=1$.} \end{array}\right.$$ And (4) is incorrect because conditions (1) and (3) are certainly guarantees for the existence of global extremes.

On the other hand, if the question is quoted correctly and is asking about conditions that are necessary (requirements) for the existence of a global maximum, and we rephrase (4) to read:

4 . None of the above conditions are required for the existence of global extremes.

then that would be the correct answer. Continuity and positivity (conditions 1 and 2) are not required; for example, the function $f(x)$ defined on $[0,2]$ $$f(x) = \left\{\begin{array}{ll} 2x-1&\mbox{if $0\leq x\leq 1$;}\\ 2x-3&\mbox{if $1\lt x\leq 2$,} \end{array}\right.$$ has global minimum $-1$ at $0$, and has global maximum $1$ at both $x=1$ and $x=2$. The function is neither continuous nor always positive. The function is also not increasing over the interval, so this example also shows that (3) is not a requirement.

Finally, note that a constant function does have global extremes. $M$ is the global maximum for $f(x)$ on the set $A$ if and only if:

  • There is a point $p\in A$ such that $f(p)=M$; and
  • For all points $q\in A$, $f(q)\leq M$.

If $f(x)=k$ for all $x$, then $k$ is a global maximum; similarly for "global minimum."

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If the phrasing was :

Which of the following must occur for a function to have a global extrema in the interval [a,b]

Then we can construct simple counter-examples to the first 3. For example, suppose that the function was identically 0 at every point except at the point $\frac{b+a}{2}$, where it was 1. Clearly, the function has a global maximum there, and the function is not continuous. Thus continuity is not required.

For number 2, consider the parabola $y = -(x - \frac{b+a}{2})^2$ - 1. Everywhere, this function is negative, but it also has a global maximum at the midpoint of a and b.

For number 3, either of the previous two functions also serves as a counter-example.

So if that was the phrasing of the question, then number 4 is correct. But I think that the question might have been phrased differently - it might have asked which of the 4 guarantees that there is a global maximum. In this case, continuity guarantees both a maximum and a minimum, as seen here.

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This is very hard to answer without the exact phrasing of the question.

The phrase "must be true in order" refers to a necessary condition.

The phrase "no guarantee" refers to the absence of a sufficient condition.

Speculation: This may be about a function that is defined on the closed interval, so the function does indeed have a global extremum if it is continuous and your line is cut off at the end of the interval. Also note that usually each point of a horizontal line is considered a global maximum and minimum.

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A horizontal line is continuous. But continuity is not necessary for a global maximum of the function.

I would have chosen 4, as it seems quite possible that the global maximum of the function occurs outside the interval.

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The question asks for the extrema of the function on the closed interval $[a,b]$. Who knows whether the function is even defined outside of $[a,b]$? – Pete L. Clark May 12 '11 at 8:41
@Pete L. Clark: There are various problems with the phrasing of the question, but as I read it, "to have global extrema along the interval?" is asking whether the global extrema occur inside or outside the interval, or perhaps nowhere. – Henry May 12 '11 at 16:17
well, I didn't read it that way, but it's certainly not worth arguing about. – Pete L. Clark May 12 '11 at 18:49

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