Every function on [0,1] attains its maximum

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SUMMARY

The statement "Every function on [0,1] attains its maximum" is false. While continuous functions on the closed interval [0,1] do attain their maximum values, the question does not specify continuity. Discontinuous functions, such as f(x) = 1/x for x ≠ 0 and f(0) = 0, do not necessarily attain a maximum on this interval. The discussion clarifies that the term "attains its maximum" implies the maximum value must exist within the function's image, which is not guaranteed for all functions defined on [0,1].

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Homework Statement


True or False: Every function on [0,1] attains its maximum

I was working on a practice midterm, and this question stumped me

The Attempt at a Solution


I know that by "every function on [0,1]" means that dom f = [0,1] but what exactly does it mean by "attains its maximum" simply that it has a limit, as in, it doesn't approach +∞?

If this is what it means, it would be false, correct? Because you could have a function that approaches ±∞ as it approaches 1 from below.

For instance, take f(x)=\frac{1}{1-x}
 
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PirateFan308 said:
but what exactly does it mean by "attains its maximum"
The question presumes the function has a maximum value, and asks if that maximum value is in the function's image.

It's a bit oddly phrased, since a maximum is, by definition, in the function's image. It's either a 'trick' question, or they meant "has a maximum" or "attains its supremum". Or, you learned a slightly non-standard definition of those terms in class.

For instance, take f(x)=\frac{1}{1-x}
That's not a function on [0,1].
 
The main point is that [0, 1] is a closed interval.

It's certainly not true that "Every function on (0, 1) attains its maximum."
 
I think the question really means every bounded function on [0,1] attains its least upper bound. But I may be wrong.
 
Funny nobody has mentioned the word 'continuous' yet. All continuous functions on [0,1] have an upper bound and attain it. The given problem didn't say 'continuous'. So it's false.
 
I would take that to mean there is a c in [0, 1] such that \forall x in [0, 1], f(x)\leq f(c).

Certainly this is true for all continuous functions (Use the fact that if K is compact, f(K) is also compact for continuous f, and [0, 1] is compact. It can be shown that compact sets have a maximum and minimum value).

For discontinuous functions though, take
\begin{align*}f(x)&=\frac{1}{x} \text{ when }x≠0\\ &=0 \text{ when } x=0\end{align*}

Then f(x) does not attain it's maximum on [0, 1]
 
Thanks guys! This helped a lot!
 

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