# Every function on [0,1] attains its maximum

## 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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Hurkyl
Staff Emeritus
Gold Member
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].

SammyS
Staff Emeritus
Homework Helper
Gold Member
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.

Dick
Homework Helper
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!