# Help providing function examples

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1. Jan 29, 2017

### doktorwho

1. The problem statement, all variables and given/known data

This is one test question we had today and it asks as to provide examples of functions and intervals. Some may be untrue so we had to identify it. The test isnt graded yet so these are my question answers. Hopefully you'll correct me where necessary and provide a true example.
2. Relevant equations
3. The attempt at a solution

a) $f(x)=x, (-1,1)$
b) $f(x)=\frac{1}{x}, (0,1)$ i had trouble with this and i gave this example.
c) $f(x)=x, (-1,1)$ this example from a) should cover this as well, right?
d) NOT POSSIBLE since it is not defined in some point it cant be differentiable there right?
e) $f(x)=|x|, (-1,1)$
f) NOT POSSIBLE I just couldnt find an example..
Could me provide feedback on each of these? It would be of massive help, thanks :)

2. Jan 29, 2017

### Buzz Bloom

Hi doktorwho:
I would suggest another look at (b) and (f).
My issue with (b) is a nit. I am not sure of the interval notation with "(" and ")". Is "0" in the interval or not? One way to avoid any ambiguity is to use the interval (-1,1).
My issue with (f) is also whether an ambiguity exists regarding "differentiable". Is f(x) differentiable if the derivative is unbounded? If not, your answer is correct. If f(x) can be differentiable with an unbounded derivative, then give your (b) example another look.

Hope this helps.

Regards,
Buzz

3. Jan 29, 2017

### Staff: Mentor

Interval notation using parentheses and brackets is pretty standard, although not universal. (0, 1) is the open interval from 0 to 1 with endpoints not included. [0, 1) is the half-open interval with 0 included. [-1, 1] is the closed interval from -1 to 1, with both endpoints included.

I have seen other notation used, with ]0, 1] being equivalent to $\{x | 0 < x \le 1\}$.

(Personally, I find this last notation with the reversed bracket ugly, but that's just my opinion.)

4. Jan 29, 2017

### Stephen Tashi

How do your course materials define "interval" ?

You used "open intervals" in your answer. There are other kinds of intervals.

The more customary terminology is to say a function is "bounded" instead of saying it is "limited".

5. Jan 29, 2017

### Ray Vickson

About (f): isn't the example in (b) differentiable but not limited on (0,1)?

6. Jan 29, 2017

### doktorwho

I used limited as i didnt know the correct term but it is equivalent to bounded. Its translated from another language.
When i thought about the f) i guess that it is not differentiable at x=0 so that cant be the example. I dont exactly understand the bounded term. For example b) continuous but not bounded. I understand continuous but what exactly does bounded mean?
One of the answers said 1/x on (-1,1) but shouldnt the function be in that interval?
Like, when i said (0,1) i said that because the fenction never goes to 0 so its bounded by that. You see my comprehension problem, could tou bring this to me a little closer by a few simple examples and explanations?

7. Jan 29, 2017

### Stephen Tashi

Which can't be the example?

The point x = 0 is not a element of the open interval (0,1). The point x = 0 is an element of the closed interval [0,1]. I assume you are permitted the use open intervals (which you did) in all your answers. The funtion f(x) = 1/x is not bounded on (0,1). The fact f is not differentiable at x = 0 is not relevant to whether it is differentiable "in" the interval (0,1) because x= 0 is not in that interval.

If you were required to use closed intervals in your answers, then the answer to f) would be "not possible".

"The function $f$ is bounded on the set $S$" means there exists numbers $L$ and $U$ such that for each $x \in S$ , $L \le f(x) \le U$. $\ L$ is called a "lower bound" and $U$ is called an "upper bound".

There is a distinction between a "bound" and "maximum" or "minimum". A function has a "minimum" on $S$ when it has a lower bound $L$ and actually attains that value at some point in $S$. Similary, a function has a "maximum" if it has an upper bound on $S$ and actually attains that value at some some point in $S$.

For example, one possible upper bound of $f(x) = x$ on $S = (0,1)$ is 500. The "least upper bound" of $f$ on $S$ is 1. The value 1 is not the "maximum" of $f$ on $(0,1)$ because $f(x)$ does not attain the value $1$ for any $x$ in $(0,1)$. (Remember that $x = 1$ is not an element of the open interval $(0,1)$. )