Question

"Let f(x) =
(x^2 - 2x + 2 if x E Q
and
1/(x-1) if x ER \ Q)

Q1: What is the domain of f?
Q2: Does f have a minimum value on the interval [2,3]?
- If so, what is that value?
- If not, why not?"

Thanks

Answer 1

How much work have you done on this problem so far? I have an idea for Q1 but doing more research on Q2.

Answer 2

not a whole lot. im struggling. at x = 1, 1/(x-1) is undefined so i figured maybe all other values but x not = 1

Answer 3

But notice that at x=1, x would be a rational / x E Q, right?

Answer 4

yes, meaning?

Answer 5

So, we'd never run into that issue because f(1) equals the value when x E Q and not x E R\Q. Thus, the domain holds for all real values of x.

Answer 6

what do you mean 'equals the value'

Answer 7

f(x) is defined for rational values x as the expression x^2 - 2x + 2. f(1) or x=1 is a rational number meaning we use that piece of the function defined for rational values of x. The function piece for irrational values of x does not matter when x is rational. Sorry, I was a bit ambiguous there.

Answer 8

So basically when you would look for the values where f(x) is undefined, you are correct in noting that 1/(x - 1) does not exist for x=1, but because that piece of the function only applies for irrational x, we don't have to worry about it ever having the value at x=1.

Answer 9

wait so what youre saying is that 1/x-1
x can only be irrational?

Answer 10

and for x^2 - 2x +1 x can be either
irrational or rational

Answer 11

Let f(x) =
(x^2 - 2x + 2 if x E Q
and
1/(x-1) if x ER \ Q)

x E Q is "x is rational"
x E R \ Q is "x is irrational, i.e. the complement of the rationals within the real numbers "

Answer 12

this comes directly from the definition you gave for f(x), those "if"conditions.

Answer 13

ok i think i understand what it means… but it confuses me further when answering the domain question

Answer 14

We have
f(x) = x^2 - 2x + 2 IF x is rational.
Polynomials do not have any places where they are undefined, so the domain is all values of x.
On the other hand,
f(x) = 1/(x - 1) IF x is irrational.
This has one point where it is undefined at x=1. However, having x=1 implies that x is rational. We are operating with the assertion that this part of the function is for irrational x ONLY. So, we ignore it entirely. We never deal with 1/(x - 1) when x=1. Everywhere else, 1/(x - 1) is defined, so it works for all values of x.

Together, x can be any value rational and any value irrational. Can you understand the logic there?

Answer 15

so the domain of x is all numbers?

Answer 16

because for the only case where it is undefined, 1/(x-1), x = 1, x cant = 1 because x is irrational for 1/(x-1)

Answer 17

Yes. x is all real numbers is your domain. :)

Answer 18

yay

Answer 19

so. Q2

Answer 20

thanks

Answer 21

I've been trying to find something on Q2 but not much luck. What class is this for?
I do know that there is a function defined similarly called the Dirichlet function, which is nowhere continuous. Not sure if the same reasoning is useful here.

Answer 22

class?

Answer 23

its for first yr chem engineering

Answer 24

maths class

Answer 25

ill be back shortly need to use bathroom

Answer 26

Alright, just curious. I could look up a more relevant source that way, but that one won't help me.
Do you recall the given definition of a minimum value from any previous lectures?

Answer 27

negative…
we only kinda got stuff like intervals… so the open interval (0,1) has no largest or smallest number.."there is no largest number in (0,1)" means that there is no real number closest to, but less than one..

On the other hand, [0,1] indicated 0 would be the smallest and 1 the largest

Answer 28

I see.
Because, I would expect on one hand, there is likely a minimum value. The problem is that this function is very unusual in its definition. I graphed to see that 1/(x - 1) < x^2 - 2x + 2 over the interval [2, 3]. So the minimum would at least be irrational. I imagine it would also be close to 1/(3-1) = 1/2 because I speculate there is a good chance an irrational number is floating nearby x=3 (the lowest point on 1/(x - 1) on [2,3].

But what I say is mostly speculating now, not very familiar with this sort of situation. You might be able to get better luck bumping it up!

Answer 29

well thank you heaps for your help so far. cleared a few things up for sure.

Answer 30

You're welcome! I like to look into these challenging problems! :)

Answer 31

So, I was looking around and have still not found too much useful information in direct support. However, indeed you could show an irrational number arbitrarily close to x=3 by a sequence such as x = 3 - sqrt(2)/10^k.. Each choice of a greater value of k gets us an irrational number closer to 3, but never truly reaching 3. But indeed, if we try to substitute this in for x:
f(3 - sqrt(2)/10^k) = 1/ (3 - sqrt(2)/10^k - 1) = 1/(2 - sqrt(2)/10^k)
If we took larger and larger values for k, you get closer and closer to 1/2 but never reach 1/2. This suggests to me a similar case of having y > 1/2. You never truly reach 1/2, so no true minimum value exists.

Answer 32

Idea was from http://mathforum.org/library/drmath/view/61617.html
I think this argument seems best from my knowledge, but you can always bump the question to see if anyone else has input. Or if anything I said was not clear, I can also try to elaborate.