Domain of Rational Function...4

[sologuitar]

College Algebra 

Section R.2

Find domain of

y = (-9x^2 - x + 1)/(x^3 + x).

See attachment. 

 

 

[studiot]

Edit ignore this I now see you have a bunch of these

Edited September 11, 2023 by studiot

[e jane aran]

4 hours ago, sologuitar said:

College Algebra 

Section R.2

Find domain of

y = (-9x^2 - x + 1)/(x^3 + x).

See attachment.

1. "x = 0"

Yes, this domain restriction is valid.

2. "Setting x^2 + 1 to be 0 leads to a domain over the function not over the real number."

I'm sorry, but I don't know what this means...?

3. "Let D = domain of function; D = {x || x not equal to 0}"

Yes, this is the correct domain.

[sologuitar]

7 hours ago, studiot said:

Edit ignore this I now see you have a bunch of these

What is wrong with having "a bunch of these"?

2 hours ago, e jane aran said:

1. "x = 0"

Yes, this domain restriction is valid.

2. "Setting x^2 + 1 to be 0 leads to a domain over the function not over the real number."

I'm sorry, but I don't know what this means...?

3. "Let D = domain of function; D = {x || x not equal to 0}"

Yes, this is the correct domain.

I simply mean that the domain for this function is a complex number or numbers when setting x^2 + 1 to zero and solving for x. Am I right? 

[e jane aran]

4 hours ago, e jane aran said:

2. "Setting x^2 + 1 to be 0 leads to a domain over the function not over the real number."

I'm sorry, but I don't know what this means...?

1 hour ago, sologuitar said:

I simply mean that the domain for this function is a complex number or numbers when setting x^2 + 1 to zero and solving for x. Am I right? 

On what basis have you concluded that the domain is "a complex number", rather than being some portion of the set of real numbers?

It should be noted, by the way, that the domain is the set of all values which *are* allowed, not the set of values which are *not* allowed.

[studiot]

12 hours ago, sologuitar said:

What is wrong with having "a bunch of these"?

Nothing at all.

I'm sorry, it was me not you.

Preparing latex formulae is difficult for me these days as I have to do it on my old computer,save it,  and then transfer it to a newer computer when I can get on one.

Whilst I was doing this for probably your first of the 'bunch' you must have added more so I only quickly posted in this one which was the wrong thread. Thus I didn't understand why the formula under discussion had 'changed'.

 

I deleted my post but had to put something in its place.

Sorry if this caused more confusion.

 

But hey, never mind, I see we have new mathematically minded member so welcome @e jane aran

FYI please note the site 5-posts-in- the- first- 24-hours limit.

This is a security measure that greatly reduce the rubbish the overworked moderators have to deal with.

After that time expires you will be unlimited.

 

Thank you for the help you offered. +1

Edited September 12, 2023 by studiot

[exchemist]

13 hours ago, sologuitar said:

What is wrong with having "a bunch of these"?

I simply mean that the domain for this function is a complex number or numbers when setting x^2 + 1 to zero and solving for x. Am I right? 

I'm not a mathematician but I think you may be confusing inputs and outputs. As I understand it the domain of a function is the set of inputs it can accept, i.e. the values that x can take. What you seem to be saying is the output, in the case of the function you have chosen, is imaginary rather than a real number. 

[studiot]

26 minutes ago, exchemist said:

I'm not a mathematician but I think you may be confusing inputs and outputs. As I understand it the domain of a function is the set of inputs it can accept, i.e. the values that x can take. What you seem to be saying is the output, in the case of the function you have chosen, is imaginary rather than a real number. 

Not really. but there is some confusion in this thread.

 

21 hours ago, sologuitar said:

Find domain of

y = (-9x^2 - x + 1)/(x^3 + x).

I see that you are looking for values of x that make the fraction undefined  ie division by zero.

If x is a real number that is correct.

But it is confusing at least and incorrect at worst to say

x = 0 works.

For the reason you have already given putting x = 0 leads to the fraction being undefined.

So x is not equal to 0.

 

We are all agreed on this.

 

You are also correct to note that there is no real number p such that (p² +1) = 0

So the correct form of reasoning goes that x cannot be equal to p, whatever p might be, since it is not a real number.

Thus the only restriction on numbers that you cannot choose for x or must exclude to be x = 0