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**harpazo1965****Member**- Registered: 2022-09-19
- Posts: 68

Find domain of h(x) = 5/(x^2 + 1).

Let me see.

Setting the denominator to 0, I get this:

x^2 + 1 = 0

x^2 = -1

When taking the square root on both sides of the equation, on the right side I get

sqrt{-1}, which is a problem here in terms of real numbers.

I further know that x^2 + 1 CANNOT be zero over the real numbers.

According to Professor Leonard, for this function there is no real number solution that will yield a zero in the denominator. However, Leonard also said there is no domain issue here and thus the domain is ALL REAL NUMBERS.

I DON'T GET IT.

Can someone explain why the domain for h(x) is ALL REAL NUMBERS?

Thank you.

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,584

hi

In questions about domain the assumption is that it will be all real numbers unless there are values of x that do not lead to values of y.

infinity is not a value that y can ever attain.

This is why you set the denominator to zero to see if x vales might lead to the graph 'zooming' up towartds infinity. You didn't find any so no reason not to have the whole set of real numbers as the domain.

The 'rule' is all reals unless there is a good reason to exclude some x values.

There are other reasons. For example the square root function has to be limited to x ≥ 0 as you cannot find a real number that is the square root of a negative.

If you wanted to devise a function to calculate the cost of theatre tickets where the input is 'n' the number of people wanting tickets you'd have to exclude everything except whole numbers . You might allow zero and you might have a top limit to n based on the size of the venue.

Try y = 5/x^2-1) for comparison.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**harpazo1965****Member**- Registered: 2022-09-19
- Posts: 68

Hi Bob. Good morning. Thank you for your reply.

A few things to say.

1. I am really 57 years old. I have no reason to lie about my age.

2. I graduated from Lehman College in June 1994. My final semester at Lehman College ended in December 1993. I was in 1965.

3. I took Precalculus in the Spring 1993 semester at Lehman College as an elective course. I got an A minus in the course. The course met twice a week for 90 minutes each class in the evening. It was a fun class.

4. In terms of ganesh, I have nothing against him or any other member. I welcome all respectful comments and suggestions.

5. Professor Leonard (You Tube) stated that the domain is ALL REAL NUMBERS for the question I posted here.

You requested for me to try an additional problem.

Find domain of y = 5/(x^2 - 1).

Set denominator to 0 and solve for x.

x^2 - 1 = 0

(x - 1) (x + 1) = 0

I get x = -1 and x = 1.

Let D = domain of function

In Set-builder Notation:

D = {x | x cannot = -1 & 1}.

Interval Notation:

(-infinity, -1) U (1, infinity)

You say?

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,584

hi harpazo1965

For my question you have correctly identified two values that are outside the domain.

The vertical lines x =-1 and x = + 1 are asymptotes. If you're not sure what that means have a look at Jade's post about a domain and range question here:

http://www.mathisfunforum.com/viewtopic.php?id=27953

My initial answer was limited because I was only working from a phone. Later I was able to add a graph that, I think, makes it clearer. I would certainly say that the graph is the key to domain and range questions.

Looking at your final answer you don't say what happens between the asymptotes ... ie between -1 and + 1. Do you think there are no x values there? Easy to show that's wrong because x = 0 gives y = -5.

I'll leave you to ponder that.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**harpazo1965****Member**- Registered: 2022-09-19
- Posts: 68

Hi Bob.

The answer should be:

D = {x| x is ALL REAL NUMBERS except for -1 and 1. }

P. S. To avoid fakers and scammers on this site, I think an entry math test should be a requirement to join our community of math enthusiasts, teachers, tutors and students. The test should be multiple-choice, 10 questions with a passing score of 70 percent. For those who fail, the test can be taken at least 3 times per year. The test should also be fair but not too easy. Math questions on the test should range from grade 5 to 8. With the permission of the site creator, I would like the opportunity to put together a math entry test for this site. You say?

*Last edited by harpazo1965 (2022-09-26 14:24:43)*

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,584

That answer is good.

I have addressed your suggestion in your other post.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**harpazo1965****Member**- Registered: 2022-09-19
- Posts: 68

Bob wrote:

That answer is good.

I have addressed your suggestion in your other post.

Bob

Moving on. I will now seek another question to post. This is all for fun. Solving math problems keeps my brain cells alive and well. No pun intended.

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