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Find the range of f(x) = (x + 2)/(x - 3) algebraically.
How we find the range of this rational function without a calculator?
A friend texted the following question to me.
Solve x^(2) = 2^(x) for x.
Any suggestions?
Visit https://imgur.com/a/MMz2Kir and tell me why there is not a defining inequality for (-00, 00).
Note: -00 = negative infinity
Note: 00 = positive infinity
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.
I am totally opposed to this. The forum is open access. Lots of our members join because they are weak at maths and want help to improve.
Bob
Copy. Understood. No test will be posted. It was just a suggestion.
harpazo1965 wrote:Thanks but you already know me, right?
I didn't know much before. Until your introduction, yes.
Let's keep our chat in terms of numbers and everything should work out. What else is a lonely middle-aged man to do? Math to me is like cross word puzzles are to many people. It's just a hobby. Revisiting Precalculus is not going to make me a math teacher or a math expert. It's just for fun and education. Come along for the ride.
There is a on-line maths course that objected because we were 'publishing' their worksheets. They threatened 'breach of copyright' action.
So you need to take care what you post. If you have a question that you think is good practice for a particular exam, post it in Exercises. Post one question at a time rather than collecting up a whole set and posting all in one go.
Bob
Copy. Understood.
You are right. There is no test requirement but I think it is a good idea.
Thanks but you already know me, right?
Would it be OK for me to post math homework questions as practice exams from courses taken back in my student days? Yes, I still have my math notebooks. Notebooks are a bit worn but readable. I took my last math class in the Spring 1993 semester at Lehman College.
You say?
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 create a math entry test for this site. You say?
I will not reveal my real name in an open forum but just wanted to say that I am happy to have found this math site. I am currently reviewing math learned long ago.
1. I am not a student. My last semester as a college student ended in December 1993. I am 57 years old. Yes, middle-aged and aging rapidly. Yes, old age is just around the corner. Life drags! What can I say? Our time on this planet is very short.
2. As you read above, I am not a classroom student and thus NO NEED to indicate in your reply to my threads that I should do my own homework. Homework? What's that?
3. Currently revisiting Precalculus on my way to Calculus. I wanted to study Calculus back in my student days. However, no need to take the course as a Sociology major in the 90s.
4. I live in NYC but I am not a fan of what the city has become in the last 30 years. I don't recognize this place anymore. I hate it here and hope to leave it all behind someday.
5. I have been playing solo guitar since the age of 9. One of my favorite hobbies when not working on math problems is playing solo guitar hymns. I read and write music notation. I also read and write guitar tablature. If there guitar players here, say hello. This is it for now.
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?
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?
Find the domain of f(x) = 5/[sqrt{x^2 + 1}].
Again, the domain is ALL REAL NUMBERS but it is not clear to me why that is the case.
According to Professor Leonard on YouTube, for this function, no matter what real numbers we let x be, the radicand cannot yield a negative number.
So, there is no domain issue here. The domain is ALL REAL NUMBERS.
However, I need a more detailed reason why the domain is ALL REAL NUMBERS. Can a graph of this function help me to see why Professor Leonard is right?
Thanks
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.
hi harpazo1965
(x - 1)^2 + (y + 2)^2 = 9 is correct. Well done!
Bob
Thank you, Bob. Can you please do me a favor? Contact member ganesh (who is overly sensitive) and tell him to please ignore my questions moving forward.
harpazo1965 wrote:Do you want me to homework at 57 years old?
I am doing a self-study of mathematics, more like a review
of material learned long ago. Understand? I am WAY PAST my student
days. I haven't done homework since my last semester of college, which ended in December 1983. Copy?That is why I had given the link. I wish to tell you, don't post your questions in toto. One or two, or a couple more at most, is the limit.
A part of the Forum rules:
No Personal Attacks or Put-Downs. This is a type of bullying, and just makes you look insecure.
This is not a place to be mean to others and these posts will not be tolerated. Light banter or constructive criticism can be allowed if it is polite and friendly. Remember, other people have feelings too. "Those who give respect shall receive it."
No Swearing or Offensive Topics. Young people use these forums, and should not be exposed to crudeness. There is a bad language filter on this website, do not try to "get around it".
No Spam. Spam includes messages that have no relevance to the topic, that are annoying, repetitious or promotional in nature. Overuse of short comments is also spam.
There is absolutely NOTHING insensitive about my reply to yours. I am educating you on who I am. You know math but don't know anything about me. You don't know my real name. You don't know my taste in music. You didn't know my age and the fact that my last college semester ended in December 1993 (not 1983) as originally posted. You don't need to feel offended.
1. I joined this forum to review math learned long ago.
2. I joined this forum because I like mathematics regardless of my age.
3. I didn't come here to argue and fight.
4. If my questions and/or replies to your comments offend you, then simply ignore my threads. Easy, right?
5. Don't put words in my mouth about what I meant or didn't mean in my replies. Why? You simply don't know ANYTHING about me. Got it? This is a lot easier than math to grasp. Just walk away from my threads and be happy in your little sensitive world.
Do you want me to do homework at 57 years old?
I am doing a self-study of mathematics, more like a review
of material learned long ago. Understand? I am WAY PAST my student
days. I haven't done homework since my last semester of college, which ended in December 1993. Copy?
I've spent 30 minutes trying to make this work; finally resorting to the Imgur help site. They say bcc codes are only available on desktop devices. So sorry
From the internet chat surrounding this difficulty there are lots of unhappy image posters out there.
You can get the https adress of your image like this:
Log in to Imgur.
Click the person icon (bottom right on my screen) and your images appear.
Select the one you want and click the right arrow.
Click copy url.
Log in to the forum and start a post.
On the text input screen press and hold to get the paste option.
This will give you something like this:
https://imgur.com/a/BURsC8w
This track car picture was needed when I was helping a guy with his car frame geometry.
Anyone can copy and paste this into a fresh browser tab to view your image.
NOTE TO OTHER MEMBERS: Can anyone recommend a good image hosting site that does support bcc codes when using a mobile.
Bob
Hi Bob.
I just posted a new question. The image for the post can be found at:
https://imgur.com/a/RC2gfJp
Thanks
Copy and paste https://imgur.com/a/RC2gfJp to see image of circle.
Here is my try:
Our radius is 3.
Our vertex is (1, -2).
I must use the general formula for circles not centered at the origin.
(x - h)^2 + (y - k)^2 = r^2
I can now plug and chug.
Let h = 1
Let k = -2
Let r = 3
(x - 1)^2 + (y -(-2))^2 = (3)^2
(x - 1)^2 + (y + 2)^2 = 9
Is this the correct equation for this circle?
Thank you.
Thank you, Bob. I will give it a try. It may or may not work out.
hi harpazo1965
Welcome to the forum.
This is the right place if you want help with your studies. There are some really strong mathematicians here; much better than me; so you should be able to get what you need.
Two such members have created their own videos on a host of topics. You'll find them here:
http://www.mathisfunforum.com/viewtopic.php?id=22490
and
http://www.mathisfunforum.com/viewtopic.php?id=27270
There is no charge to belong so the forum relies on others to provide the server space for it to run. With the old server it was possible to upload images directly but it was necessary to switch and this is no longer possible now.
I signed up to www.imgur.com (free) and what I do to get an image here is this:
Log in to imgur
Upload the image there.
Get the bcc code from that site and paste it into the post. When someone views the post their browser obtains the image from imgur and displays it in the post. There are some other sites that will also do this but I'm happy with imgur. It seems to be safe and free from undesirable stuff.
Give it a try.
Bob
How do I get the bcc code from imgur? I have an account and posted my first photo there about finding the equation of a circle. I am stuck here....
Try undoing the operations in reverse precedence order to isolate the variable.
Very good.
Hi harpazo1965,
All I say is it can be solved in fewer steps.
Ok. Thanks.