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Josue's Achievements


Lepton (1/13)



  1. I finally solved it: f(x) = (5 (x+1) (x-1))/((x+1) (x-4)) However, I'm not happy about it. I had to experiment a bit with a graphing calculator to see what happens to 1/x after applying several transformations to it, until I ended having ratios of quadratic functions that shared one factor with each other (point discontinuities; imatfaal's link to discontinuities helped me learn more about them). Among many other things, I need to learn, as soon as possible, what happens to rational functions when applying several transformation to them. No book that I have studied has touched on that subject, only on the graph transformations of quadratic, cubic, exponential, logarithmic, root, and trigonometric functions. (If anybody could suggest a page or a book that could teach me more about transformations of rational functions, I would be very grateful). I had to also use the answers to the problem to be able to figure out the mathematical mechanisms behind this beautiful problem. Nevertheless, I enjoyed it very much, and even more because I interacted with all of you. This is the very first time that I look for help in the internet, in a forum. The experience was very good, and I hope that I can keep interacting with all of you. I also hope that I can help other people who may not know something that I may. Thank you
  2. Oh wow! Thanks, imatfaal!! The page about discontinuities is helping me very much! New to my math vocabulary: POINT DISCONTINUITIES and JUMP DISCONTINUITIES. Yay!
  3. thank you daniton and imatfaal! What I learned new from you is that the hole means that the numerator and the denominator share one factor: (x+1). Until today I thought that (x+1) would be only in the denominator. Now I'm trying to understand why it is so, that when ther is a hole, we get a common factor at the numerator and denominator. Now I have new questions, how do I get a/b from the limit at the removable discontinuity. Where did a/b came from? Is there a general formula that I'm not aware of, and that it is needed to be able to solve this problem? It is a little bit frustrating that this type problem hasn't been discussed in the book at all. James Stewart also doesn't teach anything about removable discontinuities. He jus shows two graphs and their formulas as an illustration of removable discontinuities and that's it! Thank you very much for the help!
  4. Thank you so much, Studiot. I think your comments are going to be extremely helpful! I had tried, or looked at, already one or two of the things you suggested, but I need to try a bit more. I haven't gone back to it yet. But I will tomorrow night. I decided to move on to the next section, DERIVATIVES AND RATES OF CHANGE. I'll update about my progress with the problem. I feel that I'm very close. As soon I solve it I'll post my solution here. Just in case anybody would be interested in how I solved it. Thanks again! Thank you, Mooeypoo! I'm very happy to be part of this forum!
  5. Thank you, Studiot! Yes, it says that at x=-1, the function f has a removable discontinuity, that is, it is not defined at that spot. I'm used to seeing rational functions where f(x) stretches to infinity, -or+, as x>a (the vertical asymptote). But in this case, the lim f(x) as x>-1 is 2... So there is a hole in the function instead of a break. It is the 4th edition, published in 2010. Section 2.5 LIMITS INVOLVING INFINITY. Page 124
  6. Hello everybody! I need help, and would appreciate any help anybody could help me to solve a problem that I'm having difficulty with. To tell you about myself: I'm a 40+ regular Joe (not a student) who have picked up again his love for Math, and decided to teach himself Calculus and, if possible, higher Math and Physics. I'm using several Calculus books, among them Spivak, Apostol and James Stewart. The math problem that I can't figure out is from James Stewart's SINGLE VARIABLE CALCULUS:CONCEPTS AND CONTEXTS. The problem is the following: "A function f is a ratio of quadratic functions and has a vertical asymptote x=4 and just one x-intercept, x=1. It is known that f has a removable discontinuity at x=-1 and lim x> -1 f(x)=2. Evaluate (a) f(0) (b) lim x>infinity f(x)" So far, this is what I have thought the formula for function f should at least have: [(+or- x +or- n)(1-x)]/[(x+1)(x-4)] What confuses me is the condition lim x> -1 f(x)=2. It would have been more easy if instead of that condition, it would have had the conditions lim x> -1- f(x)= +or- infinity and lim x> -1+ f(x)= +or- infinity. The answers are (a)5/4, and (b)5.... I have tried for a couple of hours and haven't been able to produce a formula for f that would produce those answers while keeping the given conditions of the function f. Please, HELP!! The book is the 4th edition, published in 2010. Section 2.5 LIMITS INVOLVING INFINITY. Page 124
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