Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

iNeedhelpwithSchool Group Title

What kinds of discontinuities are possible for a rational function?

  • one year ago
  • one year ago

  • This Question is Open
  1. AKANicole Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    i think um,

    • one year ago
  2. AKANicole Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    um i forgot i know this somewhere,

    • one year ago
  3. jim_thompson5910 Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Case 1) The function is completely continuous over the entire domain, so there are no discontinuities Ex: y = 1/(x^2 + 1) The denominator is ALWAYS positive, so it can NEVER be zero ----> no vertical asymptotes ----> no discontinuities ------------------------------------------------------- Case 2) There is at least one asymptotic discontinuity, so it's discontinuous at this point or points. Ex: y = 1/(x+1) ... has a vertical asymptote at x = -1 ------------------------------------------------------- Case 3) There is a removable discontinuity (or a point discontinuity) Ex: y = ((x+1)(x+2))/(x+2) simplifies to y = x+1, but it has a removable discontinuity at x = -2

    • one year ago
  4. oldrin.bataku Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Adding on to @jim_thompson5910's last note -- when you "simplify" \(\frac{(x+1)(x+2)}{x+2}=x+1\), you're dividing both our numerator and denominator by \(x+2\). Remember that we're never allowed to divide by \(0\), so by simplifying we're essentially assuming that \(x+2\ne0\) -- and this assumption is good for almost all \(x\) -- except for \(x=-2\). We'll just be left with a little hole there, with the rest of the function looking identical to \(x+1\), but there's still that *removable* discontinuity, removable because it's just a single little hole that we could "fill in" to be \(y+1\).

    • one year ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.