A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • 5 years ago

can someone help me find the absolute and local extrema in this equation (3x-4)/(x^2+1)

  • This Question is Closed
  1. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    sure

  2. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    (x^2+1)(3) - (3x-4)(2x) = 0

  3. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    on the intervals [-2,2]

  4. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    3x^2 +3 -6x^2 +8x = 0 -3x^2 +3 +8x = 0 0 = 3x^2 -8x -3

  5. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    (x-3) (x+ 1/3) = x = x and x = -1/3 are the max and min

  6. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    x = 3 and x = -1/3......

  7. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    f(-2) = ? f(-1/3) = ? f(2) = ?

  8. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so what do u do with the denominator

  9. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    nothing; the top directs the =0 part; not the bottom

  10. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    put these x values into the original equationand determine which is higher

  11. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    awesome thanks a lot man the denominator was throwing me off

  12. amistre64
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    there is no vertical asypmtote; so you aint gotta worry about bad values :)

  13. myininaya
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    also x^2+1=0 has no real solution so this does not give you a critcal number anyways

  14. myininaya
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    right what amistre said no vertical asym

  15. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    awesome thanks a lot guys

  16. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

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.