A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing


  • 5 years ago

f(x)=(x^2 -4)^ 2/3 critical numbers and point of inflection

  • This Question is Closed
  1. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    For the critical points: Taking the derivative, you get f'(x) = (2/3)*(x^2 - 4)^(-1/3)*2x. Setting this equal to 0 and dividing out the (x^2 - 4)^(-1/3), you will see that x = 0. For the point of inflection: Take the derivative again, either by product or quotient rule and you'll get: (12(x^2-4)^(1/3) - 8*x^2*(x^2-4)^(-2/3))/(9*(x^2-4)^(2/3))=0. Get rid of the bottom by multiplying it to the other side then divide by (x^2-4)^(1/3) to get: 12-(8x^2)/(x^2-4) = 0. You can simplify further to eventually get x^2 = 12. Therefore the points of inflection are x = +-sqrt(12)

  2. 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...


  • 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.