A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

GIVEN A > 0 B > 0 f[x] = A x + B/x . Find the minima for all positive x's

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

    I got this far... and got stuck... Im not sure what to do next, or if I'm on the wrong track altogether. Find the derivative of f[x] f[x] = A x + B/x A and B are constants. = A + B x^-1 f'[x] = A -B x^-2 Find the zeros of the derivative f'[x] = 0 0 = A -B/x^2 B/ x^2 = A B = A x^2 B = A x x B/A = x x Sqrt[B/A] = Sqrt[x x] Sqrt[B]/Sqrt[A] = x Assign that x as the minima,

  2. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    looks good several suggestions: 1. for small x, f(x) = B/x, for large x f(x) = Ax. so there should be some sort of transition in the shape of the line. ie a stationary point. you could sketch it to visualise... 2. look at f''(x) and how it behaves across the range of the function. IOW you have to be sure the stationary point is a max/min and not a saddle / asymtotpe. f'' and a sketch will help.

  3. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Like Irish said you dont know if its a minima or a maxima yet

  4. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    It happens to be in this case, but in future, you need to do take the 2nd derivative, or check the behavior around your critical point

  5. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    your work looks good

  6. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Does anyone else have trouble with the backspace key in the equation editor causing a page refresh? .. sorry in advance if I dont use it for now. Im trying to feed the the Sqrt[B]/Sqrt[A] into the 2nd derivative to get a confirm on the minima, but having some issues with the algebra.. Do you guys have any tricks for Simplifying ( B)/ (Sqrt[B]/Sqrt[A]) I think it should end up as A*B But I'm not remembering the rules for making that happen.

  7. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1434879258339:dw|

  8. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    what is your f'' ?? need to get that right first

  9. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I got f''[x] = (2*B)/x^3​

  10. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    there was another part.. I had to explain what was the ratio of . x = Sqrt[B] /Sqrt[A] A x/ (B/x) and that's where I got stuck on the algebra.. Actually Dan, I understand how you got from B to sqrt[B] sqrt[B] but I'm still not getting how you cancelled out to sqrt(AB) I think I need to find a good link on the basics of dealing with radicals

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

    okay.. quotient rule for radicals I think

  12. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    The 2nd derivative. \[f''[x] = \frac{ 2B }{ (\frac{\sqrt{A}}{\sqrt{B}} )^3 }\] but perhaps given that A &B must be positive, perhaps I can just say that this function can never be negative, and so the critical point, must be a minima

  13. IrishBoy123
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    "perhaps I can just say that this function can never be negative, and so the critical point, must be a minima" you can but you should also know how to fiddle around with indices something like this: \(\frac{2B}{\frac{A^{3/2}}{B^{3/2}}}\) \(\frac{2B . B^{3/2}}{A^{3/2}}\) \(\frac{2B^{5/2}}{A^{3/2}}\) \(2 \sqrt{\frac{B^5}{A^3}}\)

  14. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    oh because

  15. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1434890067357:dw|

  16. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    you can reason it out logically you are dividing by a number that is already broken down into smaller number, if x and y are positive integers

  17. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    or , splitting 1 into parts of X, but if X is split into parts of Y then you must have Y parts of splits of X, that sounds a bit confusing to say lol

  18. 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.