anonymous
  • anonymous
GIVEN A > 0 B > 0 f[x] = A x + B/x . Find the minima for all positive x's
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
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,
IrishBoy123
  • IrishBoy123
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.
dan815
  • dan815
Like Irish said you dont know if its a minima or a maxima yet

Looking for something else?

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

More answers

dan815
  • dan815
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
dan815
  • dan815
your work looks good
anonymous
  • anonymous
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.
dan815
  • dan815
|dw:1434879258339:dw|
IrishBoy123
  • IrishBoy123
what is your f'' ?? need to get that right first
anonymous
  • anonymous
I got f''[x] = (2*B)/x^3‚Äč
anonymous
  • anonymous
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
anonymous
  • anonymous
okay.. quotient rule for radicals I think
anonymous
  • anonymous
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
IrishBoy123
  • IrishBoy123
"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: \(\frac{2B}{\frac{A^{3/2}}{B^{3/2}}}\) \(\frac{2B . B^{3/2}}{A^{3/2}}\) \(\frac{2B^{5/2}}{A^{3/2}}\) \(2 \sqrt{\frac{B^5}{A^3}}\)
dan815
  • dan815
oh because
dan815
  • dan815
|dw:1434890067357:dw|
dan815
  • dan815
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
dan815
  • dan815
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

Looking for something else?

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