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

1. 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,

2. 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.

3. dan815

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

4. 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

5. dan815

6. 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.

7. dan815

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8. IrishBoy123

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

9. anonymous

I got f''[x] = (2*B)/x^3​

10. 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

11. anonymous

okay.. quotient rule for radicals I think

12. 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

13. 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: $$\frac{2B}{\frac{A^{3/2}}{B^{3/2}}}$$ $$\frac{2B . B^{3/2}}{A^{3/2}}$$ $$\frac{2B^{5/2}}{A^{3/2}}$$ $$2 \sqrt{\frac{B^5}{A^3}}$$

14. dan815

oh because

15. dan815

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16. 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

17. 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