1
$\begingroup$

Find the minimum value of $(\frac{x^n}{n} + \frac{1}{x})$ for $n \ge 4$.

One possible approach could be by first writing $$ \left(\frac{x^n}{n} + \frac{1} {x}\right) = \left( \frac{x^n}{n} + \frac{1}{nx} + \frac{1}{nx}+ \frac{1}{nx} + \text{upto n terms}\right) $$ then using the property $\mathrm{AM} \ge \mathrm{GM}$, we would get $$\left(\frac{x^n}{n} + \frac{1}{nx} + \frac{1}{nx}+ \frac{1}{nx}+ \text{upto }n \text{ terms}\right) = \left( \frac{x^n}{n} + \frac{1}{x} \right) \ge \frac{n+1}{n}$$ But this does not hold good for negative $x$; I am inquisitive to know how to approach for that case?

  • 2
    Approaching $x=0$ from the left gives a vertical asymptote going to negative infinity, implying there will never be a global minimum. I take it you're looking for local minima then? Are you familiar with the method of using derivatives to find local extrema?2011-08-01
  • 0
    yes,I am familiar of using the derivatives for finding the maxima and minima,however I got mired while using it here :/2011-08-01
  • 0
    You shouldn't have come across any problems. Can you tell me what the derivative of $x^n/n+1/x$ is?2011-08-01
  • 0
    With respect to $x$,$\large \frac{n \times x^{n-1}}{n} - \frac{1}{x^2}$ then I have to equate this to $0$ and find the $x$ ...2011-08-01
  • 0
    You can cancel out $n/n$ as just $1$. And after setting $=0$, you can multiply everything through by $x^2$. The only real solutions will be either just $1$ or $\pm 1$, but $x=-1$ is ruled out by the second derivative test (it's a local max, not a min). But if you're not looking for local extrema anyway then you can certainly leave all this behind.2011-08-01
  • 1
    Ehhh... my fault,I somehow assumed as the minimum point as the minimum value :/,thanks a lot,however this A.M,G.M approach seems a bit more intuitive to me :)2011-08-01
  • 2
    The AM-GM approach is indeed nice. But the calculus deals just as easily, for example, with the minimum value of $$\frac{x^{\sqrt{2}}}{\sqrt{2}}+\frac{1}{x}$$ for positive $x$.2011-08-02

1 Answers 1

2

For $x \lt 0$ there is no minimum. As $x$ gets very close to $0$, the $\frac{1}{x}$ term gets very large and negative, faster than the $\frac{x^n}{n}$ can get positive (assuming $n$ is even-if $n$ is odd this term is negative, too).

  • 0
    Thanks,I should have taken the limit first,one more doubt,how $n \ge 4$ is justified here?I mean what is the problem with $n=3$?2011-08-01
  • 0
    @FoolForMath: There is no problem. The argument is the same no matter what natural number $n$ is. I encourage you to test and see for yourself using [WolframAlpha](http://www.wolframalpha.com).2011-08-01
  • 0
    @ anon:Yes,I noticed that but I was just wondering why the problem setter mentioned it explicitly.2011-08-01
  • 0
    @FoolForMath: Doesn't look like there's any meaningful reason to set $n\ge 4$. Also, you might want to repair your spacebar key. :)2011-08-01
  • 0
    @anon:Sorry,I don't understand,what is wrong with spacebar key?2011-08-01
  • 3
    It's a joke! You're supposed to put spaces after punctuation. No biggie.2011-08-01