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Finding local minimum

  1. Aug 20, 2009 #1
    1. The problem statement, all variables and given/known data

    Show that the function [tex] f(x) = \frac{x^8 + 9x^9 - 12x^{13}}{1 + x^3 - x^6} [/tex] has a local minimum at x = 0.

    2. Relevant equations



    3. The attempt at a solution

    If f is to have a local minimum at x = 0 then we must have f'(0) = 0. Without typing up the tedious calculations, is this the correct procedure:

    [tex] f(x) = (x^8 + 9x^9 - 12x^{13}) \frac{1}{1 - (x^3 +x^6)} [/tex]. Let [tex] u = x^3 + x^6 [/tex]. Then [tex] f(x) = (x^8 + 9x^9 - 12x^{13}) \frac{1}{1-u} = (x^8 + 9x^9 - 12x^{13}) (1 + u + u^2 + ...) [/tex].

    I can differentiate this using the product rule (and differentiate the expression on the right hand side term by term) or I can multiply both brackets and then differentiate term by term, remembering that u is a function of x. It seems to me for that f'(x), every term will be of the form x^i for some integer i, and so f'(0) = 0??
     
  2. jcsd
  3. Aug 20, 2009 #2
    When you wrote this: [tex]f(x) = (x^8 + 9x^9 - 12x^{13}) \frac{1}{1-u} = (x^8 + 9x^9 - 12x^{13}) (1 + u + u^2 + ...) [/tex] --- how did you know that [tex]|u| = |x^3 + x^6| < 1[/tex] in order for the convergent geometric series as you had defined holds?
     
  4. Aug 20, 2009 #3
    There is some neighborhood about x = 0 such that |x^3 + x^6| < 1.

    EDIT: To be more specific, |x^3 + x^6| <= |x^3| + |x^6|. It's possible to find x values close enough to 0 so that both |x^3| and |x^6| are less than 1/2 and so |x^3| + |x^6| < 1.
     
    Last edited: Aug 20, 2009
  5. Aug 20, 2009 #4
    Dick, if you're reading this, as I was making my reply to fmam3, I noticed your post. How come you deleted it?
     
  6. Aug 21, 2009 #5

    zcd

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    Can't you just derive once to make sure f'(0)=0 and derive a second time to check f''(0)>0? And for the record [tex]\frac{1}{1+x^{3}-x^{6}}\neq\frac{1}{1-(x^{3}+x^{6})}[/tex].
     
  7. Aug 21, 2009 #6

    Dick

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    I deleted it because it was incomplete and I was too tired to fix it. Sorry. But your function can be written as f(x)=x^8*p(x), where p(x) is positive in a neighborhood of x=0, right? Doesn't that show f(x) has a local minimum at x=0 without even worrying about any derivatives or infinite series? f(0)=0 at x=0 and is nonzero otherwise in the neighborhood.
     
  8. Aug 21, 2009 #7
    Damn, that's simple. Thanks Dick. The question said beside it "Hint: Taylor Expansion" so I figured I'd use an infinite series, but that's way more complicated.

    Just one question though, since I can write the infinite series as [tex] (x^8 +9x^9 - 12x^{13})(1 + u + u^2 + ...), u = x^6 - x^3 [/tex], then the term with the smallest exponent would be x^8 after multiplying through, and so f'(0) = f''(0) = 0. Since there is a local minimum at x = 0, shouldn't the second derivative be positive at x = 0?
     
  9. Aug 21, 2009 #8

    Dick

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    Mmm, no. If the second derivative is positive it's a minimum. That doesn't mean if it's a minimum then the second derivative is positive. Look at f(x)=x^8.
     
  10. Aug 21, 2009 #9
    Ah, I see. Once again, thanks for the help!
     
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