# Homework Help: Finding local minimum

1. Aug 20, 2009

### JG89

1. The problem statement, all variables and given/known data

Show that the function $$f(x) = \frac{x^8 + 9x^9 - 12x^{13}}{1 + x^3 - x^6}$$ 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:

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

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. Aug 20, 2009

### fmam3

When you wrote this: $$f(x) = (x^8 + 9x^9 - 12x^{13}) \frac{1}{1-u} = (x^8 + 9x^9 - 12x^{13}) (1 + u + u^2 + ...)$$ --- how did you know that $$|u| = |x^3 + x^6| < 1$$ in order for the convergent geometric series as you had defined holds?

3. Aug 20, 2009

### JG89

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
4. Aug 20, 2009

### JG89

Dick, if you're reading this, as I was making my reply to fmam3, I noticed your post. How come you deleted it?

5. Aug 21, 2009

### zcd

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 $$\frac{1}{1+x^{3}-x^{6}}\neq\frac{1}{1-(x^{3}+x^{6})}$$.

6. Aug 21, 2009

### Dick

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.

7. Aug 21, 2009

### JG89

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 $$(x^8 +9x^9 - 12x^{13})(1 + u + u^2 + ...), u = x^6 - x^3$$, 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?

8. Aug 21, 2009

### Dick

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.

9. Aug 21, 2009

### JG89

Ah, I see. Once again, thanks for the help!