# Homework Help: Comparing two functions

1. Oct 1, 2012

### the_kid

1. The problem statement, all variables and given/known data
I'm trying to show that the function f(x)=$\frac{1}{x^{2}}$ is greater than g(x)=$\frac{1}{x}$-ln(1+$\frac{1}{x}$) for all x>0.

2. Relevant equations

3. The attempt at a solution
This is really simple, but I'm having some trouble showing it rigorously. I thought about trying to compare the derivatives, but it's not that simple. Are there any good ways to do this?

2. Oct 1, 2012

### SammyS

Staff Emeritus
Well... then the function (f-g)(x) must be positive ...

the function (f/g)(x) must be greater than 1, or be negative ...

the function (g/f)(x) must be less than 1, or be negative.

3. Oct 1, 2012

### the_kid

Right. That was the other method I tried. I just can't figure out a way to do it without some "hand-waving."

4. Oct 2, 2012

### jbunniii

Not sure if this simplifies anything, but if $0 < x < 1$, then $1/x^2 > 1/x$ so clearly $f(x) > g(x)$. So you only have to worry about $x > 1$.

5. Oct 2, 2012

### HallsofIvy

Using that, a fourth method: it is easy to show this is true for x= 1 so if you can prove that f'(x)> g'(x) for all x>1 you are done.

6. Oct 2, 2012

### SammyS

Staff Emeritus
How about comparing f(1/x) and g(1/x) ?

7. Oct 13, 2012

### the_kid

Hmmm, OK, how does this help me?

8. Oct 13, 2012

### Ray Vickson

Well, derivatives are a bit simpler, etc.

RGV

9. Oct 14, 2012

### the_kid

I've been able to show that f(1/x)>g(1/x) for all x>1. From this, how exactly can I conclude that f(x)>g(x)?

10. Oct 14, 2012

### SammyS

Staff Emeritus
That shows that f(x)>g(x) for 0 < x < 1, which isn't a big help.