# How would you undo these?

• Sep 1st 2009, 04:59 PM
nautica17
How would you undo these?
1/h(x+h)^2

1/hx^2

What would you do to make the above rationals NOT be over 1?
• Sep 1st 2009, 05:22 PM
skeeter
Quote:

Originally Posted by nautica17
1/h(x+h)^2

1/hx^2

What would you do to make the above rationals NOT be over 1?

you're question is not making sense.

what is the original question in its entirety?
• Sep 1st 2009, 05:28 PM
nautica17
There is no other question. I just want to make what I listed to not be over 1.

The original equation that I am solving is:

((1/h(x+h)^2)-(1/hx^2)) / h

I just wanted to make those two parts in the numerator a little bit more simpler so that they are not fractions. How do I go about doing so? Or is there some other way I can solve.
• Sep 1st 2009, 05:48 PM
skeeter
Quote:

Originally Posted by nautica17
There is no other question. I just want to make what I listed to not be over 1.

The original equation that I am solving is:

((1/h(x+h)^2)-(1/hx^2)) / h

I just wanted to make those two parts in the numerator a little bit more simpler so that they are not fractions. How do I go about doing so? Or is there some other way I can solve.

did you get this rational expression from the difference quotient ...

$\displaystyle \frac{f(x+h) - f(x)}{h}$ , where $\displaystyle f(x) = \frac{1}{x^2}$ ?
• Sep 1st 2009, 06:01 PM
nautica17
Yea I did. Then I just multiplied the top and bottom by 1/h after I set up the equation and got to this point. Now I'm stuck. Sorry for lack of info on my part. I have a bunch of homework and I'm trying to multi-task everything right now.
• Sep 1st 2009, 06:04 PM
Stroodle
Quote:

Originally Posted by nautica17
There is no other question. I just want to make what I listed to not be over 1.

The original equation that I am solving is:

((1/h(x+h)^2)-(1/hx^2)) / h

I just wanted to make those two parts in the numerator a little bit more simpler so that they are not fractions. How do I go about doing so? Or is there some other way I can solve.

That's not an equation, and it's unclear as to what you mean by "not be over 1"
• Sep 1st 2009, 06:08 PM
nautica17
Quote:

Originally Posted by Stroodle
That's not an equation, and it's unclear as to what you mean by "not be over 1"

I mean to make the one in the numerator disappear altogether; to have no more fraction.
• Sep 1st 2009, 06:14 PM
skeeter
your difference quotient should be ...

$\displaystyle \frac{\frac{1}{(x+h)^2} - \frac{1}{x^2}}{h}$

it can be rewritten as ...

$\displaystyle \frac{1}{h}\left(\frac{1}{(x+h)^2} - \frac{1}{x^2}\right)$

work inside to get a common denominator ...

$\displaystyle \frac{1}{h}\left(\frac{x^2 - (x+h)^2}{x^2(x+h)^2}\right)$

expand the numerator ...

$\displaystyle \frac{1}{h}\left(\frac{x^2 - (x^2+2xh+h^2)}{x^2(x+h)^2}\right)$

combine like terms in the numerator ...

$\displaystyle \frac{1}{h}\left(\frac{-2xh-h^2)}{x^2(x+h)^2}\right)$

factor $\displaystyle -h$ out of the numerator ...

$\displaystyle \frac{1}{h}\left(\frac{-h(2x+h)}{x^2(x+h)^2}\right)$

divide out the h's ...

$\displaystyle \frac{-(2x+h)}{x^2(x+h)^2}$

if you have covered limits in class, take the limit as $\displaystyle h \to 0$ and simplify ...
• Sep 1st 2009, 06:29 PM
nautica17
Thank-you that explains it. :) I guess my tutor at school didn't know what they were doing.