# Math Help - How would you undo these?

1. ## 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?

2. 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?

3. 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.

4. 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 ...

$\frac{f(x+h) - f(x)}{h}$ , where $f(x) = \frac{1}{x^2}$ ?

5. 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.

6. 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"

7. 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.

8. your difference quotient should be ...

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

it can be rewritten as ...

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

work inside to get a common denominator ...

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

expand the numerator ...

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

combine like terms in the numerator ...

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

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

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

divide out the h's ...

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

if you have covered limits in class, take the limit as $h \to 0$ and simplify ...

9. Thank-you that explains it. I guess my tutor at school didn't know what they were doing.