Results 1 to 9 of 9

Math Help - How would you undo these?

  1. #1
    Member nautica17's Avatar
    Joined
    Aug 2009
    Posts
    121

    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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    Quote Originally Posted by nautica17 View Post
    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?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member nautica17's Avatar
    Joined
    Aug 2009
    Posts
    121
    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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    Quote Originally Posted by nautica17 View Post
    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} ?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member nautica17's Avatar
    Joined
    Aug 2009
    Posts
    121
    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.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member Stroodle's Avatar
    Joined
    Jun 2009
    Posts
    367
    Quote Originally Posted by nautica17 View Post
    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"
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member nautica17's Avatar
    Joined
    Aug 2009
    Posts
    121
    Quote Originally Posted by Stroodle View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    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 ...
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member nautica17's Avatar
    Joined
    Aug 2009
    Posts
    121
    Thank-you that explains it. I guess my tutor at school didn't know what they were doing.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Undo a mathematical conversion done inside a black box
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: February 10th 2009, 01:56 PM

Search Tags


/mathhelpforum @mathhelpforum