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Math Help - Rate of Change

  1. #1
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    Rate of Change

    How would I solve this?

    G(x) = 1 / x where x = 1, x = a



    So far i've gotten to :

    (1 / a - 1 / 1) / a - 1


    But after that i'm completely stuck..... help someone!!
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by JonathanEyoon View Post
    How would I solve this?

    G(x) = 1 / x where x = 1, x = a



    So far i've gotten to :

    (1 / a - 1 / 1) / a - 1


    But after that i'm completely stuck..... help someone!!
    what do you want to do exactly? find the average rate of change between x = 1 and x = a for a > 1? what form are you looking for the answer to be in. you seem to be ok, but you just need to simplify
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    Quote Originally Posted by Jhevon View Post
    what do you want to do exactly? find the average rate of change between x = 1 and x = a for a > 1? what form are you looking for the answer to be in. you seem to be ok, but you just need to simplify

    It's the simplification sorry I didn't specify
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    Quote Originally Posted by JonathanEyoon View Post
    It's the simplification sorry I didn't specify
    first combine the fractions in the numerator? there are a lot of ways to do this, do you know any? then multiply by the reciprocal of the numerator (that is, \frac {1}{a - 1} )

    what do you get?
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    Quote Originally Posted by Jhevon View Post
    first combine the fractions in the numerator? there are a lot of ways to do this, do you know any? then multiply by the reciprocal of the numerator (that is, \frac {1}{a - 1} )

    what do you get?

    Sorry could you teach me some?
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    Quote Originally Posted by JonathanEyoon View Post
    Sorry could you teach me some?
    I explained several methods here, check them out. get back to me if you are not clear

    i find that the last method i used in my last post is most useful when figuring out the LCD is a pain, or when dealing with algebra involving a lot of variables.
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    Quote Originally Posted by Jhevon View Post
    I explained several methods here, check them out. get back to me if you are not clear

    i find that the last method i used in my last post is most useful when figuring out the LCD is a pain, or when dealing with algebra involving a lot of variables.

    Ok so for the top portion of the problem, I would cross multiply and end up with

    (1-a) / a - 1


    From here i'm not certain where to go. I've been on this problem for a while and still can't come up with the answer of -1 / a. Help =(
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    Help
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    Quote Originally Posted by JonathanEyoon View Post
    Help
    Here:

    \frac { \frac {1}{a} - \frac {1}{1}}{a - 1} = \frac { \frac {1 - a}{a}}{a - 1}

    ......... = \frac {1 - a}{a(a - 1)}

    ......... = \frac {-(a - 1)}{a(a - 1)}

    ......... = - \frac {1}{a}
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  10. #10
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    Quote Originally Posted by Jhevon View Post
    Here:

    \frac { \frac {1}{a} - \frac {1}{1}}{a - 1} = \frac { \frac {1 - a}{a}}{a - 1}

    ......... = \frac {1 - a}{a(a - 1)}

    ......... = \frac {-(a - 1)}{a(a - 1)}

    ......... = - \frac {1}{a}


    The first line that you worked, wouldn't the result from

    1 / a - 1 / 1 =

    1 - a? Because you would have to find the common denominator and that being (a), the (a) would cancel out and leave 1 and for the second term, 1/1 = 1 so it'll simply be 1 x a which is a. The whole problem would be

    (1 - a) / a - 1

    Am i wrong? if so could you tell me why?
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    Quote Originally Posted by JonathanEyoon View Post
    The first line that you worked, wouldn't the result from

    1 / a - 1 / 1 =

    1 - a? Because you would have to find the common denominator and that being (a), the (a) would cancel out and leave 1 and for the second term, 1/1 = 1 so it'll simply be 1 x a which is a. The whole problem would be

    (1 - a) / a - 1

    Am i wrong? if so could you tell me why?
    yes you are wrong. You can only cancel things that way in a fraction when you have a single term in the numerator and denominator. this means we have to have a product or division of terms for that to work.

    (1 - a) are two terms. we separate terms using +'s and -'s. we cannot cancel the a without affecting the 1 in this case, and therefore, what you did was incorrect. if it was a(1 - a)/a then we could cancel the a's, since we have one term in the top and bottom, yes a(1 - a) is considered to be ONE term, since it is a product
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    Quote Originally Posted by JonathanEyoon View Post
    The first line that you worked, wouldn't the result from

    1 / a - 1 / 1 =

    1 - a? Because you would have to find the common denominator and that being (a), the (a) would cancel out and leave 1 and for the second term, 1/1 = 1 so it'll simply be 1 x a which is a. The whole problem would be

    (1 - a) / a - 1

    Am i wrong? if so could you tell me why?

    How did you go from the second line to the third line? I know how to you got (1 - a) / a(a - 1) but from there what exactly did you do? I'm so sorry for the million questions, I just want to be sure I get this down right the first time
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by JonathanEyoon View Post
    How did you go from the second line to the third line? I know how to you got (1 - a) / a(a - 1) but from there what exactly did you do? I'm so sorry for the million questions, I just want to be sure I get this down right the first time
    no, you should ask questions! i think everyone here likes that, it shows that you are not just after answers but are really trying to understand what is going on, which is refreshing.

    i noticed that 1 - a = -(a - 1)

    and we have an a - 1 in the denominator. since we have one term in the numerator and the denominator, i can cancel the like terms. so i cancel the a - 1 in the top and bottom, but it was a minus a - 1 in the top, so i had to leave the minus sign there, so i ended up with - \frac {1}{a}
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    Quote Originally Posted by Jhevon View Post
    no, you should ask questions! i think everyone here likes that, it shows that you are not just after answers but are really trying to understand what is going on, which is refreshing.

    i noticed that 1 - a = -(a - 1)

    and we have an a - 1 in the denominator. since we have one term in the numerator and the denominator, i can cancel the like terms. so i cancel the a - 1 in the top and bottom, but it was a minus a - 1 in the top, so i had to leave the minus sign there, so i ended up with - \frac {1}{a}

    Thank you so very much~~~!!! I think i might have one more question but i'll post it in a new thread. This really helped alot
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    Quote Originally Posted by JonathanEyoon View Post
    Thank you so very much~~~!!! I think i might have one more question but i'll post it in a new thread. This really helped alot
    your welcome!

    yes, post in a new thread
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