1. ## 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!!

2. Originally Posted by JonathanEyoon
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

3. Originally Posted by Jhevon
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

4. Originally Posted by JonathanEyoon
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?

5. Originally Posted by Jhevon
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?

6. Originally Posted by JonathanEyoon
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.

7. Originally Posted by Jhevon
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 =(

8. Help

9. Originally Posted by JonathanEyoon
Help
Here:

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

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

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

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

10. Originally Posted by Jhevon
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?

11. Originally Posted by JonathanEyoon
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

12. Originally Posted by JonathanEyoon
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

13. Originally Posted by JonathanEyoon
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 $- \frac {1}{a}$

14. Originally Posted by Jhevon
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

15. Originally Posted by JonathanEyoon
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