- SRQ baller24
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- Exalted Mythic Member
Posted by: ll 0wN4g3 ll
hey guys so I'm doing the instantaneous rate of change using the difference quotient and I have a test on monday and I'm studying but I don't know how to do some of them can someone help me? ok for these problems I have this equation: http://www.mathwords.com/d/difference_quotient.htm
except i'm finding the instantaneous rate of change. I have for f(x)=2/x and for the a/x-zero = 2. I just have to get the H from the bottom out and multiply it by zero to find the limit.
(2/(x+h)-2/x)/h
multiply by x(x+h)/x(x+h) which is just 1, so it is allowed. Basically this step just allows you to add the terms in the numerator by giving them the same denominator. I will expand on this now as it might be confusing. Focus on just the numerator without the h in the denominator to make it easier to read.
2/(x+h)-2/x
To add the terms, they have to have the same denominator. So the first term needs to be multiplied by x/x yielding 2x/x(x+h)
the second term needs to be multiplied by (x+h)/(x+h), yielding -2(x+h)/x(x+h). So at this point you have
2x/x(x+h)-2(x+h)/x(x+h)
Now, you can add the terms with common denominator, so
[2x-2(x+h)]/x(x+h)
Now, dividing by the h that was originally in the denominator.
get
[2x-2(x+h)]/(xh(x+h))
add terms in the numerator
-2h/(xh(x+h)
divide out the h
-2/(x(x+h))
or
-2/(x^2+hx)
take limit as h->0
-2/x^2
[Edited on 12.15.2012 8:16 AM PST]