Precalculus
posted by Casey .
Estimate the instanteous rate of change f(x) = x/(x+5) at (3, 3/2)
Could someone explain this question to me, this isn't exactly the same as polynomial functions >.<

It just says estimate so do that. add a small amount, say .1 to x and see how much f(x) changes
then the rate of change is (f(3+.1)  f(3) )/.1
[f(2.9)  f(3) ] / .1
= [1.38  (1.5)]/.1
= .12/.1
= 1.2 
Now if you knew calculus you would have said:
dy/dx = [(x+5)(1)  x(1) ] / (x+5)^2
= 5/(x+5)^2
at x = 3 that is
5/4 = 1.25 well not bad 
What did you with 3/2? You have not used it, why?

Well, I used 1.5 instead of 3/2
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