# Precalculus

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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 >.<

• Precalculus -

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

• Precalculus -

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

• Precalculus -

What did you with -3/2? You have not used it, why?

• Precalculus -

Well, I used 1.5 instead of 3/2

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