# Math (Instanttaneous rate of change)

Compute the instantaneous rate of change f'(a) for the function f(x) at x=a using:
f'(a)=lim(h->0) (f(a+h)-f(a))/h

f(x)=x^2 at x=3

My teacher didn't even go over the problem since class was over but gave us the answer being 6. How do you approach this problem to end up with 6? Any help is greatly appreciated!

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1. just do the algebra. At a=3,

f(3+h) = (3+h)^2 = 9+6h+h^2
f(3) = 3^2 = 9

subtract and the numerator becomes

6h+h^2

divide that by h and you have 6+h

Up to now, no mention of limits is made. It's just straight Algebra I. Now, to get to the magical part:

as h->0, (6+h) -> 6

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2. I see, this was a lot simpler than I expected! Thanks so much

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