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!

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

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

To find the instantaneous rate of change f'(a) for the function f(x) at x=a, you can use the formula you mentioned:

f'(a) = lim(h -> 0) (f(a + h) - f(a)) / h

In this case, you are given the function f(x) = x^2 and need to find f'(a) when x = 3 (a = 3). Let's substitute these values into the formula:

f'(3) = lim(h -> 0) (f(3 + h) - f(3)) / h

Now, let's plug in the values into the formula:

f'(3) = lim(h -> 0) ((3 + h)^2 - 3^2) / h

Expanding and simplifying:

f'(3) = lim(h -> 0) ((9 + 6h + h^2) - 9) / h
= lim(h -> 0) (6h + h^2) / h

Now, we can factor out an h from the numerator:

f'(3) = lim(h -> 0) h(6 + h) / h

Canceling out the h terms:

f'(3) = lim(h -> 0) (6 + h)

Finally, taking the limit as h approaches 0:

f'(3) = 6 + 0

Therefore, the instantaneous rate of change f'(3) for the function f(x) = x^2 at x = 3 is indeed 6.