Suppose that f(x+h)-f(x) =-4 h x^2 - 1 h x + 4 h^2 x + 3 h^2 -3 h^3.

Find: f^1(x)

I am not certain what you mean f^1(x). Can you put that in words? Is this inverse?

its f prime.

I got the answer thanks anyways:)

good. While answering, I got a virus taking me to the dark web and I was in a mess.

To find the derivative of f(x), denoted as f'(x) or f^1(x), we need to apply the limit definition of the derivative:

f^1(x) = lim(h->0) [f(x+h) - f(x)] / h

Given the expression: f(x+h) - f(x) = -4hx^2 - hx + 4h^2x + 3h^2 - 3h^3, we can substitute it in the limit definition:

f^1(x) = lim(h->0) [(-4hx^2 - hx + 4h^2x + 3h^2 - 3h^3) / h]

Now, let's simplify the expression inside the limit:

f^1(x) = lim(h->0) [-4hx^2/h - hx/h + 4h^2x/h + 3h^2/h - 3h^3/h]

f^1(x) = lim(h->0) [-4x^2 - x + 4hx + 3h - 3h^2]

Next, we can distribute the limit across the terms:

f^1(x) = lim(h->0) -4x^2 - lim(h->0) x + lim(h->0) 4hx + lim(h->0) 3h - lim(h->0) 3h^2

Since h is approaching 0, all h terms become 0:

f^1(x) = -4x^2 - x + 0 + 0 - 0

Simplifying further, we have:

f^1(x) = -4x^2 - x

Therefore, the derivative of f(x), f'(x) or f^1(x), is -4x^2 - x.