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.