Another function problem???

Question

using f(x)=x^2-x+1, find f(x+h)-f(x)/x

Answer 1

(x+h)^2 - x - h + 1 - (x^2+x+1)/x
= (x^3 +2x^2h + h^2x - x^2 - hx + x - x^2 - x - 1)/x
= (x^3 +2x^2h +h^2x - 2x^2 -hx -1)/x

Answer 2

So you have

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

And you want to solve for

[f(x+h) - f(x)]/h

(I presume you meant to say "h" instead of "x", because this looks exactly like the formula for the definition of a derivative).

Let's start. First, let's write out what f(x+h) actually is.

f(x+h) = (x+h)^2 - (x+h) + 1
f(x+h) = (x^2 + 2xh + h^2) - (x + h) + 1
f(x+h) = x^2 + 2xh + h^2 - x - h + 1

Now, we solve for [f(x+h) - f(x)]/h, plugging in what we solved for and plugging what is given.

[f(x+h) - f(x)]/h = {[x^2 + 2xh + h^2 - x - h + 1] - [x^2 - x + 1]} / h

Expanding it out, we get

[f(x+h) - f(x)]/h = {x^2 + 2xh + h^2 - x - h + 1 - x^2 + x - 1} / h

And then simplifying,

[f(x+h) - f(x)]/h = [2xh + h^2 - h] / h

We can factor out an h on the numerator, to get

[f(x+h) - f(x)]/h = {h[2x + h - 1]} / h

And, we can cancel the h on the top and bottom ***PROVIDED*** that h IS NOT EQUAL TO ZERO. This next step would be an invalid step otherwise.

[f(x+h) - f(x)]/h = 2x + h - 1