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- Oct 8th 2009, 03:00 PM #1

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## f(x+h)-f(x)

f(x)=x^2+x

f(x+h)-f(x)

I'm very confused on how I'm supposed to substitute. Can someone fill me in on the process?

b. As a consequence, f(x+h) - f(x) = x^2 + 2hx + h^2 + x + h - (x^2 + x) = 2hx + h^2 + h

I don't get where this part comes from ------------------------********

- Oct 8th 2009, 03:03 PM #2

- Oct 8th 2009, 03:07 PM #3

- Oct 8th 2009, 03:07 PM #4

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So if there are 2 x's in the equation then I just sub the f(x + h) in for the first x and - f(x) in for the second x?

I'm unsure as to how -(x^2 +x) fits into it, where does that come from? How do you put - f(x) into the equation?

- Oct 8th 2009, 03:12 PM #5
"minus ". That is, plug in into your function of , and then subtract from it the results of plugging into your function of .

"f" is just a machine that says: "Take this thing, and add its square to it."

"- f(x)" means (in this context) "... and subtract from that what you get from adding the square of x to x."

- Oct 8th 2009, 03:15 PM #6

- Oct 8th 2009, 03:24 PM #7

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I understand it a little better thanks, so in this example.

f(x) = 2x^2 + 3 and find f(x + h) - f(x)

I would first sub in (x + h) yielding f(x) = 2(x^2 + 2hx + h^2) + 3

then distribute the 2 giving me f(x) = 2x^2 + 4hx + 2h^2 + 3

what would I do next then? I don't see how I could simplify it further and now I'm supposed to subtract that from itself?

- Oct 8th 2009, 03:57 PM #8

- Oct 8th 2009, 11:19 PM #9
You are trying to calculate something (which hasn't actually been given a name as such) which is

**defined**as:

where is**defined**as .

What you've done is:

which is saying which is wrong.

Let's call it or something:

What you're trying to do is work out what is by unpacking the**two**instances of : one of those instances is , and the other (which you have to subtract from the first) is just .

So you work out what is, you've done that - it's which works out as .

Then you work out what is. You don't need to "work that out" because it's already in exactly the form you want it to be in:

So you substitute the expressions you've just worked out for and into your expression for :

Is this any clearer?