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

1. ## 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 ------------------------********

2. Originally Posted by teddyryan
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?
Wherever you see "x", put "(x+h)" there in its place.

So $f(x+h) = (x+h)^2 + (x+h)$.

It really is that simple.

The complicated bit comes with multiplying it out and then simplifying it.

3. Originally Posted by teddyryan
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 ------------------------********
$f(x+h) - f(x)$ ... is where we start

$= (x+h)^2 + (x+h)$ ... this bit's $f(x+h)$
$- (x^2 + x)$ ... this bit's $f(x)$

$= x^2 + 2hx + h^2 + x + h - (x^2 + x)$ ... multiplying it all out

$= 2hx + h^2 + h$ ... and that's what you get left with when you've simplified it.

4. 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?

5. Originally Posted by teddyryan
Sorry I think I confused you, I understand it when its f(x) = x^2 + x and x = f(x + h), you simply substitute. But I get lost when x = f(x + h) - f(x), I'm unsure of what the - f(x) is implying.
"minus $f(x)$". That is, plug in $x+h$ into your function of $x$, and then subtract from it the results of plugging $x$ into your function of $x$.

"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."

6. Originally Posted by teddyryan
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?
You put $-f(x)$ into the equation by replacing what $- f(x)$ actually is - that is, $-(x^2 + x)$.

7. 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?

8. Originally Posted by teddyryan
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+h) = 2(x^2 + 2hx + h^2) + 3
then distribute the 2 giving me f(x+h) = 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?
f(x+h) - f(x)

2x^2 + 4hx + 2h^2 + 3 - (2x^2 + 3)

combine like terms.

9. Originally Posted by teddyryan
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?
You are trying to calculate something (which hasn't actually been given a name as such) which is defined as:

$f(x+h) - f(x)$

where $f(x)$ is defined as $2x^2 + 3$.

What you've done is:

$f(x) = 2(x^2 + 2hx + h^2) + 3$

which is saying $f(x) = f(x+h)$ which is wrong.

Let's call it $z$ or something:

$z = f(x+h) - f(x)$

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

So you work out what $f(x+h)$ is, you've done that - it's $2(x+h) + 3$ which works out as $2x^2 + 4hx + 2h^2 + 3$.

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

$f(x) = 2x^2 + 3$

So you substitute the expressions you've just worked out for $f(x+h)$ and $f(x)$ into your expression for $z$:

$z = f(x+h) - f(x) = (2x^2 + 4hx + 2h^2 + 3) - (2x^2 + 3)$

Is this any clearer?