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

• Oct 8th 2009, 02:00 PM
teddyryan
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, 02:03 PM
Matt Westwood
Quote:

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 \$\displaystyle 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.
• Oct 8th 2009, 02:07 PM
Matt Westwood
Quote:

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 ------------------------********

\$\displaystyle f(x+h) - f(x)\$ ... is where we start

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

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

\$\displaystyle = 2hx + h^2 + h\$ ... and that's what you get left with when you've simplified it.
• Oct 8th 2009, 02:07 PM
teddyryan
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, 02:12 PM
Matt Westwood
Quote:

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 \$\displaystyle f(x)\$". That is, plug in \$\displaystyle x+h\$ into your function of \$\displaystyle x\$, and then subtract from it the results of plugging \$\displaystyle x\$ into your function of \$\displaystyle 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."
• Oct 8th 2009, 02:15 PM
Matt Westwood
Quote:

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 \$\displaystyle -f(x)\$ into the equation by replacing what \$\displaystyle - f(x)\$ actually is - that is, \$\displaystyle -(x^2 + x)\$.
• Oct 8th 2009, 02:24 PM
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?
• Oct 8th 2009, 02:57 PM
skeeter
Quote:

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.
• Oct 8th 2009, 10:19 PM
Matt Westwood
Quote:

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:

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

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

What you've done is:

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

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

Let's call it \$\displaystyle z\$ or something:

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

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

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

Then you work out what \$\displaystyle 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:

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

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

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

Is this any clearer?