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

  1. #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 ------------------------********
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    Quote Originally Posted by teddyryan View Post
    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.
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  3. #3
    MHF Contributor Matt Westwood's Avatar
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    Quote Originally Posted by teddyryan View Post
    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.
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  4. #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?
    Last edited by mr fantastic; Oct 9th 2009 at 06:13 AM. Reason: Merged posts
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  5. #5
    MHF Contributor Matt Westwood's Avatar
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    Quote Originally Posted by teddyryan View Post
    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."
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  6. #6
    MHF Contributor Matt Westwood's Avatar
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    Quote Originally Posted by teddyryan View Post
    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, $\displaystyle -(x^2 + x)$.
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  7. #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?
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  8. #8
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    Quote Originally Posted by teddyryan View Post
    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.
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  9. #9
    MHF Contributor Matt Westwood's Avatar
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    Quote Originally Posted by teddyryan View Post
    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?
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