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Math Help - Difference quotient

  1. #1
    Newbie
    Joined
    Jun 2009
    Posts
    6

    Exclamation Difference quotient

    Next problem:

    For each function f(x), determine this difference quotient in a simplified form.
    f
    (x + h) − f(x)

    h

    a.) f(x) = 2x + 1

    The problem I have here is that I understand all of the steps except for the first one, which is this:

    2(x + h) + 1 − (2x + 1)
    h

    The second part of the first step ((2x + 1)) makes complete sense.

    It's this I can't wrap my head around:

    2(x + h) + 1


    How does f(x) = 2x + 1 turn this f(x + h) into this
    2(x + h) + 1
    ?
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  2. #2
    Super Member
    Joined
    May 2009
    Posts
    527
    Quote Originally Posted by CaptianHawk1 View Post
    It's this I can't wrap my head around:

    2(x + h) + 1


    How does f(x) = 2x + 1 turn this f(x + h) into this
    2(x + h) + 1
    ?
    The function is f(x) = 2x + 1. When you put in a number for x, like f(4), you're essentially saying, "I want the value of f(x) when x = 4." So to find this value you plug in 4 for x:
    f(4) = 2(4) + 1 = 9

    But you don't have to just put in numbers. You could put in variables and expressions too. For example, I want to know what f(y) is. Who cares whether I know what y is equal to or not? I plug in y for x:
    f(y) = 2y + 1.

    In your case, you want to know the value of the function when x = x + h. Plug in x + h every time you see an x:
    f(x + h) = 2(x + h) + 1.

    I'll give you one more example. What if I want to know the value of the function when x = ab + c?
    f(ab + c) = 2(ab + c) + 1.


    01
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  3. #3
    Newbie
    Joined
    Jun 2009
    Posts
    6
    I get it now. I was doing it in reverse, that is taking the 2x + 1 and trying to plug it into x's in the fraction.

    Keep in mind, I'm teaching myself this stuff.

    Thanks again.
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