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Homework Help: Rate of change from 0 to x 2x^2 + 3x

  1. Apr 8, 2006 #1
    2x^2 + 3x

    I'm not really sure what to do with the "X"
    Form: f(x) - f(c)/ x-c

    -2(x)^2 +3(x) - (-2(0)^2 + 3 (0) / x-0

    I get
    = -2x^2 + 3x + 0 / x - 0

    = -2x^2 + 3x / x

    that doesn't seem right
    :grumpy:

    I'm taking a break
     
  2. jcsd
  3. Apr 8, 2006 #2
    You forgot to take the limit as x -> 0.

    Personally, I've never been too keen on that form for the derivative. I think it's more transparent to use:
    [tex]\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}[/tex]

    That might show you a bit better what happens to everything.

    -Dan
     
  4. Apr 8, 2006 #3

    VietDao29

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    topsquark, you don't need to take the limit as x -> 0. The problem does not ask for the slope of the tangent line at x = 0. It just ask for the rate of change from 0 to x.
    There should be no minus sign in front of the 2x2, and you seem to be missing some parentheses. :)
    It should read:
    [tex]\frac{2x ^ 2 + 3x}{x}[/tex] or (2x ^ 2 + 3x) / x
    Now, you can simplify that expression a little bit further, right?
    Can you go from here? :)
     
  5. Apr 9, 2006 #4
    Ok, so I just do this:

    =2x^2 + 3x/x

    =x(2x + 3) / x

    =2x + 3
     
    Last edited: Apr 9, 2006
  6. Apr 10, 2006 #5

    HallsofIvy

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    As long as x is not 0!
     
  7. Apr 10, 2006 #6
    Well 0! = 1, so I don't think that would be a problem ;)
     
  8. Apr 11, 2006 #7

    VietDao29

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    Err, I don't understand, why is factorial involved in here??? :confused:
     
  9. Apr 11, 2006 #8

    HallsofIvy

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    It was joke, son, a joke! (The ;) was a give-away)

    ksinclair13, you can get the actual :wink: by typing ": w i n k :" without the spaces or by going to "advanced" and clicking on the icon.
     
  10. Apr 11, 2006 #9
    Okay, thank you. :wink:

    I always use the quick reply, but now I know how to do it with that as well :smile:
     
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