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How to get instantaneous rate of change

  1. Sep 26, 2009 #1


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    1. The problem statement, all variables and given/known data
    6x^2 - 4
    x = -2

    2. Relevant equations

    3. The attempt at a solution

    I input -2 for x but i got the wrong answer..the answer is suppose to be -24
  2. jcsd
  3. Sep 26, 2009 #2
    You'll need to use the difference quotient
    [tex]\frac{f(x + h) - f(x)}{h}[/tex]

    Do you know why and how to use it for the problem?
  4. Sep 26, 2009 #3


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    Thanks a lot Bohrok. I got it :shy:
  5. Sep 26, 2009 #4


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    i know how to use it but not why.
  6. Sep 26, 2009 #5
    The slope m between two points (x, f(x)) and (x1, f(x1)) is given by the following, which you should be familiar with:

    [tex]m = \frac{f(x_1) - f(x)}{x_1 - x}[/tex]

    To make it easier to work with, let x1 = x + h, so h is basically the distance between the x values of the two points.

    [tex]m = \frac{f(x + h) - f(x)}{x + h - x} = \frac{f(x + h) - f(x)}{h}[/tex]

    Although this is precalc, this page should help you understand it all

    Look especially at the secant lines where you let h go to 0.
  7. Sep 26, 2009 #6


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    oh thanks Bohrok. I know have a better understanding.
  8. Sep 26, 2009 #7
    The easier way that I know is that you have to find the derivative of f(x) = 6x^2 - 4.

    Then substitute -2 to x.

    Note: Use the easier way in finding derivatives.
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