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Math Help - ...Rate of Change...

  1. #1
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    ...Rate of Change...

    Well, Finals are coming up so I'm reviewing, and I'm already in need of some help!

    This problem isn't hard, I just can't remember how to do it.

    1) In exercise 1, find the average rate of change of the function over the given interval or intervals
    f(x)=X^3+1
    a. [2.3] b. [-1,1]
    >>Answer = 19, 1

    I'll be posting in here a lot throughout the day, so I'll be checking back often. As always, any help is greatly appreciated!
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  2. #2
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    Quote Originally Posted by Super Mallow View Post
    Well, Finals are coming up so I'm reviewing, and I'm already in need of some help!

    This problem isn't hard, I just can't remember how to do it.

    1) In exercise 1, find the average rate of change of the function over the given interval or intervals
    f(x)=X^3+1
    a. [2.3] b. [-1,1]
    >>Answer = 19, 1

    I'll be posting in here a lot throughout the day, so I'll be checking back often. As always, any help is greatly appreciated!
    The average of a function f(x) on an interval [a, b] is defined as int(f(x),x=a..b)/(b-a).
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Super Mallow View Post
    Well, Finals are coming up so I'm reviewing, and I'm already in need of some help!

    This problem isn't hard, I just can't remember how to do it.

    1) In exercise 1, find the average rate of change of the function over the given interval or intervals
    f(x)=X^3+1
    a. [2.3] b. [-1,1]
    >>Answer = 19, 1

    I'll be posting in here a lot throughout the day, so I'll be checking back often. As always, any help is greatly appreciated!
    Ah! Curvature gave it to me. He (I assume curvature is a he anyway) gave you the average of the function over an interval, but you are looking for the average of the rate of change. So we have for the first one:
    \text{Average of Rate of Change} = \frac{1}{3 - 2}\int_2^3 \frac{df}{dx}dx = \frac{1}{1}(f(3) - f(2)) = 19

    You do the second one.

    -Dan
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  4. #4
    Super Member angel.white's Avatar
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    That's a cute problem, I like how it works.
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Gentlemen, you are thinking too hard (wow, i almost never get to say that, i'm always the one doing things the hard way. i feel so cool now , i almost never get to use that emoticon either). The average rate of change of a function on an interval, is the slope of the secant line on that interval, which is simply \frac {f(b) - f(a)}{b - a} (if our interval is [a,b] and our function is f(x) of course), which ends up being the same as topsquark's answer...but without entering the realm of integrals. But i guess it's nice to see how it all works out in the end and we can use an integral to confirm our answer
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  6. #6
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    Yeah, I figure dit out shortly after posting this...it was just a mental lapse..
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