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Math Help - average cost problem - simple, but having some trouble

  1. #1
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    average cost problem - simple, but having some trouble

    Hello everyone! I am having some trouble working out this question:

    If a marginal cost function nis c'(q)=6q^2+880 dollars per unit, then the cost to increase production from 6 units to 9 units is ...


    I did this question in 2 ways. One of which I may have found a fault in my steps.

    1) Take q=9 and q=6, sub into the marginal cost function and subtract q=9 from q=6. Personally, I don't see how that can give a reasonable answer, but it was worth a shot.

    2) Integrate (although this type of question was around before I learned it) and sub in q=6 and q=9 for c(q) and subtract.


    So I was wondering, if that is how you should go about doing the questions or what I should do.



    And on the lines of that, another question was bugging me.


    If for all real x we assume f(g(x)) = x and f'(x) = 1+ [f(x)]^2, then g'(0) equals .. I came up with the answer of: 1

    Thanks!
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  2. #2
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    Dear finalfantasy,

    If for all real x we assume f(g(x)) = x and f'(x) = 1+ [f(x)]^2, then g'(0) equals .. I came up with the answer of: 1
    What is the derivative of the function f(g(x)) at the point a?
    The chain rule says g'(a) * f'(g(a)).
    So g'(0) = 1 / (1 + g^2(0))
    If we knew g(0)=0 you would be right.
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  3. #3
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    increase in cost is

    C(9) - C(6) = \int_6^9 C'(q) \, dq
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  4. #4
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    Quote Originally Posted by skeeter View Post
    increase in cost is

    C(9) - C(6) = \int_6^9 C'(q) \, dq

    That's what I did, but how do you know what the C is, or in other words, fixed cost?


    I got $3666 by using FTC BUT that's assuming FC were 0.
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  5. #5
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    C'(q) = 6q^2 + 880

    antiderivative ...

    C(q) = 2q^3 + 880q + k , where k = fixed cost

    C(9) - C(6) = [2(9)^3 + 880(9) + k] - [2(6)^3 + 880(6) + k]

    what happens to the fixed cost, k ?

    doesn't matter what the fixed cost is, does it?
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