# Thread: average cost problem - simple, but having some trouble

1. ## 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!

2. 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.

3. increase in cost is

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

4. Originally Posted by skeeter
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.

5. $C'(q) = 6q^2 + 880$

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?