00:01
So given the fact that we have, or if we want to find the cost function, given that the marginal cost is, so c prime of x is equal to x plus 1 over x squared, and that two units will cost $5 .50 to make.
00:25
Well, so even if we weren't told that our marginal cost was the derivative of the cost function, where we could.
00:34
Do is recall that what marginal means is a very small change of something.
00:43
So this is saying a very small change in the cost function and the way we look at very small changes in calculus is by the by the derivative.
01:07
So and the unit cost being or to produce two units costing by 50 will be, so this means if our cost function is c, and c of two should be $5 .56.
01:34
So if i want to find one possible c or one possible cost function, i can go ahead and integrate this first, or integrate our marginal cost.
01:45
So let's go ahead and write that now.
01:46
So i have c prime of x is equal to x plus.
01:53
So i can rewrite 1 over x squared as x to the negative second power, just because we are going to end up taking the integral of this, or antiderivative.
02:06
So i'll just go ahead and rewrite it like this first.
02:09
So when we do take that anti -derivative, it will be a little bit easier for us to do so.
02:14
So now i'm going to integrate each size.
02:18
Of this, so i'll need to move my equal sign over a little bit so i can squeeze in a dx there.
02:25
So i integrate each side with respect to x and now the anti -derivative of the derivative or the marginal cost will just be our cost function...