<p>the free response questions are up!</p>
<p>I’m working out the solutions now; am I allowed to post what I come up with so that we can come to a consensus, or do we have to wait?</p>
<p>You can post them! People did for the other ap’s I’ve taken</p>
<p>What did you guys put for 3D, 4B, and 6C?</p>
<p>For 3D I put 6.4/e^2 ounces per minute. For 4B I said x=0.</p>
<p>For 6C (h prime of x = f(2x)) you had to differentiate with the chain rule. h double prime of x becomes 2 f prime of 2x, etc. I wound up with something on the order of 7 - 4x + 2x^2 - (4/9)x^3 if I remember correctly.</p>
<p>@ACT 4B said to determine the absolute minimum value, not where it occurs. That is true, it does occur at x=0, but the actual y-coordinate is -8.</p>
<p>Also for 6C, what I did was first I took the Taylor of f(x) about x=0, then I substituted ‘2x’ for every ‘x’, and then I integrated that to get the h(x) Taylor. Then +c came out to be +7, so the answer was just whatever you got from the integration + 7.</p>
<p>Oops, I read the question too quickly. Good catch, thanks.</p>
<p>EDIT: I like your way of doing 6C. I think they should produce the same result, but I’ll double check quick.</p>
<p>@ACT yeah I was unsure why you differentiated, made no sense to me o.O still cant figure it out. We’re trying to find h(x), not h"(x). I’m not 100% sure why you differentiated tbh. And maybe your answer is correct, I don’t remember either because I left the + 7 at the end, and I can’t really remember my answer clearly, only if it matches someone else’s will I remember haha.</p>
<p>Okay for the series problem, I did my pemdas wrong (I know laugh, I was under pressure) and got that f’(0) = -2.</p>
<p>I just worked with it for the rest of the problem, but I did multiple each term by 2 and then integrated for h(x). How many points will I lose for that minor error? 2 or 3 at most? I really want a 5 for BC. I thought all the AB ones were cake.</p>
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<p>They give you h(0) = 7, so your +7 should be legit. Then I did h’(0) = f(2(0)) = f(0) = -2. I think I followed up with h"(0) = 2f’(2(0)) = 2f’(0) = 4, etc.</p>
<p>It seemed like a good idea at the time. XD I wish the scoring would come out sooner.</p>
<p>@ACT, did you get the same answer as I did? (Assuming you followed my train of thought correctly)</p>
<p>You don’t just multiply the terms by two, the two needs to be plugged in and put to whatever power x is at.</p>
<p>Hahah yeah, thats what I did! Im so stupid, integrating was so much easier, Im so sure I made a math error. PLus you had to do the chain rule and it just got so complicated</p>
<p>There was a chain rule in the series? I have never seen chain rule on series problems!</p>
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<p>Well, the way I did the problem, you had to find each h^n(x). So I had to use the chain rule and product rule to find h’’(x) and h’’’(x). Simply integrating would have been much easier.</p>
<p>Oh so integrating for me was fine? Haha. I feel so dumb for only multiplying by 2 instead of 2^n :(. I shoulda remembered you plug in 2x to the x value instead of just multiplying by 2 :/.</p>
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<p>Yeah, you had to plug in 2^n. killerskullz, I just did the problem a couple times to be sure and I think we have the same answer 
your way was way easier, though. I’m lucky I didn’t mess up in the middle of mine.</p>
<p>@ACT haha great to hear!</p>
<p>But that means I will have only messed up the third degree term by a coefficient of two :/</p>
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