<p>I wrote “on the interval (0,1) and (3,4)”</p>
<p>I didn’t put the union sign do u think it’ll be fine?</p>
<p>I didn’t put the unison either I think it should be okay :)</p>
<p>My opinion of the exam was much different than most everyone else’s… I had a lot of trouble with the non-calculator multiple choice, like only got 12/28 for certain, and then 3 or so “maybes”, and the rest I just quickly bubbled in randoms at the end… So I’ll probably end up with 15/28 or something… On the calculator section, I think I aced it, like 17/17… And the FRQ I pretty much aced as well I’m pretty sure, with the exception of the last part of the 2nd question, in which I didn’t have time to use my calculator on, but went back at the end and wrote out the equations necessary, and a complete explanation of the mechanics behind the problem, so probably like 8/9 on that one. Hoping for a 5, but wouldn’t be too too mad about a 4.</p>
<p>What’s the answer for FRQ 4 absolute minimum value of f? I got 0… but i’m not sure</p>
<p>The absolute minimum occurred at x=0, where the value of the function was -8.</p>
<p>MVT in the FRQ was used today correct? </p>
<p>for 6b, what did you guys get as c?</p>
<p>f is concave down and increasing when f’ is positive and f’’ is negative, so like in (0,1) and (3,4)</p>
<p>c=2 10char</p>
<p>oh man i got c wrong -_-
we get points for the process right even though we screwed up the last step to get c?</p>
<p>i got c=1 but i am not perfect by any stretch of the imagination…and yes mean value theorem for derivatives is what it was for that one part</p>
<p>oh no… i thought c=2 but now i’m scared that i might get that wrong.</p>
<p>what was the equation for that part?</p>
<p>y=-ln(something-1)</p>
<p>collegedreams, that’s what i got! i screwed up part b though. oh well, at least i separated the variables lol</p>
<p>BTW are you guys (with MVT) talking about the one where it was asking if the graph’s derivative crossed two? for that one, i found the slopes of the two points sandwiching that point, and explained how the slopes went from like 3.5 to 1.5 or something, and since the graph is continuous it has to pass through 2…any points?</p>
<p>i got c=1. my equation turned out to be y=-ln(x^3-3x^2-1)</p>
<p>i believe i got the same thing siddg, and i remember when i plugged in 1 and 0 it worked out too</p>
<p>what really sucks is that, at my school, there are two or three people, including myself, that really know what is going on. So when we can’t agree on an answer, it is nerve racking…not to be arrogant, but i was shooting for a perfect score going in…</p>
<p>i think i sorta said that but just didn’t explicitly say MVT?? cuz i said the slopes and it was continuous soi said it was diff. and continuous? FGETARta ugh.</p>
<p>siddg…so close…but if u got c=1 then it would be a positive number instead of a minus one…</p>