<p>Hmmm, now I'm not sure lol</p>
<p>(2,3) is the correct interval for 5(b).</p>
<p>intellec7, the initial condition of (0,8) helps you to determine which one of those solutions is correct.</p>
<p>Although since you already know y=8 is part of the range, it would have to be (6,8], rather than (6,8).</p>
<p>you only need about a 65% to get a 5 on the AP exam</p>
<p>"^5b has to be never....how is it 2,3...f'(x)<0 and f''(x)>0, are never met simaltaneousy in any one given interval so...."</p>
<p>Yes they are, just take a look at 1/x on (0, oo)</p>
<p>And while we're on this topic, here's the work for that question:</p>
<p>Sorry for the messiness, writing with a mouse is pretty hard.</p>
<p>Anyways, you can see from the work that f'(x) is negative on (-oo, 3) and f''(x) is positive on (2, oo)</p>
<p>^ i had all that same work, but then I was being stupid, and I wrote that it never happens because it not in the same interval...do you think I'll get partial credit for the same work as you have above..(like maybe a 1 out of 2 points or something :) ?</p>
<p>On 3 (a) and 3 (b), I got 80+128(x-2) and 80+128(x-2)+488/6(x-2)^2+448/18(x-2)^3, and my answers were 67.8 and 68.677, did I mess up somewhere?</p>
<p>^you're probably fine dude :), just a minor rounding thing that's all!!!</p>
<p>How is 4b) 3 times?</p>
<p>diamondbacker, since the initial position x(0) is -2, x(3) = -10, x(5) = -7, x(6) = -9, and the Intermediate Value Theorem guarantees three solutions on the intervals, one on each of (0, 3), (3, 5), and (5, 6).</p>
<p>With the initial condition t=0, x=-2
at 3, t=-10 (passes 8 once)
at 5, t=-7 (passes 8 again)
at 6, t=-9 (passes 8 a third time)</p>
<p>Oh, man! I knew how to do that one, but I wrote down the wrong answer because midway through the problem I thought it asked for how many times the position as -6!</p>
<p>Oh well, I did use the IVT, so I can get partial credit.</p>
<p>On #1, I plugged in 1 and 2 as my bounds instead of 0 and 2, will I get any partial credit for 1a?</p>
<p>The way I'm seeing the points, you'd probably get 1 out of 2 for the question, but you'd probably lose another point if you never saw the bounds as being 0 and 2, as I'm projecting that seeing it in either (a), (c), or (d) is probably worth another point.</p>
<p>Ok, thank you</p>
<p>Quick Question: I read somewhere that showing a table for Euler's method is not "sufficient work" So I also defined what each of the columns in my table where like y<em>new = y</em>old + delta<em>y
and delta</em>y = dy/dx * delta_x</p>
<p>will I lose points for not showing work?</p>
<p>^you shouldn't because you clearly labeled each column but, if the grader is in a ****y mood then maybe (1 point tops though :))</p>
<p>i swear that the particle passes -8 three times. Once it each region.
Also for the taylor series for the differential at the end i got 8-2+1.25=6.25
You guys think i will get docked for such a dumb mistake?</p>
<p>would a composite score of 60 be a 4 even if this year's curve is the worst that it could possibly be?</p>