<p>For #1-What made you say the graph was concave down? I found it to be concave up…It’s possible that I made an error but just curious what your reasoning was.</p>
<p>I just realized I never put the concave down thing on my test. Just the strictly increasing. Ignore the concave down part lol my mistake</p>
<p>crap, i think i messed up 4 a and d. i got like 23 and 46 something…not sure where i messed up though</p>
<p>a- y-15=8(x-1). Plug in 1.4 and you get 18.2</p>
<p>d- f(x)+ f’(x)(x-1) + f’'x(x-1)^2/2-> 15+8(x-1)+10(x-1)^2, plug in 1.4 and you get 19.8</p>
<p>On 1d)</p>
<p>I did the integral from 0 to 5 + 71. My thinking was that since the new function given represented the rate after t=20 of the original data, its x=0 value would be the rate of the new temperatures after t=20. The answer comes out 73.924 Anyone else do this?</p>
<p>But the equation tells you that it’s written in terms of t, and that t is from 20<=t<=25. So you’d be using 20 to 25, not 0 to 5.</p>
<p>How did you do 4b? I’m fairly positive I did this right on the exam but for some reason I keep getting a different answer when I try to solve it now. I don’t know how I could have forgotten riemann sum so quickly…</p>
<p>For 4b, I did midpoints of two, so 1.1 and 1.3. The distance between was .2, so f’(1.1)<em>.2+ f’(1.3)</em>.2 or 10(.2)+13(.2). That was the value of integral of f’(x).</p>
<p>By first fundamental theorem of calculus, we know that integral is equal to f(1.4)-f(1). So you move f(1) to the other side to find f(1.4). That would be 15+2+2.6 or 19.6.</p>
<p>Colleges cannot accept 4s from this exam type. The curve is just too massive. Instead, they should leave the upper half of fives, all of the lower fives should become fours, all fours should become threes, all threes should become twos, and all twos ones. 49% of people shouldn’t be getting 5s, therefore the cutoff for five is obviously too easy to achieve. Otherwise, colleges should not take fours.</p>
<p>Well, the standard for BC is much higher. The average person taking BC is much more qualified in math in general than in students taking AB. The purpose of the AP exams is not to show you against a qualified pool of applicants but the nation in general, which is why BC students perform so well. The curve is actually harsher than AB, but because BC students are much more qualified the mean scores tend to be higher.</p>
<p>Yup got every thing right on the free response except the error which I didn’t study since they had it for the past two years. And as long as I made no stupid mistakes on the MC I only missed 1 out of 45. My goal was to be investigated my CB but I guess I can’t anymore :(</p>
<p>and I’m not asian btw</p>
<p>For free response question 3d. How are there three inflection points. I only got two. at x=-2 , and 1. The sign of the slope does not change at x=-1, therefor there is no inflection point there.</p>
<p>yeah, i agree, that one, the 2nd deriv does not change signs at all.</p>
<p>No one said anything about -1. The other inflection point is 0.</p>
<p>Oops, you’re right. That’s not an inflection point since it’s concave up over it. It should just be -2 and 1 only</p>
<p>@Kickpuncher666
The reason for the high percentage of 5s on the BC test is not because the curve is too easy, but rather there is a large number of highly qualified students taking the test.</p>
<p>For 6a., is the interval of convergence -1<x<=1 or -1<=x<=1?</p>
<p>From checking the endpoints: x=1 produces a series that converges by alternating series test while x=-1 produces a divergent series by direct comparison to p-series.</p>
<p>Various AP Calculus teachers are posting their solutions to the FRQs on this forum here:
[Math</a> Forum Discussions - ap-calculus](<a href=“Classroom Resources - National Council of Teachers of Mathematics”>Classroom Resources - National Council of Teachers of Mathematics)</p>
<p>I got 3 inflection points for 3d); -2, 0, and 1. At -2 and 0 f’(x) and therefore g’’(x) are undefined and the slope of f changes signs. At 0, f’(x) = 0 = g’’(x) and changes signs from - to +</p>