<p>vertical line</p>
<p>edit: or no line works too.</p>
<p>is integration by parts on the AB exam? :"D</p>
<p>is there an AIM chat room for AB</p>
<p>Question: Find the radius of convergence of Sum((x-3)^2n/n)?
How do you find it? Use the ratio test, and I get (x-3)^2 < 1, which doesn’t have a radius of 1… What did I do wrong? </p>
<p>It’s on the released practice multiple choice test-question 17.</p>
<p>Can we simply put the answer like the answers from collegeboard’s official answer? Like super concise but gives one line explanation + values?</p>
<p>Does anyone have any FRQ strategies for AB?</p>
<p>how do i use calculator to solve polars like -2= Θ+sin(2 Θ)?</p>
<p>There is always a rotated volume question on the free response.</p>
<p>So the question is what will it be this time?</p>
<p>Washer or shell method?</p>
<p>I am praying for washer although I just studied shell and it isn’t as nuanced.</p>
<p>I always have trouble recognizing the inverse trig forms inside the integral. Sometimes you need to do crazy things to get it to look like the inverse trig derivative…hopefully thats not a big part of it.</p>
<p>Rotation, hopefully with washer.</p>
<p>Particle problem, that’s always there too.</p>
<p>Rolle’s theorem with a chart? Maybe.</p>
<p>Interpreting graph of a derivative, that’s also a possibility.</p>
<p>@ Manatee. The only thing I can think of is that you pull the ^2 part out of the absolute value part so you have (abs(x-3))^2 < 1 (since both the absolute value and the square will make the function output positive, perhaps the order doesn’t matter?), then take the square root so you have Abs(x-3)<1, then just -1<x-3<1 so it has an interval of convergence between 2 and 4 with a radius of 1. But I’m not sure you can just pull that ^2 out of the absolute value convergence condition.</p>
<p>Don’t forget slope fields / differential equations.
High chance that one is on there.</p>
<p>Remind me what Rolle’s Thm, Squeeze Them and Intermediate Value Theorems are? (lol i only remember Mean Value)…</p>
<p>Rolle’s theorem just states that if there are 2 points on a continuously twice differentiable curve on the interval of [a,b] where the secant is 0, then there also is a point, point c, on the curve between a and b where the derivative of that point will have a slope of 0.</p>
<p>rolle’s theorem states that if there is an a and b and f(a) = f(b), then there is a c such that f’(c)=0.</p>
<p>Intermediate value theorem basically states that there is a y value between 2 different y values, such as (1,3) and (2,5) . There has to be a point that passes through (x,4)</p>
<p>Rolle’s Theorem is really just a special case of the Mean Value Theorem.</p>
<p>Ah yea yea, I remember now. I’ve been using them in problems lol, just didn’t refer to them by name. </p>
<p>And what about the Squeeze Theorem? Isn’t that for limits? (something like lim x>infinity, (sinx)/x)…</p>
<p>The Squeeze Theorem just says that if the lim x->c f(x) = L and lim x ->c g(x) = L, and f(x) <= h(x) <= g(x), then lim x->c h(x) = L also.</p>
<p>They use that one to prove that the lim x-> 0 (sin x)/x = 1 early in the course pre-L’Hopital, so that you can use that result to use the limit definition of the derivative to find the derivative of sin x.</p>
<p>will partial fractions be on the AB exam??
ie: integral of (5x-3/x^2-2x-3)dx</p>
<p>Not on AB for partial fractions.</p>
<p>Yes! Its all coming back now. Thanks a lot, MathProf.</p>
<p>Btw, the FRQs in the Barron’s practice tests are ridiculous…I’ll easily fail it if the problems are that level.</p>