<p>Hmm… 3 is bit weird. I got same answer for two different methods but it still feels “wrong.” I even know what I had to do but eh…</p>
<ol>
<li>0</li>
</ol>
<ol>
<li>Killed it with calculator and almost got it wrong (cuz I put it in t-3 for function haha). </li>
</ol>
<p>Answer should be -8.</p>
<ol>
<li>I’m guessing left hand limit = approaching from -infinity to 0, right? If that’s the case, it should be -1.</li>
</ol>
<ol>
<li>This one was intriguing. Two of the choices actually wondered me so I actually worked this problem out (one choice is evidently… right or wrong 
I’m not telling it).</li>
</ol>
<p>My answer is B.</p>
<h1>7 has really bizarre diagram. I honestly don’t feel like doing it. Let someone have this point.</h1>
<p>on #7 part a), there are no POIs for f because concavity on the original graph is the inc/dec of the first derivative, and since the first derivative is increasing from (-infinite, 2) and neither increasing or decreasing from (2, infinite), the original graph is concave up from (-infinite, 2) and neither confave up or concave down from (2, infinite), leaving no POIs.</p>
<p>i hope that explanation makes sense and i hope the answer is correct. its a tricky question.</p>
<p>maybe i’ll attempt part b) later</p>
<p>Leaderboard and the next set of questions will be up at 9 PM EST.</p>
<p>Point Winners For Each Question:
1)grayfalcon89 (+1)
2)grayfalcon89 (+1)
3)grayfalcon89 (+2) [sorry yillfog, question asked for slope, not the equation of the horizonal tangent]
4)yillfog (+2)
5)grayfalcon89 (+2)
6)grayfalcon89 (+3)
7)yillfog (+3)</p>
<p>Question Commentary
-The multiple choice question number 6 could have been worked out using two integrals, but logic would have also sufficed. Notice that f(x) = 9 - x^2 is a downward facing parabola symmetric about the y-axis, so x=0 would split f into two equal areas. Then, notice that when dy slices are taken, the bulk of the area rests towards the bottom of the parabola. Therefore, the horizontal line dividing f into two regions of equal area should at least be below y=9/2. Neither II nor III lie below that line, therefore, we can eliminate those two choices and conclude that x = 0 is the only one.</p>
<p>-FR #7 is very tricky indeed. Yillfog pretty much got the geist of part a correct; when you sketch out f, (-inf,2) is an upwards (or concave up) parabola and (2, +inf) is a line. Lines do not express concavity, therefore, (2,5), the only possible inflection point, is not actually a point of inflection. As for part b, you must know what the graph of f looks like, write two separate equations, and set up two separate integrals to find the area under the curve. I will say that the two equations are y = (x-1)^2 + 4 for the parabola on the interval (0,2), and the equation for the line on interval (2,6) is y=3x-1. Integrate with respect to x in both cases, and the area of the first region is 26/3 square units and the area of the second region is 44 square units, making the total area 158/3 square units. If anybody wants me to explain how I drew f based on the f ’ graph, feel free to ask and I’ll go further into that one.</p>
<p>Overall Leaderboard
- grayfalcon89 (19.5 pts)
- mathwiz90 (10.5 pts)
- yillfog (5 pts)</p>
<p>Leaderboard Commentary
I see a lot of views on this topic, meaning a lot of you are stopping by, but only a couple of you are actually trying these! If more people did these, the competitive spirit would be higher and this would be more exciting (not that calculus isn’t already exciting enough, heh). Will grayfalcon keep the lead, or will another person overtake him?</p>
<p> Question Set 3: Due Sunday 10:30 PM EST </p>
<p>Remember, only answer one question per post.</p>
<p>1) T/F: If f(x) = sin^2 (2x), then f '(x) = 2(sin2x)(cos2x).</p>
<p>2) T/F: The integral of x^ -2 given the lower bound -1 and upper bound 1 is -2.</p>
<p>3) Concpt: Solve the differential equation: y’ = x(1+y)</p>
<p>4) Concpt: Can Rolle’s Theorem be applied to f(x) = 1 – absval(x-1) on interval [0,2]? Why or why not?</p>
<p>5) Concpt: Find the area bounded by the graphs of f(x) = x^2 and g(x) = x+2.</p>
<p>6) MC: Which equations below represent graphs that have negative slopes at every point?
I.y-2 = -2(x+4)’
II.-x^3
III.-absval(x-2)
A) I Only
B) II Only
C) I and II Only
D) II and III Only
E) I, II, and III</p>
<p>7) FR: Let f(x) = c/x + x^2, where c is a nonzero constant.
a) Determine all values of the constant c such that f has a relative minimum, but no relative maximum.
b) Discuss the continuity of f.</p>
<ol>
<li>No</li>
</ol>
<p>Since the question asked “why or why not,” I’ll take this as “justify your response” question.</p>
<p>Rolle’s Theorem consists of three parts. The say:</p>
<ol>
<li>f is continuous on a closed interval [a,b]</li>
<li>f is differentiable on the open interval (a,b)</li>
<li>f(a) = f(b) = 0
then there exists a number c in (a,b) such that f’(c) = 0.</li>
</ol>
<p>Graph of f(x) = 1-|x-1| (I think it’s just easier to write this way than writing out absval :-D) IS continuous from negative infinity to positive infinity so #1 rule is good. f(0) = 1-|0-1| = 0 = 1-|2-1| = f(2) so #3 rule is good. BUT, #2 is NOT good because at x = 1, we get a sharp point since this is an absolute value graph. So, f is indifferentiable at x = 1 so Rolle’s Theorem cannot be applied. QED.</p>
<ol>
<li>y = C(e^(1/2*x^2) - 1</li>
</ol>
<p>Note that C indicates constant.</p>
<ol>
<li>Very neat. I’m pretty sure people can get confused on this A LOT. False.</li>
</ol>
<ol>
<li>True</li>
</ol>
<p>I’m curious about this question though. Why does when I punch this in calculator, I get error: Divide by zero?</p>
<ol>
<li>4.5</li>
</ol>
<p>Go calculator! HAHAHA…</p>
<h1>2 False,</h1>
<p>doing it algebraically i got 2, not “-2”</p>
<p>Hmm… I must’ve gotten screwed up in #2. Care to show your work? Here’s mine:</p>
<p>integral of x^-2:</p>
<p>-1*1/x] from -1 to 1 = -1[1/1-(1/-1)] = -1[1+1] = -2</p>
<p>Anyway, for #6, I get A. Not sure what I’s function is supposed to be though. It says y-2 = -2(x+4)’ (what’s (x+4)'? x+4 prime?)</p>
<p>It prob is a typo, but it has to be A. I quit this bc when I see the questions, all have bene answered already.</p>
<p>for #5, i got 15/2
i should double check that, hm</p>
<p>I got 9/2 for 5, though im prob wrong</p>
<p>((Grey Falcon, u forgot a parenthesis on the Diffeq, but it doesnt make a difference because C will be C no matter what)).</p>
<p>aw dam, in my head i knew what to write on #5, but i accidently wrote the integral as (x^2)-(x+2) instead of (x+2)-(x^2)</p>
<p>i’m so dumb. nice call on that one grayfalcon and mathwiz</p>