<p>And I won't even see my score until August 3rd. I think I'll cry. The suspense is just killing me.</p>
<p>B version as in the make- up version?!!?!...</p>
<p>What?! Isn't that supposed to be HARDER...ARG again</p>
<p>Yah, I would love the 2003 version...then I would feel actually confident like dragoneye1589</p>
<p>I don't think any of our futures are bright...</p>
<p>URG...why didn't I have a seizure during the free response section?!</p>
<p>What's really weird is that the questions make since now that I'm shot my test to h-e-double hocky sticks.</p>
<p>sense**
I can't spell</p>
<p>Ha...I'm going to be sent to "h-e-double hocky sticks" by my mom if that score sheet comes and the number is anything less than a 5 -__-....</p>
<p>But yah seriously...the questions would be so easy if we were given sometime to think and chill out after the initial panic...</p>
<p>Some time**
I can't spell either :]</p>
<p>Or rather, I didn't go far enough. I found that x = e^4, but the question asks you to solve for k. Since k rt(x) = ln x, k rt (e^4) = ln (e^4), so ke^2 = 4, and k = 4/e^2.</p>
<p>Which is the same value as that on the link.</p>
<p>Oops. It helps to read the directions.</p>
<p>AB6
a) f'(x) = (k sqr(x) -2)/(2x) f''(x)=(4 - k sqr(x))/(4x^2)</p>
<p>i'm confused I got</p>
<p>(k/2sqr(x)) - (1/x) for f'</p>
<p>I'm pretty sure finding a common denominator will not produce that answer...</p>
<p>for
k(sqr(x)) - lnx</p>
<p>because </p>
<p>(k(x^-.5)/2) -(1/x)</p>
<p>k/(2sqr(x)) - 1/x is correct for f '.</p>
<p>When you multiply the first fraction's numerator and denominator by sqr(x) and the second fraction's numerator and denominator by 2, you get:</p>
<p>k(sqr(x))/(2x) - 2/(2x) = (ksqr(x) - 2)/(2x)</p>
<p>Though either answer would be accepted for f '.</p>
<p>what you guys get for 6c?</p>
<p>yay!</p>
<p>That feels nice</p>
<p>I didn't I jsut set the second derivative equal to zero and moved on...</p>
<p>It looked to complicated... =(</p>
<p>I believe math prof has answered your question o here already.</p>
<p>Indeed. For 6c, k = 4/e^2.</p>
<p>oppsie too*</p>
<p>Wait a sec...
for 3d... i got the right answer, but now I'm confused...</p>
<p>shouldn't the inverse be </p>
<p>1 / inverse (g'(x))</p>
<p>at x= 2, g(x) = 3 but g'(x) = 1 </p>
<p>so shouldn't the slope of the inverse be 1/1</p>
<p>so the equation should be </p>
<p>(y-3) = 1(x-2)</p>
<p>can someone explain 3d cuz im so lost</p>
<p>Not g'(x) = 1, but g'(1), which is 5. :)</p>