<p>If I look above the window, I'm not going to find anything...b/c that equation has an asymptote at x=0 and y=0.</p>
<p>Maybe I'm not seeing it correctly, but prove me wrong if you can. I didn't know how to respond to this question...so I just used +-3, but setting the equation = to each other..and then got like -37 something for the area, which didn't make too much sense...but oh well...</p>
<p>so someone confirm this. the area that we were supposed to find was ABOVE the line y=2 and BELOW the graph of the function?</p>
<p>no it doesnt. It has no asymptote at all.
Listen im 100% sure.
I even checked it afterwards with an 89 and i got it right.
x^2+1 is the denomintaor (not x^2)</p>
<ol>
<li>Screwed the whole thing up by shading the wrong region...</li>
<li>
a. I integrated the formula from zero to seven to find accumulated change, but I think I messed it up because I forgot about the g stuff.
b. I didn't know what to do for this one. I just bsd the answer
c. I didn't know either...I just said at seven...sigh</li>
<li> the f(g(x)) didn't work for me...I just said put hat the intermediate value theorem proves the point's existence, and then I said that the mean value theorem for h'(-5). I tried to work it out mathematically, but I couldn't do it...</li>
<li>(e^-t)sin(t) I set the derivative equal to zero and found a minimum...I forgot what point I said...anyone?
b. WTH?</li>
<li>a. I said f(x) + f'(x)(delta x) and got like 30.4
b. implicit differentiation
c. I found (b-a/n) and then I quit because I was too frustrated, but I think I now know what I should have done
d. I said less because it is an under approximation</li>
<li>I did as much as I could. I couldn't solve c though, so I just set it up</li>
</ol>
<p>I'm sure none of you screwed up as bad as I did...
SIGH
Do you think I'll be ok if I did well on multiple choice? I was so nervous on FRQ</p>
<p>i also used an 89...</p>
<p>assume y=0
then 0 = 20/(1+x^2)
cross multiply and you get 0 = 20, which isn't true...and is an asymptote</p>
<p>please show me a picture of where the graph ends...gimme something.</p>
<p>I was looking forward to a question like 5 and 6 on the form b exam...GRRRRR</p>
<p>@ G Bush:</p>
<p>I am not sure why you are assuming y=0 in the first place, that seems like it came from nowhere. It is not an asymptote as it approaches 0, a simple substitution of 0 for X shows that y approaches 20 as x approaches 0.</p>
<p>George thats wrong. Look at this. You said you were looking for an asymptote when y = 0. THIS IS A HORIZONTAL LINE. Not vertical. What this means is that the HORIZONTAL asym. will go to 0. NOT vertical.</p>
<p>I thought 20/(1+x^2) didn't have a vertical asymptote because the denominator can never equal zero. It shouldn't have a horizontal asymptote because it is in the form of the p-series (1/n^2, which converges definitely when the exponent is greater than 1), right? I'm probably explaining it all incorrectly since math obviously isn't my forte, but I graphed the function a billion times and it definitely doesn't have any asymptotes.</p>
<p>george: it peaks at 20. it can't be indefinite b/c the bottom can't be negative</p>
<p>what's the answer?</p>
<p>i just didn't know what the heck to do...so i just did this
A = S(20/(1+x^2) -2,x,-3,3) = -37 something</p>
<p>is that right?</p>
<p>Also here is the derivative for number 4. it was long.
E^-x sin X
F`(X) = E ^ -x cos x - E^-x sin x.
Now since F(X) = E^-x sin x and F(X) = what I just said above you will end up with F(X) + F(X) as E^-x cos x. Just remember this for later.
Now do the D^2y / DX ^ 2. for E^-x cos x - E^-xsinx
Don't forget. You have an A in front so..
A (-2E^-xcosx) + E^-x cosx (remeber this if F(x) + F`(X) = 0
A = -E^-x cos x / (-2E^-x cos x).</p>
<p>So A is simply 1/2. This is how i did it.</p>
<p>if you plug that exact equation into your calculator, you get +37 not -37...</p>
<p>George just set your Y max to like 50 and you will see that it intersects.</p>
<p>IT DOES have a horiz. asymptote which is y = 0.</p>
<p>You had the answer right. Type it AGAIn in your calc. You should get a + 37 something. NOT negative. You must of done 3, -3 on the test...</p>
<p>i got it...it's all good.</p>
<p>i just went with my gut though...i thought something was fishy...i didn't set my window range correctly, but i went with my gut...and put down something</p>
<p>turns out it was right (except i threw in a negative there...probably did what gyros suggested)</p>
<p>(this is for ab right?)...im getting frustrated noww i got like 12 for the area of "R"...between y=2 and whatever the other y equaled...</p>
<p>n for the 2nd one juss integrated that answer squared times pi for revolution of x...</p>
<p>So for the second one you did</p>
<p>PI * FINT((20/(1+X^2)^2 - 4), X, -3, 3) and didn't get the sAme answer?</p>
<p>Alright for 1c. isn't the region R the radius? How the hell is it the diameter? I thought that at first but then it didn't really give a hint to whether it was the diameter.</p>
<p>ok i got that 37 first..but then when i did it with my calc it got 12 n then somehow my calculations gave me 12..i think u guys are getting negatives because u putting one curve on top of the other which u shouldnt..cuz u put the curve closest to the top first when u integrate...</p>
<p>but yeahh except for that..in 2a. i got like 8000 something..i didnt..or dont think u have to subtract 5000..do u?</p>