<p>who's planning on getting 100% tomorrow?</p>
<p>meeeee!!!!!</p>
<p>take the integral remembering that it is an indefinite one so you must add c, then plug in for y and x and you get the equation because you now have c. </p>
<p>so the integral would be x^2 + 3x + c right so then plug in 1 for x, and 2 for y so you get 2 = 1^2 +3(1) +c then simply solve for c and you get -2.</p>
<p>how hard are the princeton review practice tests compared to the real deal?</p>
<p>eating food, overachiever:P
lol</p>
<p>hereicome09.</p>
<p>use the derivative.</p>
<p>S (2x+3) = x^2+3x+C</p>
<p>2=(1)^2+3(1)C
2=1+3+C
-2=C</p>
<p>equation:
y=y= x^2 + 3x -2</p>
<p>get it now?
yes? no?</p>
<p>lol :P</p>
<p>ok I'll solve this...
A differential function f has the property that f(5) =3 and f '(5)=4. What is the estimate for f (4.8) using the local linear approximation for f at x=5?
The tangent slope at (5,3) is 4. And now that we have a point and the slope we can create a linear equation. That is: y= 4x -17. Now plug in 4.8 for x to get 2.2.</p>
<p>oh wait</p>
<p>∫ƒ(x)dx !!!!!</p>
<p>no but seriously, i don't think it will be that bad</p>
<p>my class did basically every free response from 2007-1999 and all the release multiple choice... and the 2007A FR were ridiculously stupid (so i don't think they will be this year).</p>
<p>also, apparently, they're releasing the MC questions this year - so my teacher says that they'll be fair questions for the most part</p>
<p>Oh man, the 2007 FR killed me too; it was ridiculous! My teacher said it won't be as difficult this year either.</p>
<p>I have a stupid question. To find the derivative at a point on a TI-89, can't you just type d(function,x,x-coordinate)? Because every time I type in a value for the x-cooridinate, it keeps taking higher order derivatives. For example, if I wanted to find the derivative of y=x^2 at x=2, wouldn't I just type in d(x^2,x,2,)? But it keeps showing d^2/dx^2. And for x=3, it shows, d^3/dx^3. And so on. So what am I doing wrong?</p>
<p>wait, i usually:</p>
<p>1) plug in the equation
2) 2nd -> Trace (calc)
3) #6 (dy/dx)
4) type in the value i want to find the derivative at</p>
<p>I have a TI 89 and and you have most of it right. If you have y=x^2 and want to evaluate the derivative at two you would type:d(x^2,x)|x=2. You have to end the bracket and then add a vertical pipe stating the required x value.</p>
<p>Wow, this is pretty cool. (:! I have a general question:</p>
<p>I was wondering what d/dx means when its next to an expression? </p>
<p>EX:
d/dx (e^(sin2x)) =</p>
<p>a,b,c,d mult choice answers</p>
<p>We did dy/dx etc in class, but why the lone d?</p>
<p>I see. Thank you Sixthsense. It just seems that I used to be able to do it the way I previously spoke about without any problems...but this way works so I'll obviously have to use it.</p>
<p>simply put, ohlollipop, d/dx means to take the derivative.</p>
<p>so is Newton's method on the test for AB?</p>
<p>I dont think its on the AB test</p>
<p>No, Newton's method isn't on it.</p>
<p>Another stupid question: How would you find f '(1) if the tangent line to a function contains the points (1,7) and (-2,-2)?</p>
<p>Ahh, thanks hotjava xD I can't wait to see my score in summer. I'm rooting for a 1. :/</p>
<p>well if the tangent line contains the point (1,7) then f'(1) is 7 bc f'(x) is the tangent line. if the original equation contains those points then the equation is y-7=3(x-1) and then u take derivative of that and then plug in 1 and u get the answer</p>
<p>another stupid question...</p>
<p>i'm using that study guide that someone linked to a few pages back, and it says that the limit as x-->0 of (3x^2 + x)/2x is 3/2. I'm getting 1/2, so what am I doing wrong?</p>