2/19/11

So this week we continued with practice AP's; I'll just give a few example problems like those on the multiple choice section..

Ex. 1) If f(x) = -x^3 + x + 1/x, then f'(-1) =
*So first let's determine what we're given in this problem.
*They give you the original equation, and they're asking for f'(-1)..which means they want you to take the derivative and then plug in -1 for x after.
*So first let's take the derivative and you should have this:
-3x^2 + 1 - 1/x^2
*Now all you do is plug in -1 for x:
-3(-1)^2 + 1 - 1/(-1)^2
= -3

Ex. 2) d/dx cos^2(x^3)
*So when you first take a look at this problem, you see d/dx, which means all you have to do is differentiate it
*So since this is a chain rule, first you have to derive the outside, cos^2, so you should first get this:
2(cos(x^3))
*Buttt, you're not done yet; next you have to derive cos and then recopy the rest of the equation. So now you should have this:
2(cos(x^3)) * sin(x^3)
*Last, multiply that by the derivative of the inside, x^3, which is 3x^2
*So now you have 2(cos(x^3)) * sin(x^3) * 3x^2
*Now simplifying all of that, your final answer should be:
6x^2cos(x^3)sin(x^3)

Ex. 3) The area of the region enclosed by the graph of y = x^2 + 1 and the line y = 5 is....
*Alright they're asking for AREA so that means you're going to use some kind of INTEGRAL..
*First, let's figure out what our graphs look like...y=x^2+1 is a parabola with vertex (0,1) and y=5 is a horizontal line at the point (0,5)
*So looking at the two graphs on the same coordinate plane, the line y=5 is on the top and the parabola is on the bottom, and you're looking for the area in between.
So you're going to have to use the formula: S top-bottom
*Plugging into the formula you get:
S 5-x^2-1
= S 4-x^2
Now integrating that^ you get:
4x - 1/3x^3....but we still have to find the bounds!
So to do that all you do is take the two equations and set them equal and solve for x:
x^2 + 1 = 5
x^2 = 4
x = -2,2 ...and those are your bounds-- -2 being the lower bound and 2 being the upper bound
*So now you have this:
[4x - 1/3x^3] I-2,2 ...the "I" being the bar (meaning you didn't plug in yet)
*Now Fundamental Theorem of Calculus!
4(2)-1/3(2)^3-[4(-2)-1/3(-2)^3]
Simplifying that you should end up with 32/3 and that's your area