# Review Sheet

• Feb 12th 2010, 04:51 AM
MathmaticAvatar
Review Sheet
Hello! (Doh)
I am in dire need of help! I just recieved this Review Sheet that I completely do not understand.
I believe some of the stuff is from last year, Pre-Calc. and I'm completely blank as far as that goes. I'm currently a high school students taking the class for AP-credit and for some reason I can't remember anything from class and/or review books.

Please if you can, help me out by explaining what and how I need to do these problems.
I have another review sheet but I believe if I understand this one I will be able to do the other one on my own.

• Feb 12th 2010, 05:38 AM
Pulock2009
your 1st question involves an understanding of finding critical points of a function by finding its derivative. your second question simply involves the following points:acceleration is given by the derivative of velocity and the velocity is given by the derivative of the postion.
• Feb 12th 2010, 05:53 AM
skeeter
1. (a) use the facts that $f'(-1) = 0$ and $f''(-2) = 0$ to solve for $a$ and $b$.

(b) straight forward problem, once you find $a$ and $b$.

2 (a) what tells you the direction that a particle moves ?

(b) acceleration is the derivative of velocity

(c) $y(t) = \int v(t) \, dt$.

you'll have to use substitution to find the antiderivative and the given initial condition to find the constant of integration.

(d) total distance $= \int_0^2 |v(t)| \, dt$
• Feb 12th 2010, 06:38 AM
MathmaticAvatar
Quote:

Originally Posted by skeeter
1. (a) use the facts that $f'(-1) = 0$ and $f''(-2) = 0$ to solve for $a$ and $b$.

(b) straight forward problem, once you find $a$ and $b$.

2 (a) what tells you the direction that a particle moves ?

(b) acceleration is the derivative of velocity

(c) $y(t) = \int v(t) \, dt$.

you'll have to use substitution to find the antiderivative and the given initial condition to find the constant of integration.

(d) total distance $= \int_0^2 |v(t)| \, dt$

That the problem I do not understand how to use $f'(-1) = 0$ and $f''(-2) = 0$ to find critical points.

for 2 (a) It's veolcity. When the function is bellow the x-axis then it is moving to the left but when it's above then it's moving to the right. That I know.

Like I'm it's a bit fuzzy because we had this from way back in the year and now he, the teacher, expects us to remember everything. (Headbang)
• Feb 12th 2010, 07:12 AM
skeeter
Quote:

Originally Posted by MathmaticAvatar
That the problem I do not understand how to use $f'(-1) = 0$ and $f''(-2) = 0$ to find critical points.

find f'(x) , sub in -1 for x and set = 0 ... you'll get an equation in terms of a and b.

find f''(x) , sub in -2 for x and set = 0 ... you'll get an equation that you can solve for a

once you get a , go back and find b.

find f''(x), sub in -2 for

for 2 (a) It's veolcity. When the function is bellow the x-axis then it is moving to the left but when it's above then it's moving to the right. That I know.

ok ... what is the sign of v(1.5) ? how will that answer the question ?

...