I have a list of questions that I dont understand. I can provide the study guide and answers. Willing to pay for help, we can negotiate a price before hand if you wish.
Calc 1 college level. I have the final on Monday, so any time after that this thread can be closed.

Thanks,
~Lee

i can do this. pm me

post the questions here, lot's of people can help with that

this, don't let jdota lead you astray

Study guide:
http://math.arizona.edu/~calc/studyguides/m124SGFa13.pdf

#s I need help with so far:
14, parts b-f
17
20
23
26
29
35
38
41
44
47
53, parts b-d
57
60
66
72
78

Would prefer people refrain from mocking me for asking for help on a lot of questions lol. In my defense, I have a vague understanding of how to do them, I would just rather have a clear understanding.

e/

I haven't finished the study guide yet, so there will be more questions to come.
Also, appreciate the fast replies

This post was edited by smoothy9 on Dec 13 2013 11:22pm

41:
a - the derivative it the straight line, not the curved one
b - does the picture represent the domain and range of the function f(x) completely ?

if not, g(x) will grow towards it's maximum when x goes towards -infinity

Many thanks to Dota, provided amazing help.

Updated questions I need help with:
26
29
35
38
41 part b (forgot to specify I already got a)
44
47
53, parts b-d
57
60
66
72
78

60)
Well that is an easy integral. Remember that the derivative of tan(x) is sec^2(x); therefore the integral of sec^2(x) is tan(x)

so the integral evaluates to 3*tan(pi/4) which is just 3

leaving us 1/(pi/4) * 3 or:
12/pi

38)
The area of the triangle is f(x) = (1/2)*b*h

the height is the value of the function given: sqrt{4-(x-2)^2}

the base is just x

so the area of the triangle is f(x) = (1`/2) * x * sqrt{4-(x-2)^2}

f'(x) = [sqrt{-x*(x-4}]/[2] - [x*(x-2)]/[2*sqrt{-x*(x-4}]

set f'(x) = 0 and solve for x you will find x = 3
plug that back into the original function and you find y = sqrt(3)
so:
{3,sqrt(3)}
normally you would want to test the end points as well, but since we have the picture it is easy to see the area is 0 when x = 0 or x=4; so the point above is the maximum.

29)
f'(t) = [-B*(a*t^2-1)]/[(A*t^2+1)^2]

for f(t) to have a critical point at {4,1}, f'(t) must equal 0 when t = 4, therefore:

-B*(A*4^2-1)=0
rewritten:
-B*(16A-1)=0

Therefore B = 0 or 16A-1 = 0

however B can not be zero, since if B was 0, f(t) would always evalutate to 0, and we need it to evaluate to 1 when t = 4, so we can discard that solution, leaving:

16A-1 = 0

A = 1/16

lets plug that into f(4) and cook it up so f(4) = 1

(B*4)/(1+(1*16)*4^2) = 1
(B*4)/2 = 1
4B = 2
B = 1/2