RULES MUST BE FOLLOWED:

Please...
Do PROBLEMS on a SEPARATE PIECE OF PAPER AND SHOW ALL WORK THEN TAKE PICTURE AND UPLOAD IT HERE! (Please be neat and legible).
In order for you to get 100fg. First one to answer a problem gets the fg (if correct) subsequent posts will not be awarded.

or u can feel free and help out of kindness of your heart and just post

please upload by 2/18/2015 12am PST.

200 fg please.


#5
a) Horizontal Asymptote: y=2 (just take highest power term on numerator and divide it with the x^(2) from denominator, i.e., 2x^(2)/x^(2) = 2 )

Vertical Asymptote: x=3, x=-1 (set denom=0 and solve for x)

need to use L'hopital's rule here (take deriv of numerator and denom until you get a limit you can take:

deriv of numerator: 4x
deriv of denominator: 2x-2

lim x ->3 of 4x/(2x-2) = 3

can't be differenciated at -1 because it is discontinous

no solution at x=-4, could be an equation of type f(x)=x*(x+4) /(x+4) so it is not defined at x=-4 but still is a straight line curve

the function at x=3 has to be infinity nomatter which way it comes from, so it must look as such

This post was edited by TheStealthTarget on Feb 17 2015 10:36am

part 6 A) we can simplify the k(x) to equal 2/x^(1/2) + 6*x^(3/2)

now we can go ahead and do are integration on each term

f(x)=2/x^(1/2) so f'(x)=- 1/x^(3/2)

f(x)=6*x^(3/2) so f'(x)=9*x^(1/2)

so k'(x)=9*x^(1/2) - 1/x^(3/2)

the slope of the line at x=4 is k'(4)= *4^(1/2) - 1/4^(3/2)=143/8=m

now we need the point it goes through

k(4)= 2/4^(1/2) + 6*4^(3/2)=49

the point is (4,49) = (x1,y1)

now use point slope form

y - y1 = m(x - x1)

which is y=143/8*(x-4)+49

simplified the equation is

y=143/8*x - 45/2

the graph of both are below


This post was edited by TheStealthTarget on Feb 17 2015 10:58am

#1

http://postimg.org/image/f35xc0yvx/

gold has been distributed thanks for ur help guys.. but shin, stealth beat u to #1 so i gave u 50g anyways.

Could care less about the fg but he never posted a solution to 1 unless he pmed you the graph in his last post was not for problem 1

quoted teh solution posted above

Yeah I was just trying to help him, for some reason link wasn't showing on iPhone but it does in iPad ... Odd