Thread: Calculus Help Needed please

01252002, 05:51 AM #1
Calculus Help Needed please
A friend of mine needs help on a take home calculus assignment, if you can help answer, please do, there are four questions....
find derivative:
3x^2  square root x + 1/(x1)
determine values of a and b so that this function is
continuous:
1 if x < 1
ax^2 + bx if 1 < or = x < 1
3x  1 x> or =to 1
Use intermediate value theorem to prove this has at
least 1 root.:
e^x + cos(x)  2x  4 = 0
Limit:
as x>0 (1cos6x)/x^2
as x>pie/3 (sin(x)  sin(pie/3))/(x(pie/3))
Thanks

01252002, 06:40 AM #2VET
 Join Date
 Sep 2001
 Posts
 7,435
if i hadn't dropped out of school 2 years ago i might still remember that shit. although i should b/c my mother's a math teacher.

01252002, 06:40 AM #3
Your posts sometimes shock me, but this one takes the cake sweetie
A double bump for TV.......

01252002, 07:31 AM #4
OMG....I feel like I need to go back to school....
Someone on here will know...lots of college people.

01252002, 09:41 AM #5
Re: Calculus Help Needed please
Originally posted by tripleV
A friend of mine needs help on a take home calculus assignment, if you can help answer, please do, there are four questions....
find derivative:
3x^2  square root x + 1/(x1)
f(x) = 3x^2  root x + 1/(x1)
f'(x) = 6x  1/(2 times root x)  1/(x^22x+2)

01252002, 09:47 AM #6
Re: Calculus Help Needed please
Originally posted by tripleV
determine values of a and b so that this function is
continuous:
1 if x < 1
ax^2 + bx if 1 < or = x < 1
3x  1 x> or =to 1
a(1)^2 + b(1) = 1 (A)
a(1)^2 + b(1) = 3(1)  1 (B)
from (A) a  b = 1 (C)
from (B) a + b = 2 (D)
(C)(D) 2b = 3
b = 1.5
sub into (C) a  1.5 = 1
a = 0.5
Answer:
a = 0.5, b = 1.5

01252002, 09:47 AM #7
Gotta go to the gym. I'll check out the last 2 later.

01252002, 10:01 AM #8
Ugh  ya beat me to the first 2 !!
here is #3 :
*****Use intermediate value theorem to prove this has at
least 1 root.:
e^x + cos(x)  2x  4 = 0 *****
The proof is x = 0 y = 2
x = 2pi y = +
therefore there is a root.
I will let someone else get #4...

01252002, 10:14 AM #9
Re: Re: Calculus Help Needed please
Originally posted by bad_man
f(x) = 3x^2  root x + 1/(x1)
f'(x) = 6x  1/(2 times root x)  1/(x^22x+2)
My answer is 6x1/2(square root of x) + ln (x1)

01252002, 12:50 PM #10
Thanks so much guys, really appreciate your help

01252002, 01:04 PM #11
Wow. This board is a way to cheat on math too?!?!
I love the comradery.

01252002, 01:07 PM #12
Jenna,
You may be right, I may be wrong (do I feel a Billy Joel singalong starting?).
I looked over this while I had a few minutes and then rushed, realizing I had to meet my training partner at the gym. If there are any doubts, I would probably trust Jenna more than myself.

01252002, 02:13 PM #13
Bad_man  is that you in your avatar?

01252002, 02:20 PM #14
Ain't I handsome?

01252002, 06:28 PM #15
Years of math practice...
Limit Problems
i) L Hopital's Rule
x>0 (1cos6x)/x^2
= x>0 (6sin6x)/2x
= x>0 (36cos6x)/2 = 18
ii)
x>pie/3 (sin(x)  sin(pie/3))/(x(pie/3))
= x> pie/3 (cos(x) / 1)
= cos (pie/3)
Or something like that...just sucessively take the derivatives of the top and bottom until it's something solvable.

01252002, 07:45 PM #16
the answer is 3 .

01252002, 11:36 PM #17Junior Member
 Join Date
 Sep 2001
 Location
 Miami
 Posts
 132
Hey guys,
Calculus, ha...havent done this in a while, but I can manage..
Guys, the answers are correct except for the derivate calculation
the answer should be
6x.5(x+1/(x1))^(.5) X (1(x1)^2)
Jenna your answer was intuitive but remeber that ln x would be the integration dx/dy of ln x= 1/x but not vice versa
remeber 1/x is also equal to x^1
dx/dy of 1/x is 1x^2
if you need anything else PM

01262002, 12:23 AM #18
This blast from the past is interesting, and a little bit scary!

01262002, 12:30 AM #19Junior Member
 Join Date
 Sep 2001
 Location
 Miami
 Posts
 132
dude if you still remember L'Hopital's you are in good shape. In a few years I'll be saying Hospital what????
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