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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/(x-1)

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 (1-cos6x)/x^2

as x->pie/3 (sin(x) - sin(pie/3))/(x-(pie/3))

Thanks

2. VET
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.

3. Your posts sometimes shock me, but this one takes the cake sweetie

A double bump for TV.......

4. OMG....I feel like I need to go back to school....

Someone on here will know...lots of college people.

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/(x-1)

f(x) = 3x^2 - root x + 1/(x-1)

f'(x) = 6x - 1/(2 times root x) - 1/(x^2-2x+2)

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

a = 0.5, b = 1.5

7. Gotta go to the gym. I'll check out the last 2 later.

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...

9. Re: Re: Calculus Help Needed please

f(x) = 3x^2 - root x + 1/(x-1)

f'(x) = 6x - 1/(2 times root x) - 1/(x^2-2x+2)
I have to disagree with you on #2 (yet, I could be wrong)

My answer is 6x-1/2(square root of x) + ln (x-1)

10. Thanks so much guys, really appreciate your help

11. Wow. This board is a way to cheat on math too?!?!

12. Jenna,

You may be right, I may be wrong (do I feel a Billy Joel sing-a-long 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.

14. Ain't I handsome?

15. Years of math practice...

Limit Problems

i) L Hopital's Rule

x->0 (1-cos6x)/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.

16. the answer is 3 .

17. Junior 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

6x-.5(x+1/(x-1))^(-.5) X (1-(x-1)^-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

18. This blast from the past is interesting, and a little bit scary!

19. Junior 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????