~~~Help Needed CALCULUS~~~

[tripleV]

Ok guys and gals, a friend of mine needs help on a take home calculus assignment, 4 questions, here they are:

1) find derivative:

3x^2 - square root x + 1/(x-1)


2) 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

3) Use intermediate value theorem to prove this has at
least 1 root.:

e^x + cos(x) - 2x - 4 = 0

4) Limit:

as x->0 (1-cos6x)/x^2

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


PLEASE HELP IF YOU CAN

 

[HighIntensity]

2+2=4

 

[tripleV]

Thanks for the help asswipe

 

[HighIntensity]

Thanks for the help asswipe

we don't need this on the boards, frankly im shocked at you....

 

[tripleV]

at the asswipe or the calculus, now shush

 

[HighIntensity]

shush x2 = more shushing for you

 

[Taps]

Honestly, if I wasn't half asleep right now, I'd whip out the 'ol calculator.

Sorry. There's some smart people here during the day who I'm sure will be more than willing to answer your questions.

In the meantime, enjoy this:

 

[tripleV]

omfg, THANK YOU HON

 

[Taps]

You're most welcome. The least I can do is bump this thread.

God, I think I outdid myself with that pic. It's gonna be my new desktop wallpaper.

 

[Enrico]

Ok guys and gals, a friend of mine needs help on a take home calculus assignment, 4 questions, here they are:

1) find derivative:

3x^2 - square root x + 1/(x-1)

OK this one is easy, the answer is:
6x-0.5x^-1/2 - 1/x^2

2) 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

3) Use intermediate value theorem to prove this has at
least 1 root.:

e^x + cos(x) - 2x - 4 = 0

4) Limit:

as x->0 (1-cos6x)/x^2

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


I've got to go to class, but maybe I'll try 2, 3, and 4 later. OK but I did post the answer to #1...see ya later

 

[GaryWary]

If you can get hold of them I suggest you get hold of Engineering Mathematics by K.A. Stroud. I would not have got through the maths on my maths/engineering degree if I didn't have 'Stroud'.

 

[tripleV]

Thanks enrico

 

[tripleV]

can anyone get 3 or 4?

 

[big4life]

Nope, can't help you...

But I do like the pic. taps posted.

 

[tripleV]

me too, but i need some math peeps here......come on smarty pants

 

[Enrico]

OK I'm back


1) Find the derivative:
3x^2- sq. root x + 1/(x-1)

ANSWER: this can be rewritten as 3x^2-x^1/2 + x^-1
therefore, d/dx=6x- 1/2x^-1/2- x^-2
To simplify, d/dx=6x - 1/(2 sq root x) - 1/(x^2)

2) 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

ANSWER: a=1/2 and b=3/2

3) Use intermediate value theorem to prove this has at
least 1 root.:

e^x + cos(x) - 2x - 4 = 0

To show this, we have to find where d/dx=0
d/dx=e^x - sinx - 2
e^x - sin x - 2 = 0
e^x - sin x = 2
To simplify completely, (e^x = 2+sinx)

4) Limit:

as x->0 (1-cos6x)/x^2
ANSWER: The limit does not exist. This is because as x--> 0, x^2 also approaches 0. 1/0 is undefined, and there is no way to factor x out of the denominator.

as x->pie/3 (sin(x) - sin(pie/3))/(x-(pie/3))
ANSWER: This can be rewritten as lim x--> pi/3 of (Sin x)/x - (sin pi/3)/(pi/3). Since x is approaching (pi/3), (sin x)/x will get closer to (sin pi/3)/(pi/3). Therefore, the limit is 0.

Hope this helps

 

[KAYNE]

V

I WOULD HELP YOU WITH THIS BEING THAT I DID IT ALL LAST SEMESTER AND DID PRETTY GOOD IN THE CLASS. HOWEVER, I'M ONE OF THOSE PEOPLE WHO FORGET MATH IF THEY DONT DO IT CONTINUOUSLY. AND I HAVENT DONT IT SINCE LAST SEMESTER AND I HATE THE SHIT ALSO!!! HEHE SORRY. IF YOU NEED ANY FINANCE HELP, LET ME KNOW!!!


KAYNE

 

[Warik]

as x->0 (1-cos6x)/x^2
ANSWER: The limit does not exist. This is because as x--> 0, x^2 also approaches 0. 1/0 is undefined, and there is no way to factor x out of the denominator.

Use L'Hopitale's rule. lim x-> 0 of 1 - cos6x = 0, lim x-> 0 of x^2 = 0, so take the first derivative of the top and bottom. Derivative of 1 - cos6x is either 6 x sin6x or some shit like that I can't remember, and derivative of x^2 = 2x. Still doesn't work, so take the second derivative. Top becomes some shit, and bottom becomes 2. Solve from there.

-Warik

 

[guards]

Law student here......I think I forgot how to count

 

[Enrico]

Use L'Hopitale's rule. lim x-> 0 of 1 - cos6x = 0, lim x-> 0 of x^2 = 0, so take the first derivative of the top and bottom. Derivative of 1 - cos6x is either 6 x sin6x or some shit like that I can't remember, and derivative of x^2 = 2x. Still doesn't work, so take the second derivative. Top becomes some shit, and bottom becomes 2. Solve from there.

-Warik

But if this person is not in like Calculus 2 or 3, they wouldn't know L'Hospital's rule, so maybe the prof would think something was up. All four of your questions are like basic Calculus 1 stuff, so that's what I had assumed...we didn't learn L'Hospital's rule till we studied Indeterminate Forms, so....do whatever you think is best.

 

[Warik]

Good point, Enrico. They don't teach that till Calc2... which I cannot understand, because it's such an easy concept... hell, I took Calc I and II in the same year in HS. Easiest class I ever passed.

-Warik

 

[Enrico]

Dude--and you were thinking about dropping out? You might as well stay in college, you know your shit!

 

[Warik]

You might as well stay in college, you know your shit!

Doesn't that kind of defeat the purpose of going??? Not like I get paid to go... well... I do sort of. Leftover scholarship money. =)

hehe

-Warik

 

[mattcanning99]

did tripleV get an answer yet? im too lazy to read every reply...im a math major, so if you havent got an answer yet, i'll get it for you...

 

[velvett]

Brainy guys turn me on.



Excuse me. I need a shower.

 

[Enrico]

Did triple v ever get the answers down? I don't think she ever responded again!

 

[tripleV]

yup, thanks babe

 

[OMEGA]

Warik

you are from the 24th Century,

but I am from the 25th

hope this helps