Total Posts:43|Showing Posts:1-30|Last Page
Jump to topic:

Help me with Calculus

Disquisition
Posts: 391
Add as Friend
Challenge to a Debate
Send a Message
2/2/2014 11:11:55 AM
Posted: 2 years ago
If your good at calculus can you show all your steps in working out the problem below.
You can not use L'Hopitals rule because I'm not allowed to yet.

What is the Limit of ( sin(x) / x - pi) as x goes to pi

Show all your steps thanks.
yay842
Posts: 5,680
Add as Friend
Challenge to a Debate
Send a Message
2/2/2014 12:33:26 PM
Posted: 2 years ago
At 2/2/2014 11:11:55 AM, Disquisition wrote:
If your good at calculus can you show all your steps in working out the problem below.
You can not use L'Hopitals rule because I'm not allowed to yet.

What is the Limit of ( sin(x) / x - pi) as x goes to pi

Show all your steps thanks.

is the (x-p) in the denominator, or is it: sin(x)/x - pi?
30 Important Life Lessons
http://www.debate.org...
20 Terrifying Two-Sentence Horrors
http://www.debate.org...
20 Jokes That Only Geniuses Will Understand
http://www.debate.org...
Name One Song That Can't Match This GIF
http://d24w6bsrhbeh9d.cloudfront.net...
Disquisition
Posts: 391
Add as Friend
Challenge to a Debate
Send a Message
2/2/2014 12:39:46 PM
Posted: 2 years ago
At 2/2/2014 12:33:26 PM, yay842 wrote:
At 2/2/2014 11:11:55 AM, Disquisition wrote:
If your good at calculus can you show all your steps in working out the problem below.
You can not use L'Hopitals rule because I'm not allowed to yet.

What is the Limit of ( sin(x) / x - pi) as x goes to pi

Show all your steps thanks.

is the (x-p) in the denominator, or is it: sin(x)/x - pi?

The (x -pi) is the denominator and sin(x) is the numerator.
Disquisition
Posts: 391
Add as Friend
Challenge to a Debate
Send a Message
2/2/2014 12:46:04 PM
Posted: 2 years ago
And if anyone could also help me find the derivative of

f(x) = x^3 + 3x / x^2 -1 , so basically df/dx =?

Use the quotient rule and show all your steps
SeventhProfessor
Posts: 5,085
Add as Friend
Challenge to a Debate
Send a Message
2/2/2014 12:49:28 PM
Posted: 2 years ago
At 2/2/2014 12:46:04 PM, Disquisition wrote:
And if anyone could also help me find the derivative of

f(x) = x^3 + 3x / x^2 -1 , so basically df/dx =?

Use the quotient rule and show all your steps

Is x^2-1 the denominator? If so you should put it in parentheses.
#UnbanTheMadman

#StandWithBossy

#BetOnThett

"bossy r u like 85 years old and have lost ur mind"
~mysteriouscrystals

"I've honestly never seen seventh post anything that wasn't completely idiotic in a trying-to-be-funny way."
~F-16

https://docs.google.com...
yay842
Posts: 5,680
Add as Friend
Challenge to a Debate
Send a Message
2/2/2014 12:59:28 PM
Posted: 2 years ago
At 2/2/2014 12:39:46 PM, Disquisition wrote:
At 2/2/2014 12:33:26 PM, yay842 wrote:
At 2/2/2014 11:11:55 AM, Disquisition wrote:
If your good at calculus can you show all your steps in working out the problem below.
You can not use L'Hopitals rule because I'm not allowed to yet.

What is the Limit of ( sin(x) / x - pi) as x goes to pi

Show all your steps thanks.

is the (x-p) in the denominator, or is it: sin(x)/x - pi?

The (x -pi) is the denominator and sin(x) is the numerator.

i see no way of doing this one without lepitals rule but I guess you can move the (x-pi) to the numerator and do product rule
30 Important Life Lessons
http://www.debate.org...
20 Terrifying Two-Sentence Horrors
http://www.debate.org...
20 Jokes That Only Geniuses Will Understand
http://www.debate.org...
Name One Song That Can't Match This GIF
http://d24w6bsrhbeh9d.cloudfront.net...
Dazz
Posts: 1,163
Add as Friend
Challenge to a Debate
Send a Message
2/2/2014 1:02:21 PM
Posted: 2 years ago
At 2/2/2014 11:11:55 AM, Disquisition wrote:
If your good at calculus can you show all your steps in working out the problem below.
You can not use L'Hopitals rule because I'm not allowed to yet.

What is the Limit of ( sin(x) / x - pi) as x goes to pi

Show all your steps thanks.

Limit sin (x)/ x-pi
x->pi

As we know: Sin(x-pi)= sin(x)cos(pi) - sin(pi)cos(x)
...................... = sinx(-1) - (0)cosx
....................... = -sinx- 0 = -sinx

Hence Sin(x)= -sin (x-pi)

Replacing this in Limit
Limit -sin(x-pi)/ (x-pi)
x->pi

Taking out (-1)
(-1) Limit sin (x-pi)/ (x-pi)
x->pi

(-1) . (1)*
*By applying limit formula:
Limit sin(y)/y = 1
y->0
= -1

Is it ok?
Remove the "I want", remainder is the "peace". ~Al-Ghazali~
"This time will also pass", a dose to cure both; the excitement & the grievance. ~Ayaz~
yay842
Posts: 5,680
Add as Friend
Challenge to a Debate
Send a Message
2/2/2014 1:13:35 PM
Posted: 2 years ago
f(x) = x^3 + 3x / x^2 -1

1. quotient rule: (vu'-uv')/v^2
2. [(x^2-1)(3x^2+3)-(x^3+3x)(2x)]/(x^2-1)^2
3. (3x^3+3x^2-3x-3-6x^3-6x)/(x^4-1)
4. (-3x^3+3x^2-9x-3)/(x^4-1)
5. if you want to simplify it further: 3(-x^3+x^2-3x-1)/(x^4-1)

DONE
30 Important Life Lessons
http://www.debate.org...
20 Terrifying Two-Sentence Horrors
http://www.debate.org...
20 Jokes That Only Geniuses Will Understand
http://www.debate.org...
Name One Song That Can't Match This GIF
http://d24w6bsrhbeh9d.cloudfront.net...
Dazz
Posts: 1,163
Add as Friend
Challenge to a Debate
Send a Message
2/2/2014 1:16:53 PM
Posted: 2 years ago
At 2/2/2014 12:46:04 PM, Disquisition wrote:
And if anyone could also help me find the derivative of

f(x) = x^3 + 3x / x^2 -1 , so basically df/dx =?

Use the quotient rule and show all your steps

f(x)= x^3 + 3x /(x^2-1)
Quotient Rule
dy/dx = [(x^2-1)(3x^2+3)-(x^3+3x)(2x)] / {(x^2-1)^2}
= Solve the algebra to simplify.
Remove the "I want", remainder is the "peace". ~Al-Ghazali~
"This time will also pass", a dose to cure both; the excitement & the grievance. ~Ayaz~
yay842
Posts: 5,680
Add as Friend
Challenge to a Debate
Send a Message
2/2/2014 1:32:31 PM
Posted: 2 years ago
At 2/2/2014 1:16:53 PM, Dazz wrote:
At 2/2/2014 12:46:04 PM, Disquisition wrote:
And if anyone could also help me find the derivative of

f(x) = x^3 + 3x / x^2 -1 , so basically df/dx =?

Use the quotient rule and show all your steps

f(x)= x^3 + 3x /(x^2-1)
Quotient Rule
dy/dx = [(x^2-1)(3x^2+3)-(x^3+3x)(2x)] / {(x^2-1)^2}
= Solve the algebra to simplify.

beat you to it already, try solving the 1st one without lepital's rule(which i dont see how)
30 Important Life Lessons
http://www.debate.org...
20 Terrifying Two-Sentence Horrors
http://www.debate.org...
20 Jokes That Only Geniuses Will Understand
http://www.debate.org...
Name One Song That Can't Match This GIF
http://d24w6bsrhbeh9d.cloudfront.net...
Juan_Pablo
Posts: 2,052
Add as Friend
Challenge to a Debate
Send a Message
2/2/2014 1:49:44 PM
Posted: 2 years ago
Yay, I get a different answer:

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

1. quotient rule: (vu'-uv')/v^2
2. [ ( x^2 - 1 ) ( 3x^2 + 3 ) - ( x^3 + 3x ) ( 2x ) ] / ( x^2 - 1 ) ^2
3. ( x^4 - 6x^2 - 3 ) / [ ( x^2 - ) ( x^2 - 1 ) ]

4. ( x^4 - 6x^2 - 3 ) / ( x^4 - 2x^2 + 1 )

Step 4 is the answer.
Juan_Pablo
Posts: 2,052
Add as Friend
Challenge to a Debate
Send a Message
2/2/2014 1:54:37 PM
Posted: 2 years ago
Step 3 should read ( x^4 - 6x^2 - 3 ) / [ ( x^2 - 1 ) ( x^2 - 1 ) ]

I did it twice and I still get the same answer I got in Step 4.
yay842
Posts: 5,680
Add as Friend
Challenge to a Debate
Send a Message
2/2/2014 1:56:23 PM
Posted: 2 years ago
At 2/2/2014 1:49:44 PM, Juan_Pablo wrote:
Yay, I get a different answer:

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

1. quotient rule: (vu'-uv')/v^2
2. [ ( x^2 - 1 ) ( 3x^2 + 3 ) - ( x^3 + 3x ) ( 2x ) ] / ( x^2 - 1 ) ^2
3. ( x^4 - 6x^2 - 3 ) / [ ( x^2 - ) ( x^2 - 1 ) ]

4. ( x^4 - 6x^2 - 3 ) / ( x^4 - 2x^2 + 1 )

Step 4 is the answer.

you screwed up 2 to 3. there is no power to the 4 in in the numerator
30 Important Life Lessons
http://www.debate.org...
20 Terrifying Two-Sentence Horrors
http://www.debate.org...
20 Jokes That Only Geniuses Will Understand
http://www.debate.org...
Name One Song That Can't Match This GIF
http://d24w6bsrhbeh9d.cloudfront.net...
yay842
Posts: 5,680
Add as Friend
Challenge to a Debate
Send a Message
2/2/2014 1:58:01 PM
Posted: 2 years ago
At 2/2/2014 11:11:55 AM, Disquisition wrote:
If your good at calculus can you show all your steps in working out the problem below.
You can not use L'Hopitals rule because I'm not allowed to yet.

What is the Limit of ( sin(x) / x - pi) as x goes to pi

Show all your steps thanks.

well i see no way to do this with lepitals ruel, but just to show you:

1. derivative of sinx = cosx
2. derivative of x - pi = 1
3. cos pi = 1
4. the limit as x approaches pi of (sin(x))/(x-pi)=1
30 Important Life Lessons
http://www.debate.org...
20 Terrifying Two-Sentence Horrors
http://www.debate.org...
20 Jokes That Only Geniuses Will Understand
http://www.debate.org...
Name One Song That Can't Match This GIF
http://d24w6bsrhbeh9d.cloudfront.net...
yay842
Posts: 5,680
Add as Friend
Challenge to a Debate
Send a Message
2/2/2014 1:59:01 PM
Posted: 2 years ago
At 2/2/2014 1:54:37 PM, Juan_Pablo wrote:
Step 3 should read ( x^4 - 6x^2 - 3 ) / [ ( x^2 - 1 ) ( x^2 - 1 ) ]

I did it twice and I still get the same answer I got in Step 4.

nvm i take that back, there is a power to the fourth
30 Important Life Lessons
http://www.debate.org...
20 Terrifying Two-Sentence Horrors
http://www.debate.org...
20 Jokes That Only Geniuses Will Understand
http://www.debate.org...
Name One Song That Can't Match This GIF
http://d24w6bsrhbeh9d.cloudfront.net...
yay842
Posts: 5,680
Add as Friend
Challenge to a Debate
Send a Message
2/2/2014 1:59:53 PM
Posted: 2 years ago
i cant tell sh*t in this format, i gots to right it down
30 Important Life Lessons
http://www.debate.org...
20 Terrifying Two-Sentence Horrors
http://www.debate.org...
20 Jokes That Only Geniuses Will Understand
http://www.debate.org...
Name One Song That Can't Match This GIF
http://d24w6bsrhbeh9d.cloudfront.net...
Juan_Pablo
Posts: 2,052
Add as Friend
Challenge to a Debate
Send a Message
2/2/2014 2:05:23 PM
Posted: 2 years ago
Step 2 to 3:

[ ( x^2 - 1 ) ( 3x^2 + 3 ) - ( x^3 + 3x ) ( 2x ) ] / ( x^2 - 1 ) ^2 =

[ ( 3x^4 + 3x^2 - 3x^2 - 3 ) + ( -2x^4 - 6x^2 ) ] / [ ( x^2 - 1 ) ( x^2 - 1 ) ]

Step 3: ( x^4 - 6x^2 - 3 ) / [ ( x^2 - ) ( x^2 - 1 ) ]

I still get the same answer. You made a mistake somewhere.

It's okay, Yay. I frequently make similar mistakes and have to re-check my answers too. ;)
yay842
Posts: 5,680
Add as Friend
Challenge to a Debate
Send a Message
2/2/2014 2:14:04 PM
Posted: 2 years ago
At 2/2/2014 2:05:23 PM, Juan_Pablo wrote:
Step 2 to 3:

[ ( x^2 - 1 ) ( 3x^2 + 3 ) - ( x^3 + 3x ) ( 2x ) ] / ( x^2 - 1 ) ^2 =

[ ( 3x^4 + 3x^2 - 3x^2 - 3 ) + ( -2x^4 - 6x^2 ) ] / [ ( x^2 - 1 ) ( x^2 - 1 ) ]

Step 3: ( x^4 - 6x^2 - 3 ) / [ ( x^2 - ) ( x^2 - 1 ) ]

I still get the same answer. You made a mistake somewhere.

It's okay, Yay. I frequently make similar mistakes and have to re-check my answers too. ;)

http://thesamerainbowsend.com...
30 Important Life Lessons
http://www.debate.org...
20 Terrifying Two-Sentence Horrors
http://www.debate.org...
20 Jokes That Only Geniuses Will Understand
http://www.debate.org...
Name One Song That Can't Match This GIF
http://d24w6bsrhbeh9d.cloudfront.net...
Disquisition
Posts: 391
Add as Friend
Challenge to a Debate
Send a Message
2/2/2014 2:24:22 PM
Posted: 2 years ago
At 2/2/2014 2:05:23 PM, Juan_Pablo wrote:
Step 2 to 3:

[ ( x^2 - 1 ) ( 3x^2 + 3 ) - ( x^3 + 3x ) ( 2x ) ] / ( x^2 - 1 ) ^2 =

[ ( 3x^4 + 3x^2 - 3x^2 - 3 ) + ( -2x^4 - 6x^2 ) ] / [ ( x^2 - 1 ) ( x^2 - 1 ) ]

Step 3: ( x^4 - 6x^2 - 3 ) / [ ( x^2 - ) ( x^2 - 1 ) ]

I still get the same answer. You made a mistake somewhere.

It's okay, Yay. I frequently make similar mistakes and have to re-check my answers too. ;)

Thanks I forgot to expand (x^2-1) to (x-1)(x+1) and multiply separately. After that I got the same answer you did.
Disquisition
Posts: 391
Add as Friend
Challenge to a Debate
Send a Message
2/2/2014 2:25:49 PM
Posted: 2 years ago
At 2/2/2014 1:02:21 PM, Dazz wrote:
At 2/2/2014 11:11:55 AM, Disquisition wrote:
If your good at calculus can you show all your steps in working out the problem below.
You can not use L'Hopitals rule because I'm not allowed to yet.

What is the Limit of ( sin(x) / x - pi) as x goes to pi

Show all your steps thanks.

Limit sin (x)/ x-pi
x->pi

As we know: Sin(x-pi)= sin(x)cos(pi) - sin(pi)cos(x)
...................... = sinx(-1) - (0)cosx
....................... = -sinx- 0 = -sinx

Hence Sin(x)= -sin (x-pi)

Replacing this in Limit
Limit -sin(x-pi)/ (x-pi)
x->pi

Taking out (-1)
(-1) Limit sin (x-pi)/ (x-pi)
x->pi

(-1) . (1)*
*By applying limit formula:
Limit sin(y)/y = 1
y->0
= -1

Is it ok?

I get everything until you get to the limit formula part. Could you elaborate on that, since I just started Cal 1
Juan_Pablo
Posts: 2,052
Add as Friend
Challenge to a Debate
Send a Message
2/2/2014 2:27:00 PM
Posted: 2 years ago
At 2/2/2014 2:24:22 PM, Disquisition wrote:
At 2/2/2014 2:05:23 PM, Juan_Pablo wrote:
Step 2 to 3:

[ ( x^2 - 1 ) ( 3x^2 + 3 ) - ( x^3 + 3x ) ( 2x ) ] / ( x^2 - 1 ) ^2 =

[ ( 3x^4 + 3x^2 - 3x^2 - 3 ) + ( -2x^4 - 6x^2 ) ] / [ ( x^2 - 1 ) ( x^2 - 1 ) ]

Step 3: ( x^4 - 6x^2 - 3 ) / [ ( x^2 - ) ( x^2 - 1 ) ]

I still get the same answer. You made a mistake somewhere.

It's okay, Yay. I frequently make similar mistakes and have to re-check my answers too. ;)

Thanks I forgot to expand (x^2-1) to (x-1)(x+1) and multiply separately. After that I got the same answer you did.

No problem, Diquisition.
Juan_Pablo
Posts: 2,052
Add as Friend
Challenge to a Debate
Send a Message
2/2/2014 2:48:42 PM
Posted: 2 years ago
By the way, Disquisition, for you first problem you can't use L'Hospital's rule.

The problem isn't of the indeterminate form

[ lim x->a f (x) ] / [ lim x->a g (x) ] = 0 / 0 or infinity / inifinity.

L'Hospital's rules doesn't apply.
Disquisition
Posts: 391
Add as Friend
Challenge to a Debate
Send a Message
2/2/2014 2:50:20 PM
Posted: 2 years ago
At 2/2/2014 2:27:00 PM, Juan_Pablo wrote:

No problem, Diquisition.

Here is another one if your up to it.

What is the derivative of f(x) = cos(2x), when c = (pi) / (4),

The question basically wants us to plug it into this equation [f ( c+h ) - f ( c )] / h
so I think its saying to determine the derivative of cos(2x) when x = (pi) / (4)
yay842
Posts: 5,680
Add as Friend
Challenge to a Debate
Send a Message
2/2/2014 2:57:05 PM
Posted: 2 years ago
At 2/2/2014 2:48:42 PM, Juan_Pablo wrote:
By the way, Disquisition, for you first problem you can't use L'Hospital's rule.

The problem isn't of the indeterminate form

[ lim x->a f (x) ] / [ lim x->a g (x) ] = 0 / 0 or infinity / inifinity.

L'Hospital's rules doesn't apply.

its not indeterminate?
30 Important Life Lessons
http://www.debate.org...
20 Terrifying Two-Sentence Horrors
http://www.debate.org...
20 Jokes That Only Geniuses Will Understand
http://www.debate.org...
Name One Song That Can't Match This GIF
http://d24w6bsrhbeh9d.cloudfront.net...
yay842
Posts: 5,680
Add as Friend
Challenge to a Debate
Send a Message
2/2/2014 2:59:00 PM
Posted: 2 years ago
At 2/2/2014 2:50:20 PM, Disquisition wrote:
At 2/2/2014 2:27:00 PM, Juan_Pablo wrote:

No problem, Diquisition.

Here is another one if your up to it.

What is the derivative of f(x) = cos(2x), when c = (pi) / (4),

The question basically wants us to plug it into this equation [f ( c+h ) - f ( c )] / h
so I think its saying to determine the derivative of cos(2x) when x = (pi) / (4)

1. cos(2x), x=pi/4
2. derive: -2sin(2x)
3. -2sine(2pi/4)
4. -2sin(pi/2)=1 x -2 = -2
30 Important Life Lessons
http://www.debate.org...
20 Terrifying Two-Sentence Horrors
http://www.debate.org...
20 Jokes That Only Geniuses Will Understand
http://www.debate.org...
Name One Song That Can't Match This GIF
http://d24w6bsrhbeh9d.cloudfront.net...
Juan_Pablo
Posts: 2,052
Add as Friend
Challenge to a Debate
Send a Message
2/2/2014 3:03:29 PM
Posted: 2 years ago
At 2/2/2014 2:57:05 PM, yay842 wrote:
At 2/2/2014 2:48:42 PM, Juan_Pablo wrote:
By the way, Disquisition, for you first problem you can't use L'Hospital's rule.

The problem isn't of the indeterminate form

[ lim x->a f (x) ] / [ lim x->a g (x) ] = 0 / 0 or infinity / inifinity.

L'Hospital's rules doesn't apply.

its not indeterminate?

It's not of an indeterminant form so that L'Hospital's rule can be used. The reason is that the lim sin(x) as x -> pi is neither 0 or infinity. Both the limits of the expression in the numerator and the denominator need equal 0 xor infinity.
Disquisition
Posts: 391
Add as Friend
Challenge to a Debate
Send a Message
2/2/2014 3:24:27 PM
Posted: 2 years ago
At 2/2/2014 2:59:00 PM, yay842 wrote:
At 2/2/2014 2:50:20 PM, Disquisition wrote:
At 2/2/2014 2:27:00 PM, Juan_Pablo wrote:

No problem, Diquisition.

Here is another one if your up to it.

What is the derivative of f(x) = cos(2x), when c = (pi) / (4),

The question basically wants us to plug it into this equation [f ( c+h ) - f ( c )] / h
so I think its saying to determine the derivative of cos(2x) when x = (pi) / (4)

1. cos(2x), x=pi/4
2. derive: -2sin(2x)
3. -2sine(2pi/4)
4. -2sin(pi/2)=1 x -2 = -2

We haven't come to the point where we can use derivations of trig functions so I have to plug into formula :( but thanks anyways.
Juan_Pablo
Posts: 2,052
Add as Friend
Challenge to a Debate
Send a Message
2/2/2014 3:25:52 PM
Posted: 2 years ago
At 2/2/2014 2:59:00 PM, yay842 wrote:
At 2/2/2014 2:50:20 PM, Disquisition wrote:
At 2/2/2014 2:27:00 PM, Juan_Pablo wrote:

No problem, Diquisition.

Here is another one if your up to it.

What is the derivative of f(x) = cos(2x), when c = (pi) / (4),

The question basically wants us to plug it into this equation [f ( c+h ) - f ( c )] / h
so I think its saying to determine the derivative of cos(2x) when x = (pi) / (4)

1. cos(2x), x=pi/4
2. derive: -2sin(2x)
3. -2sine(2pi/4)
4. -2sin(pi/2)=1 x -2 = -2

Yay, I get a different answer. I get:

1. cos ( 2x), x = pi / 4
2. [ cos ( 2x) ]' = - 2 sin ( 2 x )
3. - 2 sin ( 2 pi / 4 )
4. - 2 sin ( pi / 2 )

5. -1.972588

That can probably be cleanly restated with pi as a factor.