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Calculus I help?

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Kaulus7
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« on: November 20, 2009, 10:57:20 pm »

Anyone who remembers this?? If so, I'll start listing my problems tomorrow...

Section is Derivatives and the Shapes of Curves.
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« Reply #1 on: November 21, 2009, 10:46:37 am »

Ooh, Calculas... I like, but I think I've heard Derivatives before...
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« Reply #2 on: November 21, 2009, 11:10:39 am »

I can help you with calc 1,2, 3 (multivariate), and differential equations.
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Kaulus7
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« Reply #3 on: November 21, 2009, 02:30:55 pm »

Thanks torsion!

I'll edit this in about 10 min. with a problem I have trouble with.
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« Reply #4 on: November 21, 2009, 02:56:19 pm »

No problem.  As long as it's not a taylor series, I can help.
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« Reply #5 on: November 21, 2009, 03:34:56 pm »

Woohoo! We skipped that stuff! (I think we did...)
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« Reply #6 on: November 21, 2009, 03:38:16 pm »

So, what do you need help with?
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« Reply #7 on: November 21, 2009, 09:59:25 pm »

I'll have it by tomorrow-it's more complex than I thought.  :-\
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« Reply #8 on: November 22, 2009, 01:43:31 am »

Kaulus, Is this a homework question?
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« Reply #9 on: November 22, 2009, 07:37:17 am »

Kaulus, Is this a homework question?

I was about to mention this.  If it is, I wont solve it, I will just give you some tips, or solve a problem similar to it.
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« Reply #10 on: November 22, 2009, 11:42:51 am »

Yeah, don't worry, I can do most of this by myself, it's easier than it looks.

Just to confirm is f"(x)=7x3*(x-1)2+(7x-4)(3x2(x-1)2+2x3(x-1)2) if f(x)=x4(x-1)3?

Give me a yes/no answer and I'll check my work.
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« Reply #11 on: November 22, 2009, 11:52:24 am »

No.  It's something along the lines of "x^2(x-c)*(c*x^2-8*x+c)

C is a constant, that varies over terms...
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« Reply #12 on: November 22, 2009, 10:10:47 pm »

I could probably do this myself but my mind is addled and I need to go to bed early so...

For what values of c does the polynomial P(x)=x4+cx3+x2 have two inflection points? One inflection point? None? Illustrate by graphing P for several values of c. How does the graph change as c decreases?

I know that second derivative is the first step (I hope it is...). But how would I go about to determining the values of c? Guess and check is an option, but I don't think that's what my instructors had in mind...
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« Reply #13 on: November 24, 2009, 10:22:20 am »

Inflection?  As in max or min?
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« Reply #14 on: November 25, 2009, 01:08:38 am »

Inflection as in point of change between the concavity.
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« Reply #15 on: November 25, 2009, 09:04:34 am »

Here's a link to some examples of finding inflection points:

http://www.clas.ucsb.edu/staff/lee/Inflection%20Points.htm

The basic steps are to take the second derivative and determine the roots (where the equation is 0).  Then you have to check that the concavity is changing at that point of the function.

In this example the 2nd derivative will have c as a variable.  You need to determine for what values of c the function (the 2nd derivative) has 0, 1, 2, etc. roots.
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