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LerraLove
Toomuchtimeonhands
Joined: 22 Oct 2002
Posts: 827
Location: Redmond, Washington
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 Posted: 11/18/04 - 19:14 Post subject: Help me pls calculus wizards!!
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Ok, So I am in calculus right now. I skipped pre calc 2 which probably was a bad idea. but I need a LOT of help doing this problem since my trig sucks a bunch.
K this may be really easy for some of you and thats what I am hoping for...
Problem:
f(x)= x-2sinx ,0<x<3pi
a. Find the intervals on which f is increasing or decreasing
b. Find the local max and min values of f.
c) Find the intervals of concavity and the inflection points
Any help at all would be appreciated ! <3
I know f'(x) is 1-2cosx
Where do I go from here ? 
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quotison
RealPoor Sensei
Joined: 13 Oct 2002
Posts: 1594
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Posted: 11/18/04 - 19:35 Post subject:
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F is increasing whenever the derivative is positive.
F is decreasing whenever the derivative is negative.
To test for the min and max values, I think you test the end points (f(0) and f(3pi)) and where F' is equal to zero (where F'(x) = 0 couldl be a min or a max).
Concavity and inflection points deals with the second derivative. If I remember correctly, the inflection point is where the second derivative equals zero. Concave up is where the second derivative > 0 and concave down is where the second derivative < 0. But check your book on that- it might be the other way around.
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Desaitar
RealPoor Guru
Joined: 30 Apr 2003
Posts: 2641
Location: whore island
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Posted: 11/18/04 - 20:07 Post subject:
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Quot is right for once, you need to also get f''(x) to determine the concavity, quot explained how to determine whether it's up or down and where the inflection points are.
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Nictathan
RealPoor Master of Posts
Joined: 11 Oct 2002
Posts: 5531
Location: here... where I am... not with you
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Posted: 11/18/04 - 20:14 Post subject:
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Too many years since I did calc I suck
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Desaitar
RealPoor Guru
Joined: 30 Apr 2003
Posts: 2641
Location: whore island
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Posted: 11/18/04 - 20:17 Post subject:
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Oh, if f'(x)=1-2cosx then f'(x) = 0 when x = pi/3, 5pi/3, 7pi/3 on [0,3pi] I think =p
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lauren000
RealPoor Guru
Joined: 21 Oct 2002
Posts: 3510
Location: colorado springs
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Posted: 11/18/04 - 22:26 Post subject:
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no you suck
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Devook
RealPoor Guru
Joined: 31 Mar 2004
Posts: 2373
Location: Ypsilanti or Troy, MI
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Posted: 11/18/04 - 22:57 Post subject:
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f(x)= x-2sinx, 0<x<3pi
a) Find the intervals on which f is increasing or decreasing
b) Find the local max and min values of f.
c) Find the intervals of concavity and the inflection points
a) f'(x) = 1 - 2cos(x)
f'(x) is positive when 2cos(x) < 1
cos(x) < 1/2
pi/3 < x < 7pi/3
f'(x) is negative when 2cos(x) > 1
cos(x) > 1/2
0 < x < pi/3 or 7pi/3 < x < 3pi
b) Put the values pi/3 and 7pi/3 into the original equation.
pi/3 - 2 * (3)^(1/2)/2 ~ -0.685
7pi/3 - 2 * (3)^(1/2)/2 ~ 5.60
Critical points: (pi/3, -0.685) and (7pi/3, 5.60)
f(x) changes from decreasing to increasing at (pi/3, -0.685) so it's a min.
f(x) changes from increasing to decreasing at (7pi/3, -0.685) so it's a min.
c) f''(x) = 2sin(x)
Inflection points occur when f''(x) = 0
So, points at (0, 0) and (2pi, 2pi).
f''(x) is positive on the interval [0, 2pi] so the graph is concave up.
f''(x) is negative on the interval [2pi, 3pi] so the graph is concave down.
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Desaitar
RealPoor Guru
Joined: 30 Apr 2003
Posts: 2641
Location: whore island
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Posted: 11/18/04 - 23:35 Post subject:
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1-2cos (5pi/3) = 0. :p
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Desaitar
RealPoor Guru
Joined: 30 Apr 2003
Posts: 2641
Location: whore island
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Posted: 11/18/04 - 23:37 Post subject:
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Devook you were really close bro, you just missed one critical point (but it was important for intervals of increasing/decreasing and local extrema), you can always check by graphing.
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LerraLove
Toomuchtimeonhands
Joined: 22 Oct 2002
Posts: 827
Location: Redmond, Washington
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Posted: 11/19/04 - 11:10 Post subject:
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yaay thanks so much guys, I knew someone would be able to help me out 
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Occulis
RealPoor Jedi
Joined: 11 Oct 2002
Posts: 13293
Location: Moral Relativity Central
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Posted: 11/19/04 - 11:14 Post subject:
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f*g
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BlingBling
Toomuchtimeonhands
Joined: 20 Dec 2002
Posts: 944
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Posted: 11/19/04 - 12:18 Post subject:
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easy way without having to worry about intevals of concavity
they already talked about inflection poinets but if you take your critical values which you got from f'(x)=0 and you plug those into your f''(x) you can determine weather the function at that point is concave up or down. From there you can scetch what the graph wil look like simply by knowing points where the graph is a mountain or a valley
of course you need to plug your critical points including yoru boundary conditions o fyour problem into the origonal f(x) to evaluate which are the max and min points
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Devook
RealPoor Guru
Joined: 31 Mar 2004
Posts: 2373
Location: Ypsilanti or Troy, MI
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Posted: 11/19/04 - 16:18 Post subject:
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| Desaitar wrote: | | Devook you were really close bro, you just missed one critical point (but it was important for intervals of increasing/decreasing and local extrema), you can always check by graphing. |
/shrug, I wasn't using a calculator.
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ATM Banana
RealPoor Master of Posts
Joined: 02 Jan 2003
Posts: 8575
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Posted: 11/19/04 - 19:01 Post subject:
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the answer is yes.
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