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Math Help - Having trouble with graph of trig. function

  1. #1
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    Having trouble with graph of trig. function

    I'm doing a graph sketching of f(x)=sin^4(x) + cos^4(x)
    And I'm having trouble with two things:

    1)
    I get f''(x)=-4sin^4(x) -4cos^4(x) + 24sin^2(x)cos^2(x)
    Then, I checked on WolframAlpha and gives me an alternate form :
    f''(x)=-4cos(4x) which is equivalent to what i got. But I have no idea how to go from the first form to the latter.

    2)
    When taking limits of f(x) as x-> infinity and -infinity, I understand:
    lim (sin(x)) as x->+-inifinty ; i get y=1 and y=-1 as the horizontal asymptotes.
    Also:
    lim (sin(x))^4 as x->+-infinity; i get y=1 and y=0 as the horizontal asymptotes.

    Then for lim of cos^4(x) as x->+-infinity I get the same thing, but when those two are added, the graph is between the lines y=.5 and y=1

    Any help would be greatly appreciated.
    Last edited by Arturo_026; November 1st 2009 at 02:28 PM.
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  2. #2
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    Quote Originally Posted by Arturo_026 View Post
    I'm doing a graph sketching of f(x)=sin^4(x) + cos^4(x)
    And I'm having trouble with two things:

    1)
    I get f''(x)=-4sin^4(x) -4cos^4(x) + 24sin^2(x)cos^2(x)
    Then, I checked on WolframAlpha and gives me an alternate form :
    f''(x)=-4cos(4x) which is equivalent to what i got. But I have no idea how to go from the first form to the latter.

    [snip]
    -4[\sin^4(x) + \cos^4(x) - 6 \sin^2(x) \cos^2(x)]

    = -4 [(\sin^2 (x) + \cos^2 (x))^2 - 8\sin^2(x) \cos^2(x)]

    = -4 [1 - 4\sin^2(2x)]

    which won't quite give -4 cos(4x). Is there a typo somewhere?
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    -4[\sin^4(x) + \cos^4(x) - 6 \sin^2(x) \cos^2(x)]

    = -4 [(\sin^2 (x) + \cos^2 (x))^2 - 8\sin^2(x) \cos^2(x)]

    = -4 [1 - 4\sin^2(2x)]

    which won't quite give -4 cos(4x). Is there a typo somewhere?
    I apologize for jumping a while ago.

    after: = -4 [(\sin^2 (x) + \cos^2 (x))^2 - 8\sin^2(x) \cos^2(x)]
    shouldn't it be:

    = -4[1-2sin^2(2x)]
    =-4cos(4x)
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  4. #4
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    Quote Originally Posted by Arturo_026 View Post
    I apologize for jumping a while ago.

    after: = -4 [(\sin^2 (x) + \cos^2 (x))^2 - 8\sin^2(x) \cos^2(x)]
    shouldn't it be:

    = -4[1-2sin^2(2x)]
    =-4cos(4x)
    Right you are. I was in a hurry. I'm glad to see you got the idea of what to do.
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