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Math Help - On what interval is f increasing/decreasing and concave up/down

  1. #1
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    On what interval is f increasing/decreasing and concave up/down

    f (x)= x - 2sinx, 0<x<2pi (less than or equal to)

    Regarding this problem, I think that I have it solved. I just need someone to check my work to make sure I got the correct answer. Here is my answer:

    f ' (x)= 1 - 2cosx ; 1-2cosx=0; cosx= 1/2
    cosx > 1/2
    f is increasing from (0, pi/3) and (5pi/3, 2pi)
    f is decreasing from (pi/3, 5pi/3)

    f ''(x) = 2sinx; sinx = 0; sinx>0
    (0 , 2pi)
    concave up: (pi, 2pi)
    concave down: (0, pi)

    Feedback would be appreciated, thanks.
    Last edited by Gitano; October 20th 2009 at 11:05 AM. Reason: typo
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  2. #2
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    Quote Originally Posted by Gitano View Post
    f (x)= x + 2sinx, 0<x<2pi (less than or equal to)

    Regarding this problem, I think that I have it solved. I just need someone to check my work to make sure I got the correct answer. Here is my answer:

    f ' (x)= 1 - 2cosx ; 1-2cosx=0; cosx= 1/2
    cosx > 1/2
    f is increasing from (0, pi/3) and (5pi/3, 2pi)
    f is decreasing from (pi/3, 5pi/3)

    f ''(x) = 2sinx; sinx = 0; sinx>0
    (0 , 2pi)
    concave up: (pi, 2pi)
    concave down: (0, pi)

    Feedback would be appreciated, thanks.
    You made a mistake with the first derivative. The derivative of sin(x) is cos(x), not -cos(x).
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  3. #3
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    Quote Originally Posted by WhoCares357 View Post
    You made a mistake with the first derivative. The derivative of sin(x) is cos(x), not -cos(x).
    That was a typo, it is actually x - 2sinx.
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  4. #4
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    f(x)=x-2sin(x) given 0<x<2\Pi
    f'(x)=1-2cos(x) => 1-2cos(x)=0 => \frac{1}{2}=cos(x)
    cos^{-1}(\frac{1}{2})=x
    x=\frac{\Pi}{3} and x=\frac{5\Pi}{3}
    <-----a--neg--b--pos--c--neg--d---->
    a = 0;b = \frac{\Pi}{3};c = \frac{5\Pi}{3};d = 2\Pi

    f is decreasing from 0 to \frac{\Pi}{3} and from \frac{5\Pi}{3} to 2\Pi
    if is increasing from \frac{\Pi}{3} to \frac{5\Pi}{3}

    f''(x)=2sin(x)
    [tex]0=2sin(x) => sin(x)=0 => sin^{-1}(0)=x
    x=\Pi and x=2\Pi
    <---a--pos--b--neg--c--->
    a = 0;b =          \Pi;c =        2\Pi
    f is concaved up from 0 to \Pi
    f is concaved down from \Pi to 2\Pi

    Edit: You can see the graph here:
    http://www.wolframalpha.com/input/?i=plot+x-2sin%28x%29

    Look at it and you will see that my claims are verified.
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  5. #5
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    Thanks man, that was a very nice and thorough explanation.
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