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Math Help - Inflection And Concavity.

  1. #1
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    Inflection And Concavity.

    The directions say to find the points of inflection and discuss the concavity of the graph of the function.
    The function is
     f(x)= sinx+cosx the intervals [0,2 pi]
    I know you supposed to take the derivative, then the 2nd then set the equation to zero
    so i have up to
    0= cosx-sinx
    Is this right?
    if so what do i do next?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Red350z View Post
    The directions say to find the points of inflection and discuss the concavity of the graph of the function.
    The function is
     f(x)= sinx+cosx the intervals [0,2 pi]
    I know you supposed to take the derivative, then the 2nd then set the equation to zero
    so i have up to
    0= cosx-sinx
    Is this right?
    if so what do i do next?
    That's only the first derivative!!

    The second derivative would be f^{\prime\prime}\!\left(x\right)=-\sin x-\cos x

    To find inflection points, solve for x:

    -\sin x-\cos x=0\implies \sin x=-\cos x

    This is the case where x=\frac{3\pi}{4},\,\frac{7\pi}{4}.

    Can you determine where the function is concave up or down now?
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  3. #3
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    To get inflection points you have to solve f ''(x) = 0

    You are solving f '(x) = 0 so you need to find f '' (x).
    -sinx - cos x = 0

    -sinx = cosx

    This is only true when x =  \frac{3\pi}{4}
    x =  \frac{7\pi}{4}


    To the left of x = \frac{3\pi}{4} f '' (x) < 0

    To the right of x = \frac{3\pi}{4} f '' (x) > 0

    Therefore we have an ip( inflection point ) at x = \frac{3\pi}{4}

    When f''(x) <0 Then f is concaving down
    When f''(x) >0 then f is concaving up.

    Do this for x =  \frac{7\pi}{4}
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  4. #4
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    Quote Originally Posted by Chris L T521 View Post
    That's only the first derivative!!

    The second derivative would be f^{\prime\prime}\!\left(x\right)=-\sin x-\cos x

    To find inflection points, solve for x:

    -\sin x-\cos x=0\implies \sin x=-\cos x

    This is the case where x=\frac{3\pi}{4},\,\frac{7\pi}{4}.

    Can you determine where the function is concave up or down now?
    I always get this wrong but hen f''(x) <0 Then f is concaving down, and when f''(x) >0 then f is concaving up.
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  5. #5
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    Exactly.
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