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Math Help - Functions

  1. #1
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    Functions

    Answer the following questions for the function
    defined on the interval . Enter points, such as inflection points in ascending order, i.e. smallest x values first.
    Remember that you can enter "pi" for as part of your answer.

    A. is concave down on the region to
    B. A global minimum for this function occurs at
    C. A local maximum for this function which is not a global maximum occurs at
    D. The function is increasing on to and on to .
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Hmm

    Quote Originally Posted by tennisgirl View Post
    Answer the following questions for the function

    defined on the interval . Enter points, such as inflection points in ascending order, i.e. smallest x values first.
    Remember that you can enter "pi" for as part of your answer.

    A. is concave down on the region to

    B. A global minimum for this function occurs at

    C. A local maximum for this function which is not a global maximum occurs at

    D. The function is increasing on to and on to .
    This seems like a homework question or a test question...so I will not answer for you since that is frowned upon but I will give you hint find where f'(x)=0.... f''(x)=0....and use taht to set up test intervals for each...and for helping with the derivative maybe this set up will be nicer for you \bigg[sin\bigg(\frac{x}{4}\bigg)\bigg]^2
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  3. #3
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    Please help

    Actually, it's not a test question. It's a homework question (only worth one point). I'm more interested in the method of solving and I already found answers to part of it.

    I have found the global minimum to be zero. I found the function to be increasing on the interval 0 to 2.6415926, but I don't know the other interval (the first part is -12.46, but I don't know the endpoint of increase).
    I also am having trouble finding where the function is concave down. I took the second derivative, but can't find the local maximum. I think it might be -12.466. Can you give me some hints?

    Thanks.
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Ok

    Quote Originally Posted by tennisgirl View Post
    Actually, it's not a test question. It's a homework question (only worth one point). I'm more interested in the method of solving and I already found answers to part of it.

    I have found the global minimum to be zero. I found the function to be increasing on the interval 0 to 2.6415926, but I don't know the other interval (the first part is -12.46, but I don't know the endpoint of increase).
    I also am having trouble finding where the function is concave down. I took the second derivative, but can't find the local maximum. I think it might be -12.466. Can you give me some hints?

    Thanks.
    if f(x)=sin\bigg(\frac{x}{4}\bigg)^2...then f'(x)=2sin\bigg(\frac{x}{4}\bigg)\cdot{cos\bigg(\f  rac{x}{4}\bigg)}\cdot\frac{1}{4}=\frac{1}{4}\cdot\  sin\bigg(\frac{x}{2}\bigg)...now setting this equal to zero we get x=-6.283,x=0,6.28 now setting up the test intervals we get (-\infty,-6.283),(-6.283,0)(0,6.238)(6.238\infty)...now just test a number in each interval in the derivative...positive values are increasing intervals and negative values are decreasing intervals
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  5. #5
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Quote Originally Posted by tennisgirl View Post
    Actually, it's not a test question. It's a homework question (only worth one point). I'm more interested in the method of solving and I already found answers to part of it.

    I have found the global minimum to be zero. I found the function to be increasing on the interval 0 to 2.6415926, but I don't know the other interval (the first part is -12.46, but I don't know the endpoint of increase).
    I also am having trouble finding where the function is concave down. I took the second derivative, but can't find the local maximum. I think it might be -12.466. Can you give me some hints?

    Thanks.

    This may help:

    f(x)=\sin^{2}(\frac{x}{4})

    taking the derivative we get

    \frac{df}{dx}=\frac{1}{2}\sin(\frac{x}{4})\cos(\fr  ac{x}{4})

    using the double angle formula we get

    \frac{df}{dx}=\frac{1}{4}\sin(\frac{x}{2})

    now taking the derivatvie again we get

    \frac{d^2f}{dx^2}=\frac{1}{8}\cos(\frac{x}{2})

    for max and min use the zero's of the 1st derivative and the end points of the interval

    See what you can do from here.
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Ok

    Quote Originally Posted by TheEmptySet View Post
    This may help:

    f(x)=\sin^{2}(\frac{x}{4})

    taking the derivative we get

    \frac{df}{dx}=\frac{1}{2}\sin(\frac{x}{4})\cos(\fr  ac{x}{4})

    using the double angle formula we get

    \frac{df}{dx}=\frac{1}{4}\sin(\frac{x}{2})

    now taking the derivatvie again we get

    \frac{d^2f}{dx^2}=\frac{1}{8}\cos(\frac{x}{2})

    for max and min use the zero's of the 1st derivative and the end points of the interval

    See what you can do from here.
    But you should set f''(x)=0 and set up test intervals based on those numbers...where f''(x) changes signs you have an inflection point
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