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Math Help - concavity

  1. #1
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    concavity

    If n is a positive integer, how many times does the function f(x) = x^2 + 5cosx change concavity in the interval 0 <= x <= 2*pi*n?

    A) 0
    B) 1
    C) 2
    D) n
    E) 2n
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  2. #2
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    Quote Originally Posted by DINOCALC09 View Post
    If n is a positive integer, how many times does the function f(x) = x^2 + 5cosx change concavity in the interval 0 <= x <= 2*pi*n?

    A) 0
    B) 1
    C) 2
    D) n
    E) 2n
    The concavity of a function is its second derivative.

    So find the second derivative and set it equal to 0. (Is this last step even possible?)

    -Dan
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  3. #3
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    1st derivative = 2x - 5sinx
    2nd derivative = 2 - 5cosx

    set 2nd derivative = 0

    so:

    2 - 5cosx = 0
    cosx = -2 / -5
    x = arccos(-2/5)

    This is only satisfied once?

    So the answer should be (B) ?
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  4. #4
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    or is the answer E?

    thats what my friend got. we still lack a definent answer though.
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  5. #5
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    Quote Originally Posted by DINOCALC09 View Post
    If n is a positive integer, how many times does the function f(x) = x^2 + 5cosx change concavity in the interval 0 <= x <= 2*pi*n?

    A) 0
    B) 1
    C) 2
    D) n
    E) 2n
    Look for points of inflexion (by solving f''(x) = 0) over the given interval.
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  6. #6
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    i believe i did that up above... not sure if i did it right or not, but yeah =/
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  7. #7
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    Quote Originally Posted by DINOCALC09 View Post
    1st derivative = 2x - 5sinx
    2nd derivative = 2 - 5cosx

    set 2nd derivative = 0

    so:

    2 - 5cosx = 0
    cosx = -2 / -5
    x = arccos(-2/5) Mr F says: This should be arccos(2/5).

    This is only satisfied once? Mr F says: There are two solutions over the interval 0 < x < 2pi. To see this, draw the graphs of y = cos x and y = 2/5 over 0 < x < 2pi. You should see two intersection points ......

    And since the period of cos x is 2pi, you'll have two solutions over every interval of size 2 pi ..... So four solutions over 0 < x < 4pi, six solutions over 0 < x < 6pi ......, 2n solutions over 0 < x< 2npi


    So the answer should be (B) ?
    The answer is E.
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  8. #8
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    Quote Originally Posted by mr fantastic View Post
    The answer is E.
    I have of course skipped over the business of proving that the solutions to f''(x) = 0 are stationary points. This can be done with a small bit of effort, but i don't think that it be expected for a multiple choice question like this .....
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