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Math Help - Concavity

  1. #1
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    Concavity

    Short problem. Determine where the given function is concave up and where it is concave down.

    f(x)= 10x^3 - 3x^5

    Thanks.
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  2. #2
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    Quote Originally Posted by akuczma86 View Post
    Short problem. Determine where the given function is concave up and where it is concave down.

    f(x)= 10x^3 - 3x^5

    Thanks.
    Concavity is shown by the second derivative. Find the second derivative and go from there.

    If

    f^{\prime \prime}(x) > 0 on the interval I, then f is concave up on I

    f^{\prime \prime}(x) < 0 on the interval I, then f is concave down on I
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  3. #3
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    first derivative = 30x^2 - 15x^4.
    2nd deriv = 60x - 60x^3.
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  4. #4
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by akuczma86 View Post
    first derivative = 30x^2 - 15x^4.
    2nd deriv = 60x - 60x^3.
    f(x)= 10x^3 - 3x^5

    f^\prime = 30x^2 - 15x^4

    f^{\prime \prime} = 60x - 60x^3

    Yup that is correct.

    So now, if you want to find where it is concaving up and down. Find the points of inflection, i.e. when

    f^{\prime \prime} = 0

    Then simply plug in a point inbetween the points of inflection to see if it's increasing or decreasing. Then you know if it's concaving up or down!
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  5. #5
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    I could be wrong on this, but let's say I plug in 1 and -1. 60(1) - 60(1)^3 = 60-60=0. 60(-1) - 60(-1)^3 = -60 - (-60) = 0. Am I on the right track here?
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  6. #6
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by akuczma86 View Post
    I could be wrong on this, but let's say I plug in 1 and -1. 60(1) - 60(1)^3 = 60-60=0. 60(-1) - 60(-1)^3 = -60 - (-60) = 0. Am I on the right track here?
    Yup,

    so let us check

    f^{\prime \prime} = 60x - 60x^3 = 0

    This of course becomes

    1 - x^2 = 0

    which has the roots

     x=1 and  x=-1

    EDIT-- I didn't factor correctly. What we have here is

    f^{\prime \prime} = 60x - 60x^3 = 0 = x(1-x^2)

    This introduces the root x=0

    So we know at these points the graph has an inflextion point. What you need to do now is sub in a point inbetween these bounds and outside of them.

    In other words, there are 4 parts of this graph. We seperate them into the bounds like

    x < -1 is part 1

    -1 < x < 0 is part 2

     0 < x < 1 is part 3

    x > 1 is part 4

    Sub in a value of x that falls within each of these parts into your second derivative. This will produce the concavity of all 3 intervals
    Last edited by AllanCuz; April 13th 2010 at 11:14 AM.
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  7. #7
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    So for part 1, 60(-2) - 60(-2)^3 = 60.
    Part 2 i re-plugged in 1 and -1.
    Part 3 60(2) - 60(2)^3) = -60.

    So function is concave up hen x < -1 and the function is concave down when x > 1?
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  8. #8
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by akuczma86 View Post
    So for part 1, 60(-2) - 60(-2)^3 = 60.
    Part 2 i re-plugged in 1 and -1.
    Part 3 60(2) - 60(2)^3) = -60.

    So function is concave up hen x < -1 and the function is concave down when x > 1?
    You can't plug in 1 and -1 for part 2. We already know the function is 0 here! We want to know how the function behaves inbetween these bounds.

    Also, I believe I messed up, we have 4 segments not 3! If you notice

    <br />
f^{\prime \prime} = 60x - 60x^3 = 0<br />

    Is the same as saying

     f^{\prime \prime} = x(1-x^2) = 0

    This equation is also satisfied for x=0

    Therefore our segments become

     x< -1 Part 1
     -1 < x < 0 Part 2
      0 < x < 1 Part 3
     x>1 Part 4

    My bad!

    But the same thing applies here. We must plug numbers inbetween these points. So for example, plug in -100 for part 1, -.5 for part 2, .5 for part 3 and 100 for part 4
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