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Math Help - Family of Functions!

  1. #1
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    Family of Functions!

    find the following formula

    a curve of the form y = e^(-(x-a)^2/b) b>0

    local max: x=32
    inflection points: x=36 x=28

    find a and b

    im not sure how to use the values given to find a and b...any help
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    Quote Originally Posted by mathlete View Post
    find the following formula

    a curve of the form y = e^(-(x-a)^2/b) b>0

    local max: x=32
    inflection points: x=36 x=28

    find a and b

    im not sure how to use the values given to find a and b...any help
    Well what is the condition for a local maximum? What is the condition for a point to be an inflection point?

    -Dan
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  3. #3
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    ok, local max would be like the first derivative and i think for the inflection points you would need the second derivative, right?

    so would i just find those, then set them equal to the given values, maybe
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  4. #4
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    Quote Originally Posted by mathlete View Post
    find the following formula

    a curve of the form y = e^(-(x-a)^2/b) b>0

    local max: x=32
    inflection points: x=36 x=28

    find a and b

    im not sure how to use the values given to find a and b...any help
    y=e^{-\frac{(x-a)^2}{b}}

    Find the 1st derivative and set it equal to zero to find max/mins.

    y'=e^{-\frac{(x-a)^2}{b}} \frac{2x-2a}{b}

    0=-e^{-\frac{(32-a)^2}{b}} \frac{2(32)-2a}{b}

    You have one equation, two unknowns. To find inflection points, set the 2nd derivative equal to zero.

    y''=-e^{-\frac{(x-a)^2}{b}} \left(\frac{2x-2a}{b}\right)^2 -e^{-\frac{(x-a)^2}{b}} \frac{2}{b}

    0=-e^{-\frac{(x-a)^2}{b}} \left(\frac{2x-2a}{b}\right)^2 -e^{-\frac{(x-a)^2}{b}} \frac{2}{b}

    Plug in for x = 36 and x = 28 and you have three equations with two unknowns. Happy solving!
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