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Math Help - Families of curves... what is b?

  1. #1
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    Families of curves... what is b?

    Find a formula for a curve of the form for with a local maximum at and points of inflection at and .

    I know what a is, just not b, please help!

    thank you in advance
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  2. #2
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    Hello, purplegirl1818!

    \text{Find a formula for  a curve of the form: }\:y \:=\:e^{-\frac{(x-a)^2}{b}}\;\text{ for }b > 0

    with a local maximum at x = 3 and points of inflection at x = -1\text{ and }x = 7.

    Max/min points: . y' \;=\;\left(e^{-\frac{(x-a)^2}{b}}\right)\cdot\left(\text{-}2\,\frac{x-a}{b}\right) \;=\;0 \quad\Rightarrow\quad x \:=\:a

    Since there is a maximum at x = 3, then: \boxed{a\:=\:3}

    \text{The function is: }\;y \:=\:e^{-\frac{(x-3)^2}{b}}

    \text{The derivative is: }\;y' \;=\;-\frac{2}{b}\cdot(x-3)\cdot e^{-\frac{(x-3)^2}{b}}


    \text{The second derivative is: }\;y'' \;=\;-\frac{2}{b}\bigg[(x-3)e^{-\frac{x-3)^2}{b}}\cdot\frac{-2(x-3)}{b} + 1\cdot e^{-\frac{x-3)^2}{b}}\bigg]

    . . y'' \;=\;-\frac{2}{b}\,e^{-\frac{x-3)^2}{b}}\left[\frac{-2(x-3)^2}{b} + 1\right] \;=\;-\frac{2}{b}\,e^{-\frac{(x-3)^2}{b}}\,\left[\frac{-2(x-3)^2 + b}{b}\right]


    Inflection points: . -2(x-3)^2 + b \:=\:0 \quad\Rightarrow\quad 2(x-3)^2 \:=\:b \quad\Rightarrow\quad (x-3)^2 \:=\:\frac{b}{2}

    . . . . x-3 \:=\:\pm\sqrt{\frac{b}{2}} \quad\Rightarrow\quad x \:=\:3\pm\sqrt{\frac{b}{2}}


    Since the inflection points are at x = 7\text{ and }x = -1

    . . we have: . \begin{Bmatrix}3 + \sqrt{\frac{b}{2}} \:=\:7\\ \\[-3mm]<br />
3 - \sqrt{\frac{b}{2}} \:=\:\text{-}1 \end{Bmatrix}\quad\Rightarrow\quad \boxed{b \:=\:32}


    \text{Therefore, the function is: }\;y \;=\;e^{-\frac{(x-3)^2}{32}}

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  3. #3
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    thank you so much, you are my hero lol
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