# Thread: Families of curves... what is b?

1. ## 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 .

2. Hello, purplegirl1818!

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

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

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

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

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

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

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

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

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

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

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

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

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

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