Family of Functions!

**mathlete** · March 27th 2008, 10:45 AM

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

**topsquark** · March 27th 2008, 10:46 AM

Originally Posted by mathlete

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

**mathlete** · March 27th 2008, 10:49 AM

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

**colby2152** · March 27th 2008, 10:54 AM

Originally Posted by mathlete

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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