Find the local extrema and inflection points of g(x)=e^(-x^2/2).

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- anonymous

Find the local extrema and inflection points of g(x)=e^(-x^2/2).

- katieb

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- amistre64

is that (-x^2)/2 or is it -x^(2/2) ? its probably the first, but the first rule of mathematics is NEVER trust a mathmetician :)

- amistre64

dy/dx = dy/du * du/dx
y(u(x)) = e^((-x^2)/2) where y=e^u and u=-x^2/2.

- amistre64

dy/du = e^u
du/dx = -2x/2 = -x
y' = -x * e^((-x^2)/2) thats the first derivative.

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## More answers

- amistre64

y'' the second derivative tells us y' is max or min; and also were the inflection point are:
y'' = (-) MAX
y'' = (+) MIN
y'' = 0 inflection

- anonymous

it is (-x^2)/2~

- amistre64

y'' you will need to find by the product rule:
fl' + f'l

- amistre64

y'' = [-x][-x * e^((-x^2)/2)] + [-1][e^((-x^2)/2)]

- anonymous

for g'(x)=-xe^(-x^2/2), how to i find x?

- amistre64

make g'(x) equal to 0 to find your critical points...the bends of the graph.

- anonymous

i just don;t know how to find the value of x in e^(-x^2/2) :(

- anonymous

when i equated the equation to zero

- amistre64

x=0 I am thinking
there is no value for e^x that will make it 0. Think about the graph of the exponent funtion, it never reaches 0

- amistre64

do you agree?

- anonymous

yes, coz i cant think of any value that can make e to zero

- amistre64

since e^x cannot ever reach 0 there is no value of "x" that will make it = 0....so it doesnt matter :)

- anonymous

so how would the graph look like? would it be like concave up?

- amistre64

so we have a point at 0 that tells us there is something going on at x=0. lets use the second derivative to get more information

- anonymous

concave up from (-INF,-1), concave down (-1,1), concave up (1,INF)?

- anonymous

and the local max. is at x=0?

- amistre64

There is a bend at 0.... but we dont really know what it is doing until we use the g''(x) to tell us more about what is happening at x=0
y'' = [-x][-x * e^((-x^2)/2)] + [-1][e^((-x^2)/2)]
we can clean this up some:
y'' = x^2 * e^((-x^2)/2) - e^((-x^2)/2)

- amistre64

when x = 0; y'' = -e^(...)
y'' is (-) at x=0 which tells us that it is a MAX

- amistre64

when is y'' = 0? for our inflections. At x=1, cause e^(...) - e(..) = 0
so, at x=1 we have a change of direction.

- amistre64

so, heres what we know:
at g(0) there is a Max; and at g(1) there is an inflection. plug in those value for x in the original to get your y spots.
Does this make sense?

- anonymous

ah~yes! thanks!

- anonymous

f(x)=x^3+y^3-3xy=0 (this is not a function)
so do i need to use implicit differentiation?

- anonymous

or i just differentiate it normally? it will be like =f'(x)=3x^2-3x=0

- amistre64

sure. is it "y" is a function of "x" implicitly? cause it could go the other way around; or even be implied by an external variable like time

- amistre64

you show me how you think the steps go; and I will let you kow if you are right...

- anonymous

f(x)=x^3+y^3-3xy=0
f'(x)=3x^2+0-3x

- amistre64

explain to me:
(d/dx)(y^3) = 0 cause thats an error.

- anonymous

for this, i treated y as a constant and i didn't use implicit differentiation. should i use it? if i use it, f'(x)=3x^2+3y^2*y'-3y*y'*x

- amistre64

yes it; but your last term is in error... look at it again.

- anonymous

woops! wrong

- anonymous

yes haha lemme correct it
3(y+y'x)

- amistre64

YES!! - now solve this equation for y'

- amistre64

f'(x) = y'
So get all you y' to one side and everything else to the other side; factor out the y' and then get y' by itself
basic algebra stuff

- anonymous

y'=3x^2+y'(3y^2-3x)-3y
y'-y'(3y^2-3x)=3x^2-3y
y'(1-3y^2+3x)=3x^2-3y
y'=(3x^2-3y)/(1-3y^2+3x)

- amistre64

as long as you kept track of yourself; you got it :)

- anonymous

so do i need to find the x values in terms of y?

- anonymous

equate the whole equation to zero?

- amistre64

nope; they will give you values for x and y; or values for y' and x; or values for y' and y. you just plug in what they give you and solve for the unknown that they dont.

- anonymous

but they did nt give me any values..: sketch the curve x^3+y^3-3xy=0, indicating all extrema and points of inflection. Be sure to show all of your work.

- amistre64

you do the same thing for "with respect to time" but the thing is (d/dt) doesnt vanish, so you keep your (dx/dt) and (dy/dt).
ahh, you solve the original for y. that can be done explicitly becasue you can solve for y.

- amistre64

or use what we got :)

- amistre64

little hint; y' = 0 when that top part = 0...try that

- anonymous

x=+/- sqrty?

- amistre64

3x^2 -3y = 0 when?

- amistre64

3x^2 = 3y
y=x^2
when does n^2 = n?

- anonymous

when n=-n or +n?

- anonymous

sqrt

- anonymous

so x=+/- sqrty?

- amistre64

(0)^2 = 0 and (1)^2 = 1

- amistre64

do these make the bottom go to 0? if so, toss whichever one does.

- amistre64

but as is, we have bends at (0,0 and (1,1)
does that make sense to you?

- anonymous

yes!

- amistre64

to find the y'' just derive each side again with respect to x

- amistre64

your gonna have to use the quotient rule; and keep track of yourself tho :)

- anonymous

thats such a longgggggggggggggggggg equation...

- amistre64

lol..... do it, and eat your brussels sprouts :)

- amistre64

top is:
(1 -3y^2 +3x)(6x -3y')-(3x^2 -3y)(-6yy' +3)

- amistre64

the bottom is:
(1 -3y^2 +3x)(1 -3y^2 +3x)

- amistre64

im going with you work...so if its wrong... :)

- anonymous

y"= [-3y'(1+5y^2+3x+6x^2y)+3x(2-6y^2+9x)-9y]/(1-3y^2+3x^2)^2

- anonymous

uhhh i guess its right coz my eyes even got dried when i was differentiating it lol

- amistre64

we know y'=0 when we plug in(0,0) or (1,1) right? so plug all that in and see it we are (-),(+), or (0)

- amistre64

it might be easier to work with the unfactored forms :)

- amistre64

it is it is..... at (0,0) y'' = -3 we got a max at (0,0)

- amistre64

at (1,1) we get y''=+6 we have a MIN at (1,1)

- anonymous

i got y"=-3y' when (0,0)

- anonymous

but if i plug in 0 for y', it will be zero!

- amistre64

use the stuff I put up, its alot easier to work it in chunks that way

- amistre64

less degree for error

- anonymous

okay! but why it would be max when y"=-3? wouldnt it be min?

- anonymous

comparing to y"=+6

- amistre64

(1)(0) - (0)(3)
----------
(1)(1)

- amistre64

we good at (0,0) = y'' = -3?

- anonymous

did you plug in y'=0?

- amistre64

thin of it like this:
y = -(x^2) is an upside down bowl:
y' = -(2x)
y'' = -(2) = MAX

- amistre64

yes, I plug in (0,0) and y'=0 into what "I" did :)

- anonymous

but if i also plug in y'=0, it would be like: [(0)(1)+(0)(2)-0]/ (1)(1), so wouldnt it be y"=0?

- anonymous

i undersand the max thing now btw:) thanks

- amistre64

.... i hate to say this; but I trust my version of it better. You might have made an error when trying to squash everything together. In fact, I am sure of it :)

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