anonymous
  • anonymous
Find the local extrema and inflection points of g(x)=e^(-x^2/2).
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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amistre64
  • 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
  • amistre64
dy/dx = dy/du * du/dx y(u(x)) = e^((-x^2)/2) where y=e^u and u=-x^2/2.
amistre64
  • amistre64
dy/du = e^u du/dx = -2x/2 = -x y' = -x * e^((-x^2)/2) thats the first derivative.

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amistre64
  • 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
  • anonymous
it is (-x^2)/2~
amistre64
  • amistre64
y'' you will need to find by the product rule: fl' + f'l
amistre64
  • amistre64
y'' = [-x][-x * e^((-x^2)/2)] + [-1][e^((-x^2)/2)]
anonymous
  • anonymous
for g'(x)=-xe^(-x^2/2), how to i find x?
amistre64
  • amistre64
make g'(x) equal to 0 to find your critical points...the bends of the graph.
anonymous
  • anonymous
i just don;t know how to find the value of x in e^(-x^2/2) :(
anonymous
  • anonymous
when i equated the equation to zero
amistre64
  • 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
  • amistre64
do you agree?
anonymous
  • anonymous
yes, coz i cant think of any value that can make e to zero
amistre64
  • amistre64
since e^x cannot ever reach 0 there is no value of "x" that will make it = 0....so it doesnt matter :)
anonymous
  • anonymous
so how would the graph look like? would it be like concave up?
amistre64
  • 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
  • anonymous
concave up from (-INF,-1), concave down (-1,1), concave up (1,INF)?
anonymous
  • anonymous
and the local max. is at x=0?
amistre64
  • 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
  • amistre64
when x = 0; y'' = -e^(...) y'' is (-) at x=0 which tells us that it is a MAX
amistre64
  • 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
  • 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
  • anonymous
ah~yes! thanks!
anonymous
  • anonymous
f(x)=x^3+y^3-3xy=0 (this is not a function) so do i need to use implicit differentiation?
anonymous
  • anonymous
or i just differentiate it normally? it will be like =f'(x)=3x^2-3x=0
amistre64
  • 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
  • amistre64
you show me how you think the steps go; and I will let you kow if you are right...
anonymous
  • anonymous
f(x)=x^3+y^3-3xy=0 f'(x)=3x^2+0-3x
amistre64
  • amistre64
explain to me: (d/dx)(y^3) = 0 cause thats an error.
anonymous
  • 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
  • amistre64
yes it; but your last term is in error... look at it again.
anonymous
  • anonymous
woops! wrong
anonymous
  • anonymous
yes haha lemme correct it 3(y+y'x)
amistre64
  • amistre64
YES!! - now solve this equation for y'
amistre64
  • 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
  • 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
  • amistre64
as long as you kept track of yourself; you got it :)
anonymous
  • anonymous
so do i need to find the x values in terms of y?
anonymous
  • anonymous
equate the whole equation to zero?
amistre64
  • 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
  • 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
  • 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
  • amistre64
or use what we got :)
amistre64
  • amistre64
little hint; y' = 0 when that top part = 0...try that
anonymous
  • anonymous
x=+/- sqrty?
amistre64
  • amistre64
3x^2 -3y = 0 when?
amistre64
  • amistre64
3x^2 = 3y y=x^2 when does n^2 = n?
anonymous
  • anonymous
when n=-n or +n?
anonymous
  • anonymous
sqrt
anonymous
  • anonymous
so x=+/- sqrty?
amistre64
  • amistre64
(0)^2 = 0 and (1)^2 = 1
amistre64
  • amistre64
do these make the bottom go to 0? if so, toss whichever one does.
amistre64
  • amistre64
but as is, we have bends at (0,0 and (1,1) does that make sense to you?
anonymous
  • anonymous
yes!
amistre64
  • amistre64
to find the y'' just derive each side again with respect to x
amistre64
  • amistre64
your gonna have to use the quotient rule; and keep track of yourself tho :)
anonymous
  • anonymous
thats such a longgggggggggggggggggg equation...
amistre64
  • amistre64
lol..... do it, and eat your brussels sprouts :)
amistre64
  • amistre64
top is: (1 -3y^2 +3x)(6x -3y')-(3x^2 -3y)(-6yy' +3)
amistre64
  • amistre64
the bottom is: (1 -3y^2 +3x)(1 -3y^2 +3x)
amistre64
  • amistre64
im going with you work...so if its wrong... :)
anonymous
  • anonymous
y"= [-3y'(1+5y^2+3x+6x^2y)+3x(2-6y^2+9x)-9y]/(1-3y^2+3x^2)^2
anonymous
  • anonymous
uhhh i guess its right coz my eyes even got dried when i was differentiating it lol
amistre64
  • 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
  • amistre64
it might be easier to work with the unfactored forms :)
amistre64
  • amistre64
it is it is..... at (0,0) y'' = -3 we got a max at (0,0)
amistre64
  • amistre64
at (1,1) we get y''=+6 we have a MIN at (1,1)
anonymous
  • anonymous
i got y"=-3y' when (0,0)
anonymous
  • anonymous
but if i plug in 0 for y', it will be zero!
amistre64
  • amistre64
use the stuff I put up, its alot easier to work it in chunks that way
amistre64
  • amistre64
less degree for error
anonymous
  • anonymous
okay! but why it would be max when y"=-3? wouldnt it be min?
anonymous
  • anonymous
comparing to y"=+6
amistre64
  • amistre64
(1)(0) - (0)(3) ---------- (1)(1)
amistre64
  • amistre64
we good at (0,0) = y'' = -3?
anonymous
  • anonymous
did you plug in y'=0?
amistre64
  • amistre64
thin of it like this: y = -(x^2) is an upside down bowl: y' = -(2x) y'' = -(2) = MAX
amistre64
  • amistre64
yes, I plug in (0,0) and y'=0 into what "I" did :)
anonymous
  • 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
  • anonymous
i undersand the max thing now btw:) thanks
amistre64
  • 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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